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Hi
I have a follow up question, relating to subsetting to list items. After using the list and min(sapply()) method to adjust the length of the variables, I specify a dynamic regression equation using the variables in the list. My list looks like this:
Dcr<- list(Dcre1=DCred1,Dcre2=DCred2,Dcre3=DCred3,Dbobc1=DBoBC1,Dbobc2=DBoBC2,Dbobc3=DBoBC3,...)
By specifying the list items with names, I thought I could end by referencing them (or subsetting the list) as, eg., Dcr$Dcre1 and get DCred1, Dcr$Dbobc1 and get DBoBC1, etc so that the explanatory variables of the equation can be easily associated with their respective original names. This way, I would avoid specifying the list as Dcr<-list(Dcr1, Dcr2, Dcr, 3..., Dcr15) and then subsetting the list using Dcr[[1]][1:29], Dcr[[[2]][1:29], ..., Dcr[[15]][1:29] because the list has many variables (15) and referencing the variables with numbers makes them lose their original names.
When I specify the list as Dcr<- list(Dcr1, Dcr2, ..., Dcr15), then the regression equation specified as:
# Regression
regCred<- lm(Dcr[[1]][1:29]~Dcr[[2]][1:29]+Dcr[[3]][1:29]+Dcr[[4]][1:29]+Dcr[[5]][1:29]+Dcr[[6]][1:29]+...)
runs without problems - the results are shown here below:
Call:
lm(formula = Dcr[[1]][1:29] ~ Dcr[[2]][1:29] + Dcr[[3]][1:29] +
Dcr[[4]][1:29] + Dcr[[5]][1:29] + Dcr[[6]][1:29])
Residuals:
Min 1Q Median 3Q Max
-86.293 -33.586 -9.969 40.147 117.965
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 81.02064 13.28632 6.098 3.21e-06 ***
Dcr[[2]][1:29] -0.97407 0.11081 -8.791 8.20e-09 ***
Dcr[[3]][1:29] -0.27950 0.05899 -4.738 8.95e-05 ***
Dcr[[4]][1:29] -0.07961 0.04856 -1.639 0.115
Dcr[[5]][1:29] -0.07180 0.05515 -1.302 0.206
Dcr[[6]][1:29] -0.01562 0.02086 -0.749 0.462
But when I specify the list with names as shown above, then the equation does not run - as shown by the following error message
> # Regression
> regCred<- lm(Dcr[[1]][1:29]~Dcr[[2]][1:29]+Dcr[[3]][1:29]+Dcr[[4]][1:29]+
+ Dcr[[5]][1:29]+Dcr$Dbobc3)
Error in model.frame.default(formula = Dcr[[1]][1:29] ~ Dcr[[2]][1:29] + :
variable lengths differ (found for 'Dcr$Dbobc3')
> Dcr[[5]][1:29]+Dcr$Dbobc3[1:29])
Error: unexpected ')' in "Dcr[[5]][1:29]+Dcr$Dbobc3[1:29])"
NB: In the equation with error message, only the last term is specified by referencing its name (ie., Dcr$Dbobc3[1:29]. Also note that the error occurs whether I append '[1:29]' to Dcr$Dbobc or not.
How do I resolve this?
Thanks. Lexi
NB: I tried typing the above in the same email Petr used to reply me, but the email could not be delivered due to size problems
[[alternative HTML version deleted]]
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On Jun 30, 2012, at 6:04 PM, Lekgatlhamang, lexi Setlhare wrote:
> Hi
> I have a follow up question, relating to subsetting to list items.
> After using the list and min(sapply()) method to adjust the length
> of the variables, I specify a dynamic regression equation using the
> variables in the list. My list looks like this:
> Dcr<-
> list
> (Dcre1
> =
> DCred1
> ,Dcre2
> =DCred2,Dcre3=DCred3,Dbobc1=DBoBC1,Dbobc2=DBoBC2,Dbobc3=DBoBC3,...)
This should ahve been done like this:
Dcr<- data.frame(Dcre1=DCred1, Dcre2=DCred2, Dcre3=DCred3,
Dbobc1=DBoBC1, Dbobc2=DBoBC2, Dbobc3=DBoBC3)
> By specifying the list items with names, I thought I could end by
> referencing them (or subsetting the list) as, eg., Dcr$Dcre1 and get
> DCred1, Dcr$Dbobc1 and get DBoBC1, etc so that the explanatory
> variables of the equation can be easily associated with their
> respective original names. This way, I would avoid specifying the
> list as Dcr<-list(Dcr1, Dcr2, Dcr, 3..., Dcr15) and then subsetting
> the list using Dcr[[1]][1:29], Dcr[[[2]][1:29], ..., Dcr[[15]][1:29]
> because the list has many variables (15) and referencing the
> variables with numbers makes them lose their original names.
> When I specify the list as Dcr<- list(Dcr1, Dcr2, ..., Dcr15), then
> the regression equation specified as:
> # Regression
> regCred<- lm(Dcr[[1]][1:29]~Dcr[[2]][1:29]+Dcr[[3]][1:29]+Dcr[[4]]
> [1:29]+Dcr[[5]][1:29]+Dcr[[6]][1:29]+...)
And the you could have done
> regCred<- lm(Dcre1 ~ . , data=Dcr [ , 1:29] )
(Leaving out the , ...)
> runs without problems - the results are shown here below:
> Call:
> lm(formula = Dcr[[1]][1:29] ~ Dcr[[2]][1:29] + Dcr[[3]][1:29] +
> Dcr[[4]][1:29] + Dcr[[5]][1:29] + Dcr[[6]][1:29])
> Residuals:
> Min 1Q Median 3Q Max
> -86.293 -33.586 -9.969 40.147 117.965
> Coefficients:
> Estimate Std. Error t value Pr(>|t|)
> (Intercept) 81.02064 13.28632 6.098 3.21e-06 ***
> Dcr[[2]][1:29] -0.97407 0.11081 -8.791 8.20e-09 ***
> Dcr[[3]][1:29] -0.27950 0.05899 -4.738 8.95e-05 ***
> Dcr[[4]][1:29] -0.07961 0.04856 -1.639 0.115
> Dcr[[5]][1:29] -0.07180 0.05515 -1.302 0.206
> Dcr[[6]][1:29] -0.01562 0.02086 -0.749 0.462
>
> But when I specify the list with names as shown above, then the
> equation does not run - as shown by the following error message
>> # Regression
>> regCred<- lm(Dcr[[1]][1:29]~Dcr[[2]][1:29]+Dcr[[3]][1:29]+Dcr[[4]]
>> [1:29]+
> + Dcr[[5]][1:29]+Dcr$Dbobc3)
> Error in model.frame.default(formula = Dcr[[1]][1:29] ~ Dcr[[2]]
> [1:29] + :
> variable lengths differ (found for 'Dcr$Dbobc3')
>> Dcr[[5]][1:29]+Dcr$Dbobc3[1:29])
> Error: unexpected ')' in "Dcr[[5]][1:29]+Dcr$Dbobc3[1:29])"
>
> NB: In the equation with error message, only the last term is
> specified by referencing its name (ie., Dcr$Dbobc3[1:29]. Also note
> that the error occurs whether I append '[1:29]' to Dcr$Dbobc or not.
> How do I resolve this?
You should have offered str(Dcr)
David Winsemius, MD
West Hartford, CT
______________________________________________
[hidden email] mailing list
https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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On Jun 30, 2012, at 8:47 PM, David Winsemius wrote:
>
> On Jun 30, 2012, at 6:04 PM, Lekgatlhamang, lexi Setlhare wrote:
>
>> Hi
>> I have a follow up question, relating to subsetting to list items.
>> After using the list and min(sapply()) method to adjust the length
>> of the variables, I specify a dynamic regression equation using the
>> variables in the list. My list looks like this:
>> Dcr<-
>> list
>> (Dcre1
>> =
>> DCred1
>> ,Dcre2
>> =DCred2,Dcre3=DCred3,Dbobc1=DBoBC1,Dbobc2=DBoBC2,Dbobc3=DBoBC3,...)
>
> This should ahve been done like this:
>
> Dcr<- data.frame(Dcre1=DCred1, Dcre2=DCred2, Dcre3=DCred3,
> Dbobc1=DBoBC1, Dbobc2=DBoBC2, Dbobc3=DBoBC3)
>
>> By specifying the list items with names, I thought I could end by
>> referencing them (or subsetting the list) as, eg., Dcr$Dcre1 and
>> get DCred1, Dcr$Dbobc1 and get DBoBC1, etc so that the explanatory
>> variables of the equation can be easily associated with their
>> respective original names. This way, I would avoid specifying the
>> list as Dcr<-list(Dcr1, Dcr2, Dcr, 3..., Dcr15) and then subsetting
>> the list using Dcr[[1]][1:29], Dcr[[[2]][1:29], ..., Dcr[[15]]
>> [1:29] because the list has many variables (15) and referencing the
>> variables with numbers makes them lose their original names.
>> When I specify the list as Dcr<- list(Dcr1, Dcr2, ..., Dcr15), then
>> the regression equation specified as:
>> # Regression
>> regCred<- lm(Dcr[[1]][1:29]~Dcr[[2]][1:29]+Dcr[[3]][1:29]+Dcr[[4]]
>> [1:29]+Dcr[[5]][1:29]+Dcr[[6]][1:29]+...)
>
> And the you could have done
>
>> regCred<- lm(Dcre1 ~ . , data=Dcr [ , 1:29] )
Oh, Nuts! I meant to type:
regCred<- lm(Dcre1 ~ . , data=Dcr [ 1:29, ] )
> (Leaving out the , ...)
>
>
>> runs without problems - the results are shown here below:
>> Call:
>> lm(formula = Dcr[[1]][1:29] ~ Dcr[[2]][1:29] + Dcr[[3]][1:29] +
>> Dcr[[4]][1:29] + Dcr[[5]][1:29] + Dcr[[6]][1:29])
>> Residuals:
>> Min 1Q Median 3Q Max
>> -86.293 -33.586 -9.969 40.147 117.965
>> Coefficients:
>> Estimate Std. Error t value Pr(>|t|)
>> (Intercept) 81.02064 13.28632 6.098 3.21e-06 ***
>> Dcr[[2]][1:29] -0.97407 0.11081 -8.791 8.20e-09 ***
>> Dcr[[3]][1:29] -0.27950 0.05899 -4.738 8.95e-05 ***
>> Dcr[[4]][1:29] -0.07961 0.04856 -1.639 0.115
>> Dcr[[5]][1:29] -0.07180 0.05515 -1.302 0.206
>> Dcr[[6]][1:29] -0.01562 0.02086 -0.749 0.462
>>
>> But when I specify the list with names as shown above, then the
>> equation does not run - as shown by the following error message
>>> # Regression
>>> regCred<- lm(Dcr[[1]][1:29]~Dcr[[2]][1:29]+Dcr[[3]][1:29]+Dcr[[4]]
>>> [1:29]+
>> + Dcr[[5]][1:29]+Dcr$Dbobc3)
>> Error in model.frame.default(formula = Dcr[[1]][1:29] ~ Dcr[[2]]
>> [1:29] + :
>> variable lengths differ (found for 'Dcr$Dbobc3')
>>> Dcr[[5]][1:29]+Dcr$Dbobc3[1:29])
>> Error: unexpected ')' in "Dcr[[5]][1:29]+Dcr$Dbobc3[1:29])"
>>
>> NB: In the equation with error message, only the last term is
>> specified by referencing its name (ie., Dcr$Dbobc3[1:29]. Also note
>> that the error occurs whether I append '[1:29]' to Dcr$Dbobc or not.
>> How do I resolve this?
>
This still applies:
You should have offered str(Dcr)
>
David Winsemius, MD
West Hartford, CT
______________________________________________
[hidden email] mailing list
https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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In reply to this post by Lekgatlhamang, lexi Setlhare
Hello,
Try this:
Dcr<-lapply(1:5,function(x) rnorm(10,15))
names(Dcr)<- c("Dcre1","Dcre2","Dcre3","Dcre4","Dcre5")
#Works
regCred<-lm(Dcr[[1]]~Dcr[[2]]+Dcr[[3]])
summary(regCred)
#Works
regCred2<-lm(Dcre1~Dcre2+Dcre3,data=Dcr)
summary(regCred)
# Do not work
regCred3<-lm(Dcr[[1]][1:5]~Dcr[[4]][1:5]+Dcre5,data=Dcr)
Error in model.frame.default(formula = Dcr[[1]][1:5] ~ Dcr[[4]][1:5] + :
variable lengths differ (found for 'Dcre5')
#I guess this is what happened in your example, when different variable lengths are used
#If you had used,
regCred3<-lm(Dcr[[1]][1:5]~Dcr[[4]][1:5]+Dcre5[1:5],data=Dcr)
summary(regCred3)
#it works
#this also works
regCred4<-lm(Dcre1[1:5]~Dcre2[1:5]+Dcre3[6:10],data=Dcr)
Or you could convert the list to dataframe
Dcr2<-data.frame(Dcre1=Dcr$Dcre1,Dcre2=Dcr$Dcre2,Dcre3=Dcr$Dcre3)
#testing whether list and dataframe converted results are same
#From dataframe
regCred5<-lm(Dcre1~Dcre2+Dcre3,data=Dcr2[1:5,])
> summary(regCred5)
Call:
lm(formula = Dcre1 ~ Dcre2 + Dcre3, data = Dcr2[1:5, ])
Residuals:
1 2 3 4 5
-0.01262 0.09888 0.07133 -0.08494 -0.07265
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 16.53707 0.99604 16.603 0.00361 **
Dcre2 -0.27890 0.04185 -6.665 0.02178 *
Dcre3 0.21874 0.04643 4.711 0.04222 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#Same model using list
regCred6<-lm(Dcre1[1:5]~Dcre2[1:5]+Dcre3[1:5],data=Dcr)
> summary(regCred6)
Call:
lm(formula = Dcre1[1:5] ~ Dcre2[1:5] + Dcre3[1:5], data = Dcr)
Residuals:
1 2 3 4 5
-0.01262 0.09888 0.07133 -0.08494 -0.07265
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 16.53707 0.99604 16.603 0.00361 **
Dcre2[1:5] -0.27890 0.04185 -6.665 0.02178 *
Dcre3[1:5] 0.21874 0.04643 4.711 0.04222 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.1173 on 2 degrees of freedom
Multiple R-squared: 0.9739, Adjusted R-squared: 0.9478
F-statistic: 37.28 on 2 and 2 DF, p-value: 0.02612
The difference is in the names of the coefficients.
names(coef(regCred6))
[1] "(Intercept)" "Dcre2[1:5]" "Dcre3[1:5]"
which you can change by,
names(regCred6$coef)<-names(regCred5$coef)
regCred6$coef
(Intercept) Dcre2 Dcre3
16.5370694 -0.2788988 0.2187360
Though, it won't change the names of coefficients in the summary. I tried several ways, but so far not successful. I think in that case, the easiest way is to assign the subset to a new variable and run the analysis.
e.g.
Dcre1new<-Dcre1[1:5]
Hope this was helpful.
A.K.
----- Original Message -----
From: "Lekgatlhamang, lexi Setlhare" < [hidden email]>
To: " [hidden email]" < [hidden email]>
Cc:
Sent: Saturday, June 30, 2012 6:04 PM
Subject: [R] Adjusting length of series
Hi
I have a follow up question, relating to subsetting to list items. After using the list and min(sapply()) method to adjust the length of the variables, I specify a dynamic regression equation using the variables in the list. My list looks like this:
Dcr<- list(Dcre1=DCred1,Dcre2=DCred2,Dcre3=DCred3,Dbobc1=DBoBC1,Dbobc2=DBoBC2,Dbobc3=DBoBC3,...)
By specifying the list items with names, I thought I could end by referencing them (or subsetting the list) as, eg., Dcr$Dcre1 and get DCred1, Dcr$Dbobc1 and get DBoBC1, etc so that the explanatory variables of the equation can be easily associated with their respective original names. This way, I would avoid specifying the list as Dcr<-list(Dcr1, Dcr2, Dcr, 3..., Dcr15) and then subsetting the list using Dcr[[1]][1:29], Dcr[[[2]][1:29], ..., Dcr[[15]][1:29] because the list has many variables (15) and referencing the variables with numbers makes them lose their original names.
When I specify the list as Dcr<- list(Dcr1, Dcr2, ..., Dcr15), then the regression equation specified as:
# Regression
regCred<- lm(Dcr[[1]][1:29]~Dcr[[2]][1:29]+Dcr[[3]][1:29]+Dcr[[4]][1:29]+Dcr[[5]][1:29]+Dcr[[6]][1:29]+...)
runs without problems - the results are shown here below:
Call:
lm(formula = Dcr[[1]][1:29] ~ Dcr[[2]][1:29] + Dcr[[3]][1:29] +
Dcr[[4]][1:29] + Dcr[[5]][1:29] + Dcr[[6]][1:29])
Residuals:
Min 1Q Median 3Q Max
-86.293 -33.586 -9.969 40.147 117.965
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 81.02064 13.28632 6.098 3.21e-06 ***
Dcr[[2]][1:29] -0.97407 0.11081 -8.791 8.20e-09 ***
Dcr[[3]][1:29] -0.27950 0.05899 -4.738 8.95e-05 ***
Dcr[[4]][1:29] -0.07961 0.04856 -1.639 0.115
Dcr[[5]][1:29] -0.07180 0.05515 -1.302 0.206
Dcr[[6]][1:29] -0.01562 0.02086 -0.749 0.462
But when I specify the list with names as shown above, then the equation does not run - as shown by the following error message
> # Regression
> regCred<- lm(Dcr[[1]][1:29]~Dcr[[2]][1:29]+Dcr[[3]][1:29]+Dcr[[4]][1:29]+
+ Dcr[[5]][1:29]+Dcr$Dbobc3)
Error in model.frame.default(formula = Dcr[[1]][1:29] ~ Dcr[[2]][1:29] + :
variable lengths differ (found for 'Dcr$Dbobc3')
> Dcr[[5]][1:29]+Dcr$Dbobc3[1:29])
Error: unexpected ')' in "Dcr[[5]][1:29]+Dcr$Dbobc3[1:29])"
NB: In the equation with error message, only the last term is specified by referencing its name (ie., Dcr$Dbobc3[1:29]. Also note that the error occurs whether I append '[1:29]' to Dcr$Dbobc or not.
How do I resolve this?
Thanks. Lexi
NB: I tried typing the above in the same email Petr used to reply me, but the email could not be delivered due to size problems
[[alternative HTML version deleted]]
______________________________________________
[hidden email] mailing list
https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
--
______________________________________________
[hidden email] mailing list
https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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Thanks a lot David and A.K. for the detailed and helpful suggestions.
Lexi
________________________________
Cc: R help < [hidden email]>
Sent: Sunday, July 1, 2012 3:39 AM
Subject: Re: [R] Adjusting length of series
Hello,
Try this:
Dcr<-lapply(1:5,function(x) rnorm(10,15))
names(Dcr)<- c("Dcre1","Dcre2","Dcre3","Dcre4","Dcre5")
#Works
regCred<-lm(Dcr[[1]]~Dcr[[2]]+Dcr[[3]])
 summary(regCred)
#Works
 regCred2<-lm(Dcre1~Dcre2+Dcre3,data=Dcr)
 summary(regCred)
# Do not work
regCred3<-lm(Dcr[[1]][1:5]~Dcr[[4]][1:5]+Dcre5,data=Dcr)
Error in model.frame.default(formula = Dcr[[1]][1:5] ~ Dcr[[4]][1:5] +Â :
 variable lengths differ (found for 'Dcre5')
#I guess this is what happened in your example, when different variable lengths are used
#If you had used,
regCred3<-lm(Dcr[[1]][1:5]~Dcr[[4]][1:5]+Dcre5[1:5],data=Dcr)
 summary(regCred3)
#it works
#this also works
regCred4<-lm(Dcre1[1:5]~Dcre2[1:5]+Dcre3[6:10],data=Dcr)
Or you could convert the list to dataframe
 Dcr2<-data.frame(Dcre1=Dcr$Dcre1,Dcre2=Dcr$Dcre2,Dcre3=Dcr$Dcre3)
#testing whether list and dataframe converted results are same
#From dataframe
regCred5<-lm(Dcre1~Dcre2+Dcre3,data=Dcr2[1:5,])
> summary(regCred5)
Call:
lm(formula = Dcre1 ~ Dcre2 + Dcre3, data = Dcr2[1:5, ])
Residuals:
      1       2       3       4       5
-0.01262Â 0.09888Â 0.07133 -0.08494 -0.07265
Coefficients:
           Estimate Std. Error t value Pr(>|t|) Â
(Intercept) 16.53707Â Â Â 0.99604Â 16.603Â 0.00361 **
Dcre2Â Â Â Â Â Â -0.27890Â Â Â 0.04185Â -6.665Â 0.02178 *
Dcre3Â Â Â Â Â Â Â 0.21874Â Â Â 0.04643Â Â 4.711Â 0.04222 *
---
Signif. codes:Â 0 â***â 0.001 â**â 0.01 â*â 0.05 â.â 0.1 â â 1
Â
#Same model using list
regCred6<-lm(Dcre1[1:5]~Dcre2[1:5]+Dcre3[1:5],data=Dcr)
> summary(regCred6)
Call:
lm(formula = Dcre1[1:5] ~ Dcre2[1:5] + Dcre3[1:5], data = Dcr)
Residuals:
      1       2       3       4       5
-0.01262Â 0.09888Â 0.07133 -0.08494 -0.07265
Coefficients:
           Estimate Std. Error t value Pr(>|t|) Â
(Intercept) 16.53707Â Â Â 0.99604Â 16.603Â 0.00361 **
Dcre2[1:5]Â -0.27890Â Â Â 0.04185Â -6.665Â 0.02178 *
Dcre3[1:5]Â Â 0.21874Â Â Â 0.04643Â Â 4.711Â 0.04222 *
---
Signif. codes:Â 0 â***â 0.001 â**â 0.01 â*â 0.05 â.â 0.1 â â 1
Residual standard error: 0.1173 on 2 degrees of freedom
Multiple R-squared: 0.9739,   Adjusted R-squared: 0.9478
F-statistic: 37.28 on 2 and 2 DF, p-value: 0.02612
The difference is in the names of the coefficients.Â
names(coef(regCred6))
[1] "(Intercept)" "Dcre2[1:5]"Â "Dcre3[1:5]"Â
which you can change by,
names(regCred6$coef)<-names(regCred5$coef)
 regCred6$coef
(Intercept)Â Â Â Â Â Â Dcre2Â Â Â Â Â Â Dcre3
 16.5370694 -0.2788988  0.2187360
Though, it won't change the names of coefficients in the summary. I tried several ways, but so far not successful. I think in that case, the easiest way is to assign the subset to a new variable and run the analysis.
e.g.
Dcre1new<-Dcre1[1:5]
Hope this was helpful.
A.K.
----- Original Message -----
To: " [hidden email]" < [hidden email]>
Cc:
Sent: Saturday, June 30, 2012 6:04 PM
Subject: [R]Â Adjusting length of series
Hi
I have a follow up question, relating to subsetting to list items. After using the list and min(sapply()) method to adjust the length of the variables, I specify a dynamic regression equation using the variables in the list. My list looks like this:
Dcr<- list(Dcre1=DCred1,Dcre2=DCred2,Dcre3=DCred3,Dbobc1=DBoBC1,Dbobc2=DBoBC2,Dbobc3=DBoBC3,...)
By specifying the list items with names, I thought I could end by referencing them (or subsetting the list) as, eg., Dcr$Dcre1 and get DCred1, Dcr$Dbobc1 and get DBoBC1, etc so that the explanatory variables of the equation can be easily associated with their respective original names. This way, I would avoid specifying the list as Dcr<-list(Dcr1, Dcr2, Dcr, 3..., Dcr15) and then subsetting the list using Dcr[[1]][1:29], Dcr[[[2]][1:29], ..., Dcr[[15]][1:29] because the list has many variables (15) and referencing the variables with numbers makes them lose their original names.
When I specify the list as Dcr<- list(Dcr1, Dcr2, ..., Dcr15), then the regression equation specified as:
# Regression
regCred<- lm(Dcr[[1]][1:29]~Dcr[[2]][1:29]+Dcr[[3]][1:29]+Dcr[[4]][1:29]+Dcr[[5]][1:29]+Dcr[[6]][1:29]+...)
runs without problems - the results are shown here below:
Call:
lm(formula = Dcr[[1]][1:29] ~ Dcr[[2]][1:29] + Dcr[[3]][1:29] +
Dcr[[4]][1:29] + Dcr[[5]][1:29] + Dcr[[6]][1:29])
Residuals:
Min   1Q Median   3Q   Max
-86.293 -33.586Â -9.969Â 40.147 117.965
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)Â Â 81.02064Â Â 13.28632Â Â 6.098 3.21e-06 ***
Dcr[[2]][1:29] -0.97407Â Â 0.11081Â -8.791 8.20e-09 ***
Dcr[[3]][1:29] -0.27950Â Â 0.05899Â -4.738 8.95e-05 ***
Dcr[[4]][1:29] -0.07961Â Â 0.04856Â -1.639Â Â 0.115
Dcr[[5]][1:29] -0.07180Â Â 0.05515Â -1.302Â Â 0.206
Dcr[[6]][1:29] -0.01562Â Â 0.02086Â -0.749Â Â 0.462
But when I specify the list with names as shown above, then the equation does not run - as shown by the following error message
> # Regression
> regCred<- lm(Dcr[[1]][1:29]~Dcr[[2]][1:29]+Dcr[[3]][1:29]+Dcr[[4]][1:29]+
+ Dcr[[5]][1:29]+Dcr$Dbobc3)
Error in model.frame.default(formula = Dcr[[1]][1:29] ~ Dcr[[2]][1:29] +Â :
variable lengths differ (found for 'Dcr$Dbobc3')
> Dcr[[5]][1:29]+Dcr$Dbobc3[1:29])
Error: unexpected ')' in "Dcr[[5]][1:29]+Dcr$Dbobc3[1:29])"
NB: In the equation with error message, only the last term is specified by referencing its name (ie., Dcr$Dbobc3[1:29]. Also note that the error occurs whether I append '[1:29]' to Dcr$Dbobc or not.
How do I resolve this?
Thanks. Lexi
NB: I tried typing the above in the same email Petr used to reply me, but the email could not be delivered due to size problems
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Hi David and AK,
I have been trying to implement your suggestions since yesterday, but I encountered some challenges.
Â
As for David's suggestions, I could only implement it after some modifications. Using an abridged version of my data, I dpud my dataset and then show my steps below.
Â
> dput(ydata)
structure(c(68.1000000000004, -34.8000000000002, 90.3999999999996,
54.6000000000004, -172.3, 51.8000000000002, 175, 79.8000000000002,
-35.7000000000007, 130.5, 116.8, -67.5, 164.5, 514.8, -326.1,
98.4000000000005, 160.2, 53.1999999999998, 283.6, -111.6, 127.8,
-17.3000000000002, 286.3, NA, NA, -102.900000000001, 125.2, -35.7999999999993,
-226.900000000001, 224.1, 123.2, -95.1999999999998, -115.500000000001,
166.200000000001, -13.6999999999998, -184.3, 232, 350.3, -840.900000000001,
424.500000000001, 61.7999999999993, -107, 230.400000000001, -395.200000000001,
239.400000000001, -145.1, 303.6, NA, NA, NA, 228.1, -160.999999999999,
-191.100000000001, 451.000000000001, -100.900000000001, -218.4,
-20.3000000000011, 281.700000000002, -179.900000000001, -170.6,
416.3, 118.3, -1191.2, 1265.4, -362.700000000002, -168.799999999999,
337.400000000001, -625.600000000001, 634.600000000001, -384.500000000001,
448.700000000001, NA, NA, -164.457840999999, 17.0793539999995,
95.9767880000009, 680.238166999999, -491.348690999999, -274.694009,
-256.332907, 469.62296, -146.431891, -41.0772019999995, -106.970104,
757.688263999999, -1689.214533, 2320.098952, -1446.97942, 516.384521,
-375.277650999999, 293.867029999999, 417.845195, 278.198807,
-968.592033999999, -314.195986, NA, NA, NA, 181.537194999999,
78.8974340000013, 584.261378999998, -1171.586858, 216.654681999999,
18.3611019999998, 725.955867, -616.054851, 105.354689000001,
-65.8929020000005, 864.658367999999, -2446.902797, 4009.313485,
-3767.078372, 1963.363941, -891.662171999999, 669.144680999999,
123.978165, -139.646388, -1246.790841, 654.396048, NA, 4937,
5005.1, 4970.3, 5060.7, 5115.3, 4943, 4994.8, 5169.8, 5249.6,
5213.9, 5344.4, 5461.2, 5393.7, 5558.2, 6073, 5746.9, 5845.3,
6005.5, 6058.7, 6342.3, 6230.7, 6358.5, 6341.2, 6627.5, 4187.5,
4296.004835, 4240.051829, 4201.178177, 4258.281313, 4995.622616,
5241.615228, 5212.913831, 4927.879527, 5112.468183, 5150.624948,
5147.704511, 5037.81397, 5685.611693, 4644.194883, 5922.877025,
5754.579747, 6102.66699, 6075.476582, 6342.153204, 7026.675021,
7989.395645, 7983.524235, 7663.456839), .Dim = c(24L, 7L), .Dimnames = list(
   NULL, c("DCred1", "DCred2", "DCred3", "DBoBC2", "DBoBC3",
   "CredL1", "BoBCL1")), .Tsp = c(2001.08333333333, 2003, 12
), class = c("mts", "ts"))
Â
NB: the NAs in the dataset emanated from lagging or differencing the series
Â
David's suggestion
 df<-data.frame(DCred1,DCred2,DCred3,DBoBC2,DBoBC3,CredL1,BoBCL1)
Error in data.frame(DCred1, DCred2, DCred3, DBoBC2, DBoBC3, CredL1, BoBCL1) :
 arguments imply differing number of rows: 23, 22, 21, 24
So I modified as follows:
length(DCred3)Â # finding the minimum length of various series
[1] 21
# Then dataframe construction
dframe<- data.frame(Dcre1=DCred1[1:21],Dcre2=DCred2[1:21],Dcre3=DCred3[1:21],
+ Dbobc2=DBoBC2[1:21],Dbobc3=DBoBC3[1:21],CredL=CredL1[1:21],BoBCL=BoBCL1[1:21])
# Then estimated regression
> regCred<- lm(Dcre1~Dcre2+Dcre3+Dbobc2+Dbobc3+CredL+BoBCL, data=dframe)
> summary(regCred)
# Worked well as shown by results below
Call:
lm(formula = Dcre1 ~ Dcre2 + Dcre3 + Dbobc2 + Dbobc3 + CredL +
   BoBCL, data = dframe)
Residuals:
   Min     1Q Median     3Q    Max
-69.516 -27.695Â -8.085Â 13.851 107.276
Coefficients:
            Estimate Std. Error t value Pr(>|t|)  Â
(Intercept) 159.32304Â 157.15209Â Â 1.014 0.327873Â Â Â
Dcre2Â Â Â Â Â Â Â -0.75527Â Â Â 0.17262Â -4.375 0.000634 ***
Dcre3Â Â Â Â Â Â Â -0.21006Â Â Â 0.08656Â -2.427 0.029329 *Â
Dbobc2Â Â Â Â Â Â Â 0.05111Â Â Â 0.06565Â Â 0.779 0.449197Â Â Â
Dbobc3Â Â Â Â Â Â Â 0.03106Â Â Â 0.03510Â Â 0.885 0.391108Â Â Â
CredLÂ Â Â Â Â Â Â -0.10967Â Â Â 0.04933Â -2.223 0.043177 *Â
BoBCLÂ Â Â Â Â Â Â Â 0.09756Â Â Â 0.03097Â Â 3.150 0.007087 **
---
Signif. codes:Â 0 â***â 0.001 â**â 0.01 â*â 0.05 â.â 0.1 â â 1
Residual standard error: 52.3 on 14 degrees of freedom
Multiple R-squared: 0.9331,    Adjusted R-squared: 0.9044
F-statistic: 32.55 on 6 and 14 DF, p-value: 1.911e-07
Â
This is good, but couldn't I code the process for my 15 variable model?
Perhaps that is where the use of
Dcr<- lapply(..., function(x) ...)
comes in?
Â
AK, if you spare some minutes, please use my dput data to illustrate the suggestion you made, I searched the lapply function (using ??lapply) but could not get a handle of how to use it in my case. My dput data is as shown below.
Â
        DCred1 DCred2 DCred3     DBoBC2     DBoBC3 CredL1  BoBCL1
Feb 2001Â Â 68.1Â Â Â Â NAÂ Â Â Â Â NAÂ Â Â Â Â Â Â Â Â NAÂ Â Â Â Â Â Â Â Â NA 4937.0 4187.500
Mar 2001Â -34.8 -102.9Â Â Â Â Â NAÂ -164.45784Â Â Â Â Â Â Â Â Â NA 5005.1 4296.005
Apr 2001Â Â 90.4Â 125.2Â Â 228.1Â Â Â 17.07935Â Â 181.53719 4970.3 4240.052
May 2001Â Â 54.6Â -35.8Â -161.0Â Â Â 95.97679Â Â Â 78.89743 5060.7 4201.178
Jun 2001 -172.3 -226.9Â -191.1Â Â 680.23817Â Â 584.26138 5115.3 4258.281
Jul 2001Â Â 51.8Â 224.1Â Â 451.0Â -491.34869 -1171.58686 4943.0 4995.623
Aug 2001Â 175.0Â 123.2Â -100.9Â -274.69401Â Â 216.65468 4994.8 5241.615
Sep 2001Â Â 79.8Â -95.2Â -218.4Â -256.33291Â Â Â 18.36110 5169.8 5212.914
Oct 2001Â -35.7 -115.5Â Â -20.3Â Â 469.62296Â Â 725.95587 5249.6 4927.880
Nov 2001Â 130.5Â 166.2Â Â 281.7Â -146.43189Â -616.05485 5213.9 5112.468
Dec 2001Â 116.8Â -13.7Â -179.9Â Â -41.07720Â Â 105.35469 5344.4 5150.625
Jan 2002Â -67.5 -184.3Â -170.6Â -106.97010Â Â -65.89290 5461.2 5147.705
Feb 2002Â 164.5Â 232.0Â Â 416.3Â Â 757.68826Â Â 864.65837 5393.7 5037.814
Mar 2002Â 514.8Â 350.3Â Â 118.3 -1689.21453 -2446.90280 5558.2 5685.612
Apr 2002 -326.1 -840.9 -1191.2Â 2320.09895Â 4009.31348 6073.0 4644.195
May 2002Â Â 98.4Â 424.5Â 1265.4 -1446.97942 -3767.07837 5746.9 5922.877
Jun 2002Â 160.2Â Â 61.8Â -362.7Â Â 516.38452Â 1963.36394 5845.3 5754.580
Jul 2002Â Â 53.2 -107.0Â -168.8Â -375.27765Â -891.66217 6005.5 6102.667
Aug 2002Â 283.6Â 230.4Â Â 337.4Â Â 293.86703Â Â 669.14468 6058.7 6075.477
Sep 2002 -111.6 -395.2Â -625.6Â Â 417.84519Â Â 123.97817 6342.3 6342.153
Oct 2002Â 127.8Â 239.4Â Â 634.6Â Â 278.19881Â -139.64639 6230.7 7026.675
Nov 2002Â -17.3 -145.1Â -384.5Â -968.59203 -1246.79084 6358.5 7989.396
Dec 2002Â 286.3Â 303.6Â Â 448.7Â -314.19599Â Â 654.39605 6341.2 7983.524
Jan 2003Â Â Â Â NAÂ Â Â Â NAÂ Â Â Â Â NAÂ Â Â Â Â Â Â Â Â NAÂ Â Â Â Â Â Â Â Â NA 6627.5 7663.457
Thanks kindly. Lexi
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On Jul 2, 2012, at 5:13 AM, Lekgatlhamang, lexi Setlhare wrote:
> Hi David and AK,
> I have been trying to implement your suggestions since yesterday,
> but I encountered some challenges.
>
> As for David's suggestions, I could only implement it after some
> modifications. Using an abridged version of my data, I dpud my
> dataset and then show my steps below.
Well, your initial question (why the $ referencing did not work) is
now answered. This is not a dataframe but rather a 'ts' classed object
and there is no `$` method for such objects. They are really matrices
with some extra attributes.
> ydata$BoBCL1
Error in ydata$BoBCL1 : $ operator is invalid for atomic vectors
As I understood it you were able to get useful analyses using the
formula methods for lm on these objects, but were just having
difficulty with the "$" operator. So the answer is ..... don't do that.
--
David.
>
>> dput(ydata)
> structure(c(68.1000000000004, -34.8000000000002, 90.3999999999996,
> 54.6000000000004, -172.3, 51.8000000000002, 175, 79.8000000000002,
> -35.7000000000007, 130.5, 116.8, -67.5, 164.5, 514.8, -326.1,
> 98.4000000000005, 160.2, 53.1999999999998, 283.6, -111.6, 127.8,
> -17.3000000000002, 286.3, NA, NA, -102.900000000001, 125.2,
> -35.7999999999993,
> -226.900000000001, 224.1, 123.2, -95.1999999999998, -115.500000000001,
> 166.200000000001, -13.6999999999998, -184.3, 232, 350.3,
> -840.900000000001,
> 424.500000000001, 61.7999999999993, -107, 230.400000000001,
> -395.200000000001,
> 239.400000000001, -145.1, 303.6, NA, NA, NA, 228.1, -160.999999999999,
> -191.100000000001, 451.000000000001, -100.900000000001, -218.4,
> -20.3000000000011, 281.700000000002, -179.900000000001, -170.6,
> 416.3, 118.3, -1191.2, 1265.4, -362.700000000002, -168.799999999999,
> 337.400000000001, -625.600000000001, 634.600000000001,
> -384.500000000001,
> 448.700000000001, NA, NA, -164.457840999999, 17.0793539999995,
> 95.9767880000009, 680.238166999999, -491.348690999999, -274.694009,
> -256.332907, 469.62296, -146.431891, -41.0772019999995, -106.970104,
> 757.688263999999, -1689.214533, 2320.098952, -1446.97942, 516.384521,
> -375.277650999999, 293.867029999999, 417.845195, 278.198807,
> -968.592033999999, -314.195986, NA, NA, NA, 181.537194999999,
> 78.8974340000013, 584.261378999998, -1171.586858, 216.654681999999,
> 18.3611019999998, 725.955867, -616.054851, 105.354689000001,
> -65.8929020000005, 864.658367999999, -2446.902797, 4009.313485,
> -3767.078372, 1963.363941, -891.662171999999, 669.144680999999,
> 123.978165, -139.646388, -1246.790841, 654.396048, NA, 4937,
> 5005.1, 4970.3, 5060.7, 5115.3, 4943, 4994.8, 5169.8, 5249.6,
> 5213.9, 5344.4, 5461.2, 5393.7, 5558.2, 6073, 5746.9, 5845.3,
> 6005.5, 6058.7, 6342.3, 6230.7, 6358.5, 6341.2, 6627.5, 4187.5,
> 4296.004835, 4240.051829, 4201.178177, 4258.281313, 4995.622616,
> 5241.615228, 5212.913831, 4927.879527, 5112.468183, 5150.624948,
> 5147.704511, 5037.81397, 5685.611693, 4644.194883, 5922.877025,
> 5754.579747, 6102.66699, 6075.476582, 6342.153204, 7026.675021,
> 7989.395645, 7983.524235, 7663.456839), .Dim = c(24L, 7L), .Dimnames
> = list(
> NULL, c("DCred1", "DCred2", "DCred3", "DBoBC2", "DBoBC3",
> "CredL1", "BoBCL1")), .Tsp = c(2001.08333333333, 2003, 12
> ), class = c("mts", "ts"))
>
> NB: the NAs in the dataset emanated from lagging or differencing the
> series
>
> David's suggestion
> df<-data.frame(DCred1,DCred2,DCred3,DBoBC2,DBoBC3,CredL1,BoBCL1)
> Error in data.frame(DCred1, DCred2, DCred3, DBoBC2, DBoBC3, CredL1,
> BoBCL1) :
> arguments imply differing number of rows: 23, 22, 21, 24
>
> So I modified as follows:
> length(DCred3) # finding the minimum length of various series
> [1] 21
>
> # Then dataframe construction
> dframe<-
> data.frame(Dcre1=DCred1[1:21],Dcre2=DCred2[1:21],Dcre3=DCred3[1:21],
> +
> Dbobc2
> =
> DBoBC2
> [1:21],Dbobc3=DBoBC3[1:21],CredL=CredL1[1:21],BoBCL=BoBCL1[1:21])
> # Then estimated regression
>> regCred<- lm(Dcre1~Dcre2+Dcre3+Dbobc2+Dbobc3+CredL+BoBCL,
>> data=dframe)
>> summary(regCred)
> # Worked well as shown by results below
> Call:
> lm(formula = Dcre1 ~ Dcre2 + Dcre3 + Dbobc2 + Dbobc3 + CredL +
> BoBCL, data = dframe)
> Residuals:
> Min 1Q Median 3Q Max
> -69.516 -27.695 -8.085 13.851 107.276
> Coefficients:
> Estimate Std. Error t value Pr(>|t|)
> (Intercept) 159.32304 157.15209 1.014 0.327873
> Dcre2 -0.75527 0.17262 -4.375 0.000634 ***
> Dcre3 -0.21006 0.08656 -2.427 0.029329 *
> Dbobc2 0.05111 0.06565 0.779 0.449197
> Dbobc3 0.03106 0.03510 0.885 0.391108
> CredL -0.10967 0.04933 -2.223 0.043177 *
> BoBCL 0.09756 0.03097 3.150 0.007087 **
> ---
> Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
> Residual standard error: 52.3 on 14 degrees of freedom
> Multiple R-squared: 0.9331, Adjusted R-squared: 0.9044
> F-statistic: 32.55 on 6 and 14 DF, p-value: 1.911e-07
>
> This is good, but couldn't I code the process for my 15 variable
> model?
> Perhaps that is where the use of
> Dcr<- lapply(..., function(x) ...)
> comes in?
>
> AK, if you spare some minutes, please use my dput data to illustrate
> the suggestion you made, I searched the lapply function (using ??
> lapply) but could not get a handle of how to use it in my case. My
> dput data is as shown below.
>
> DCred1 DCred2 DCred3 DBoBC2 DBoBC3 CredL1 BoBCL1
> Feb 2001 68.1 NA NA NA NA 4937.0 4187.500
> Mar 2001 -34.8 -102.9 NA -164.45784 NA 5005.1 4296.005
> Apr 2001 90.4 125.2 228.1 17.07935 181.53719 4970.3 4240.052
> May 2001 54.6 -35.8 -161.0 95.97679 78.89743 5060.7 4201.178
> Jun 2001 -172.3 -226.9 -191.1 680.23817 584.26138 5115.3 4258.281
> Jul 2001 51.8 224.1 451.0 -491.34869 -1171.58686 4943.0 4995.623
> Aug 2001 175.0 123.2 -100.9 -274.69401 216.65468 4994.8 5241.615
> Sep 2001 79.8 -95.2 -218.4 -256.33291 18.36110 5169.8 5212.914
> Oct 2001 -35.7 -115.5 -20.3 469.62296 725.95587 5249.6 4927.880
> Nov 2001 130.5 166.2 281.7 -146.43189 -616.05485 5213.9 5112.468
> Dec 2001 116.8 -13.7 -179.9 -41.07720 105.35469 5344.4 5150.625
> Jan 2002 -67.5 -184.3 -170.6 -106.97010 -65.89290 5461.2 5147.705
> Feb 2002 164.5 232.0 416.3 757.68826 864.65837 5393.7 5037.814
> Mar 2002 514.8 350.3 118.3 -1689.21453 -2446.90280 5558.2 5685.612
> Apr 2002 -326.1 -840.9 -1191.2 2320.09895 4009.31348 6073.0 4644.195
> May 2002 98.4 424.5 1265.4 -1446.97942 -3767.07837 5746.9 5922.877
> Jun 2002 160.2 61.8 -362.7 516.38452 1963.36394 5845.3 5754.580
> Jul 2002 53.2 -107.0 -168.8 -375.27765 -891.66217 6005.5 6102.667
> Aug 2002 283.6 230.4 337.4 293.86703 669.14468 6058.7 6075.477
> Sep 2002 -111.6 -395.2 -625.6 417.84519 123.97817 6342.3 6342.153
> Oct 2002 127.8 239.4 634.6 278.19881 -139.64639 6230.7 7026.675
> Nov 2002 -17.3 -145.1 -384.5 -968.59203 -1246.79084 6358.5 7989.396
> Dec 2002 286.3 303.6 448.7 -314.19599 654.39605 6341.2 7983.524
> Jan 2003 NA NA NA NA NA 6627.5 7663.457
>
> Thanks kindly. Lexi
> [[alternative HTML version deleted]]
>
> ______________________________________________
> [hidden email] mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html> and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
______________________________________________
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https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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In reply to this post by Lekgatlhamang, lexi Setlhare
Hello,
The class of your data is not dataframe.
Suppose I call your data as ydat1
str(ydat1)
mts [1:24, 1:7] 68.1 -34.8 90.4 54.6 -172.3 ...
- attr(*, "dimnames")=List of 2
..$ : NULL
..$ : chr [1:7] "DCred1" "DCred2" "DCred3" "DBoBC2" ...
- attr(*, "tsp")= num [1:3] 2001 2003 12
- attr(*, "class")= chr [1:2] "mts" "ts"
ydat2<-data.frame(ydat1)
str(ydat2)
'data.frame': 24 obs. of 7 variables:
$ DCred1: num 68.1 -34.8 90.4 54.6 -172.3 ...
$ DCred2: num NA -102.9 125.2 -35.8 -226.9 ...
$ DCred3: num NA NA 228 -161 -191 ...
$ DBoBC2: num NA -164.5 17.1 96 680.2 ...
$ DBoBC3: num NA NA 181.5 78.9 584.3 ...
$ CredL1: num 4937 5005 4970 5061 5115 ...
$ BoBCL1: num 4188 4296 4240 4201 4258 ...
#Since you wanted only to do lm for these columns, I guess it doesn't really matter whether you have month and year in the dataset.
#With NAs
regCred<-lm(DCred1~DCred2+DCred3+DBoBC2+DBoBC3+CredL1+BoBCL1,data=ydat2)
> summary(regCred)
Call:
lm(formula = DCred1 ~ DCred2 + DCred3 + DBoBC2 + DBoBC3 + CredL1 +
BoBCL1, data = ydat2)
Residuals:
Min 1Q Median 3Q Max
-124.988463 -33.133975 7.971083 23.607953 76.813601
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -538.61375718 205.91179535 -2.61575 0.020344 *
DCred2 0.96401908 0.15623660 6.17025 2.4337e-05 ***
DCred3 -0.25720355 0.08983607 -2.86303 0.012524 *
DBoBC2 -0.11222347 0.07828182 -1.43358 0.173646
DBoBC3 0.04564621 0.03825169 1.19331 0.252578
CredL1 0.18499925 0.06565456 2.81777 0.013693 *
BoBCL1 -0.07682710 0.03406916 -2.25503 0.040666 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 54.44479 on 14 degrees of freedom
(3 observations deleted due to missingness)
Multiple R-squared: 0.9324472, Adjusted R-squared: 0.903496
F-statistic: 32.20757 on 6 and 14 DF, p-value: 2.046024e-07
Without NAs
> ydat3<-na.omit(ydat2)
> regCred<-lm(DCred1~DCred2+DCred3+DBoBC2+DBoBC3+CredL1+BoBCL1,data=ydat3)
> summary(regCred)
Call:
lm(formula = DCred1 ~ DCred2 + DCred3 + DBoBC2 + DBoBC3 + CredL1 +
BoBCL1, data = ydat3)
Residuals:
Min 1Q Median 3Q Max
-124.988463 -33.133975 7.971083 23.607953 76.813601
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -538.61375718 205.91179535 -2.61575 0.020344 *
DCred2 0.96401908 0.15623660 6.17025 2.4337e-05 ***
DCred3 -0.25720355 0.08983607 -2.86303 0.012524 *
DBoBC2 -0.11222347 0.07828182 -1.43358 0.173646
DBoBC3 0.04564621 0.03825169 1.19331 0.252578
CredL1 0.18499925 0.06565456 2.81777 0.013693 *
BoBCL1 -0.07682710 0.03406916 -2.25503 0.040666 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 54.44479 on 14 degrees of freedom
Multiple R-squared: 0.9324472, Adjusted R-squared: 0.903496
F-statistic: 32.20757 on 6 and 14 DF, p-value: 2.046024e-
#Same result
Not sure what you meant by ("This is good, but couldn't I code the process for my 15 variable model?")
A.K.
________________________________
From: "Lekgatlhamang, lexi Setlhare" < [hidden email]>
To: arun < [hidden email]>
Cc: R help < [hidden email]>
Sent: Monday, July 2, 2012 5:13 AM
Subject: Re: [R] Adjusting length of series
Hi David and AK,
I have been trying to implement your suggestions since yesterday, but I encountered some challenges.
As for David's suggestions, I could only implement it after some modifications. Using an abridged version of my data, I dpud my dataset and then show my steps below.
> dput(ydata)
structure(c(68.1000000000004, -34.8000000000002, 90.3999999999996,
54.6000000000004, -172.3, 51.8000000000002, 175, 79.8000000000002,
-35.7000000000007, 130.5, 116.8, -67.5, 164.5, 514.8, -326.1,
98.4000000000005, 160.2, 53.1999999999998, 283.6, -111.6, 127.8,
-17.3000000000002, 286.3, NA, NA, -102.900000000001, 125.2, -35.7999999999993,
-226.900000000001, 224.1, 123.2,
-95.1999999999998, -115.500000000001,
166.200000000001, -13.6999999999998, -184.3, 232, 350.3, -840.900000000001,
424.500000000001, 61.7999999999993, -107, 230.400000000001, -395.200000000001,
239.400000000001, -145.1, 303.6, NA, NA, NA, 228.1, -160.999999999999,
-191.100000000001, 451.000000000001, -100.900000000001, -218.4,
-20.3000000000011, 281.700000000002, -179.900000000001, -170.6,
416.3, 118.3, -1191.2, 1265.4, -362.700000000002, -168.799999999999,
337.400000000001, -625.600000000001, 634.600000000001, -384.500000000001,
448.700000000001, NA, NA, -164.457840999999, 17.0793539999995,
95.9767880000009, 680.238166999999, -491.348690999999, -274.694009,
-256.332907, 469.62296, -146.431891, -41.0772019999995, -106.970104,
757.688263999999, -1689.214533, 2320.098952, -1446.97942, 516.384521,
-375.277650999999, 293.867029999999, 417.845195, 278.198807,
-968.592033999999, -314.195986, NA, NA, NA,
181.537194999999,
78.8974340000013, 584.261378999998, -1171.586858, 216.654681999999,
18.3611019999998, 725.955867, -616.054851, 105.354689000001,
-65.8929020000005, 864.658367999999, -2446.902797, 4009.313485,
-3767.078372, 1963.363941, -891.662171999999, 669.144680999999,
123.978165, -139.646388, -1246.790841, 654.396048, NA, 4937,
5005.1, 4970.3, 5060.7, 5115.3, 4943, 4994.8, 5169.8, 5249.6,
5213.9, 5344.4, 5461.2, 5393.7, 5558.2, 6073, 5746.9, 5845.3,
6005.5, 6058.7, 6342.3, 6230.7, 6358.5, 6341.2, 6627.5, 4187.5,
4296.004835, 4240.051829, 4201.178177, 4258.281313, 4995.622616,
5241.615228, 5212.913831, 4927.879527, 5112.468183, 5150.624948,
5147.704511, 5037.81397, 5685.611693, 4644.194883, 5922.877025,
5754.579747, 6102.66699, 6075.476582, 6342.153204, 7026.675021,
7989.395645, 7983.524235, 7663.456839), .Dim = c(24L, 7L), .Dimnames = list(
NULL, c("DCred1", "DCred2",
"DCred3", "DBoBC2", "DBoBC3",
"CredL1", "BoBCL1")), .Tsp = c(2001.08333333333, 2003, 12
), class = c("mts", "ts"))
NB: the NAs in the dataset emanated from lagging or differencing the series
David's suggestion
df<-data.frame(DCred1,DCred2,DCred3,DBoBC2,DBoBC3,CredL1,BoBCL1)
Error in data.frame(DCred1, DCred2, DCred3, DBoBC2, DBoBC3, CredL1, BoBCL1) :
arguments imply differing number of rows: 23, 22, 21, 24
So I modified as follows:
length(DCred3) # finding the minimum length of various series
[1] 21
# Then dataframe construction
dframe<- data.frame(Dcre1=DCred1[1:21],Dcre2=DCred2[1:21],Dcre3=DCred3[1:21],
+ Dbobc2=DBoBC2[1:21],Dbobc3=DBoBC3[1:21],CredL=CredL1[1:21],BoBCL=BoBCL1[1:21])
# Then estimated regression
> regCred<- lm(Dcre1~Dcre2+Dcre3+Dbobc2+Dbobc3+CredL+BoBCL, data=dframe)
> summary(regCred)
# Worked well as shown by results below
Call:
lm(formula = Dcre1 ~ Dcre2 + Dcre3 + Dbobc2 + Dbobc3 + CredL +
BoBCL, data = dframe)
Residuals:
Min 1Q Median 3Q Max
-69.516 -27.695 -8.085 13.851 107.276
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 159.32304 157.15209 1.014 0.327873
Dcre2 -0.75527 0.17262 -4.375 0.000634 ***
Dcre3 -0.21006 0.08656 -2.427 0.029329 *
Dbobc2 0.05111 0.06565 0.779 0.449197
Dbobc3 0.03106 0.03510 0.885 0.391108
CredL -0.10967 0.04933 -2.223 0.043177 *
BoBCL 0.09756 0.03097 3.150 0.007087 **
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 52.3 on 14 degrees of freedom
Multiple R-squared: 0.9331, Adjusted R-squared: 0.9044
F-statistic: 32.55 on 6 and 14 DF, p-value: 1.911e-07
This is good, but couldn't I code the process for my 15 variable model?
Perhaps that is where the use of
Dcr<- lapply(..., function(x) ...)
comes in?
AK, if you spare some minutes, please use my dput data to illustrate the suggestion you made, I searched the lapply function (using ??lapply) but could not get a handle of how to use it in my case. My dput data is as shown below.
DCred1 DCred2 DCred3 DBoBC2 DBoBC3 CredL1 BoBCL1
Feb 2001 68.1 NA NA NA NA 4937.0 4187.500
Mar 2001 -34.8 -102.9 NA -164.45784 NA 5005.1 4296.005
Apr 2001 90.4 125.2
228.1 17.07935 181.53719 4970.3 4240.052
May 2001 54.6 -35.8 -161.0 95.97679 78.89743 5060.7 4201.178
Jun 2001 -172.3 -226.9 -191.1 680.23817 584.26138 5115.3 4258.281
Jul 2001 51.8 224.1 451.0 -491.34869 -1171.58686 4943.0 4995.623
Aug 2001 175.0 123.2 -100.9 -274.69401 216.65468 4994.8 5241.615
Sep 2001 79.8 -95.2 -218.4 -256.33291 18.36110 5169.8 5212.914
Oct 2001 -35.7 -115.5 -20.3 469.62296 725.95587 5249.6 4927.880
Nov 2001 130.5 166.2 281.7 -146.43189 -616.05485 5213.9 5112.468
Dec 2001 116.8 -13.7 -179.9 -41.07720 105.35469 5344.4 5150.625
Jan 2002 -67.5
-184.3 -170.6 -106.97010 -65.89290 5461.2 5147.705
Feb 2002 164.5 232.0 416.3 757.68826 864.65837 5393.7 5037.814
Mar 2002 514.8 350.3 118.3 -1689.21453 -2446.90280 5558.2 5685.612
Apr 2002 -326.1 -840.9 -1191.2 2320.09895 4009.31348 6073.0 4644.195
May 2002 98.4 424.5 1265.4 -1446.97942 -3767.07837 5746.9 5922.877
Jun 2002 160.2 61.8 -362.7 516.38452 1963.36394 5845.3 5754.580
Jul 2002 53.2 -107.0 -168.8 -375.27765 -891.66217 6005.5 6102.667
Aug 2002 283.6 230.4 337.4 293.86703 669.14468 6058.7 6075.477
Sep 2002 -111.6 -395.2 -625.6 417.84519 123.97817 6342.3 6342.153
Oct 2002 127.8 239.4 634.6 278.19881
-139.64639 6230.7 7026.675
Nov 2002 -17.3 -145.1 -384.5 -968.59203 -1246.79084 6358.5 7989.396
Dec 2002 286.3 303.6 448.7 -314.19599 654.39605 6341.2 7983.524
Jan 2003 NA NA NA NA NA 6627.5 7663.457
Thanks kindly. Lexi
______________________________________________
[hidden email] mailing list
https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
|
|
|
Noted David, and thanks very much.
Â
Lexi
________________________________
From: David Winsemius < [hidden email]>
Sent: Monday, July 2, 2012 4:26 PM
Subject: Re: [R] Adjusting length of series
On Jul 2, 2012, at 5:13 AM, Lekgatlhamang, lexi Setlhare wrote:
> Hi David and AK,
> I have been trying to implement your suggestions since yesterday, but I encountered some challenges.
>
> As for David's suggestions, I could only implement it after some modifications. Using an abridged version of my data, I dpud my dataset and then show my steps below.
Well, your initial question (why the $ referencing did not work) is now answered. This is not a dataframe but rather a 'ts' classed object and there is no `$` method for such objects. They are really matrices with some extra attributes.
> ydata$BoBCL1
Error in ydata$BoBCL1 : $ operator is invalid for atomic vectors
As I understood it you were able to get useful analyses using the formula methods for lm on these objects, but were just having difficulty with the "$" operator. So the answer is ..... don't do that.
--David.
>
>> dput(ydata)
> structure(c(68.1000000000004, -34.8000000000002, 90.3999999999996,
> 54.6000000000004, -172.3, 51.8000000000002, 175, 79.8000000000002,
> -35.7000000000007, 130.5, 116.8, -67.5, 164.5, 514.8, -326.1,
> 98.4000000000005, 160.2, 53.1999999999998, 283.6, -111.6, 127.8,
> -17.3000000000002, 286.3, NA, NA, -102.900000000001, 125.2, -35.7999999999993,
> -226.900000000001, 224.1, 123.2, -95.1999999999998, -115.500000000001,
> 166.200000000001, -13.6999999999998, -184.3, 232, 350.3, -840.900000000001,
> 424.500000000001, 61.7999999999993, -107, 230.400000000001, -395.200000000001,
> 239.400000000001, -145.1, 303.6, NA, NA, NA, 228.1, -160.999999999999,
> -191.100000000001, 451.000000000001, -100.900000000001, -218.4,
> -20.3000000000011, 281.700000000002, -179.900000000001, -170.6,
> 416.3, 118.3, -1191.2, 1265.4, -362.700000000002, -168.799999999999,
> 337.400000000001, -625.600000000001, 634.600000000001, -384.500000000001,
> 448.700000000001, NA, NA, -164.457840999999, 17.0793539999995,
> 95.9767880000009, 680.238166999999, -491.348690999999, -274.694009,
> -256.332907, 469.62296, -146.431891, -41.0772019999995, -106.970104,
> 757.688263999999, -1689.214533, 2320.098952, -1446.97942, 516.384521,
> -375.277650999999, 293.867029999999, 417.845195, 278.198807,
> -968.592033999999, -314.195986, NA, NA, NA, 181.537194999999,
> 78.8974340000013, 584.261378999998, -1171.586858, 216.654681999999,
> 18.3611019999998, 725.955867, -616.054851, 105.354689000001,
> -65.8929020000005, 864.658367999999, -2446.902797, 4009.313485,
> -3767.078372, 1963.363941, -891.662171999999, 669.144680999999,
> 123.978165, -139.646388, -1246.790841, 654.396048, NA, 4937,
> 5005.1, 4970.3, 5060.7, 5115.3, 4943, 4994.8, 5169.8, 5249.6,
> 5213.9, 5344.4, 5461.2, 5393.7, 5558.2, 6073, 5746.9, 5845.3,
> 6005.5, 6058.7, 6342.3, 6230.7, 6358.5, 6341.2, 6627.5, 4187.5,
> 4296.004835, 4240.051829, 4201.178177, 4258.281313, 4995.622616,
> 5241.615228, 5212.913831, 4927.879527, 5112.468183, 5150.624948,
> 5147.704511, 5037.81397, 5685.611693, 4644.194883, 5922.877025,
> 5754.579747, 6102.66699, 6075.476582, 6342.153204, 7026.675021,
> 7989.395645, 7983.524235, 7663.456839), .Dim = c(24L, 7L), .Dimnames = list(
>Â Â NULL, c("DCred1", "DCred2", "DCred3", "DBoBC2", "DBoBC3",
>Â Â "CredL1", "BoBCL1")), .Tsp = c(2001.08333333333, 2003, 12
> ), class = c("mts", "ts"))
>
> NB: the NAs in the dataset emanated from lagging or differencing the series
>
> David's suggestion
>Â df<-data.frame(DCred1,DCred2,DCred3,DBoBC2,DBoBC3,CredL1,BoBCL1)
> Error in data.frame(DCred1, DCred2, DCred3, DBoBC2, DBoBC3, CredL1, BoBCL1) :
>Â arguments imply differing number of rows: 23, 22, 21, 24
>
> So I modified as follows:
> length(DCred3)Â # finding the minimum length of various series
> [1] 21
>
> # Then dataframe construction
> dframe<- data.frame(Dcre1=DCred1[1:21],Dcre2=DCred2[1:21],Dcre3=DCred3[1:21],
> + Dbobc2=DBoBC2[1:21],Dbobc3=DBoBC3[1:21],CredL=CredL1[1:21],BoBCL=BoBCL1[1:21])
> # Then estimated regression
>> regCred<- lm(Dcre1~Dcre2+Dcre3+Dbobc2+Dbobc3+CredL+BoBCL, data=dframe)
>> summary(regCred)
> # Worked well as shown by results below
> Call:
> lm(formula = Dcre1 ~ Dcre2 + Dcre3 + Dbobc2 + Dbobc3 + CredL +
>Â Â BoBCL, data = dframe)
> Residuals:
>  Min   1Q Median   3Q  Max
> -69.516 -27.695Â -8.085Â 13.851 107.276
> Coefficients:
>Â Â Â Â Â Â Â Estimate Std. Error t value Pr(>|t|)
> (Intercept) 159.32304Â 157.15209Â 1.014 0.327873
> Dcre2Â Â Â Â -0.75527Â Â 0.17262Â -4.375 0.000634 ***
> Dcre3Â Â Â Â -0.21006Â Â 0.08656Â -2.427 0.029329 *
> Dbobc2Â Â Â Â 0.05111Â Â 0.06565Â 0.779 0.449197
> Dbobc3Â Â Â Â 0.03106Â Â 0.03510Â 0.885 0.391108
> CredLÂ Â Â Â -0.10967Â Â 0.04933Â -2.223 0.043177 *
> BoBCLÂ Â Â Â 0.09756Â Â 0.03097Â 3.150 0.007087 **
> ---
> Signif. codes:Â 0 â***â 0.001 â**â 0.01 â*â 0.05 â.â 0.1 â â 1
> Residual standard error: 52.3 on 14 degrees of freedom
> Multiple R-squared: 0.9331,  Adjusted R-squared: 0.9044
> F-statistic: 32.55 on 6 and 14 DF, p-value: 1.911e-07
>
> This is good, but couldn't I code the process for my 15 variable model?
> Perhaps that is where the use of
> Dcr<- lapply(..., function(x) ...)
> comes in?
>
> AK, if you spare some minutes, please use my dput data to illustrate the suggestion you made, I searched the lapply function (using ??lapply) but could not get a handle of how to use it in my case. My dput data is as shown below.
>
>Â Â Â Â Â DCred1 DCred2Â DCred3Â Â Â DBoBC2Â Â Â DBoBC3 CredL1Â BoBCL1
> Feb 2001Â 68.1Â Â NAÂ Â Â NAÂ Â Â Â Â NAÂ Â Â Â Â NA 4937.0 4187.500
> Mar 2001Â -34.8 -102.9Â Â Â NAÂ -164.45784Â Â Â Â Â NA 5005.1 4296.005
> Apr 2001Â 90.4Â 125.2Â 228.1Â Â 17.07935Â 181.53719 4970.3 4240.052
> May 2001Â 54.6Â -35.8Â -161.0Â Â 95.97679Â Â 78.89743 5060.7 4201.178
> Jun 2001 -172.3 -226.9Â -191.1Â 680.23817Â 584.26138 5115.3 4258.281
> Jul 2001Â 51.8Â 224.1Â 451.0Â -491.34869 -1171.58686 4943.0 4995.623
> Aug 2001Â 175.0Â 123.2Â -100.9Â -274.69401Â 216.65468 4994.8 5241.615
> Sep 2001Â 79.8Â -95.2Â -218.4Â -256.33291Â Â 18.36110 5169.8 5212.914
> Oct 2001Â -35.7 -115.5Â -20.3Â 469.62296Â 725.95587 5249.6 4927.880
> Nov 2001Â 130.5Â 166.2Â 281.7Â -146.43189Â -616.05485 5213.9 5112.468
> Dec 2001Â 116.8Â -13.7Â -179.9Â -41.07720Â 105.35469 5344.4 5150.625
> Jan 2002Â -67.5 -184.3Â -170.6Â -106.97010Â -65.89290 5461.2 5147.705
> Feb 2002Â 164.5Â 232.0Â 416.3Â 757.68826Â 864.65837 5393.7 5037.814
> Mar 2002Â 514.8Â 350.3Â 118.3 -1689.21453 -2446.90280 5558.2 5685.612
> Apr 2002 -326.1 -840.9 -1191.2Â 2320.09895Â 4009.31348 6073.0 4644.195
> May 2002Â 98.4Â 424.5Â 1265.4 -1446.97942 -3767.07837 5746.9 5922.877
> Jun 2002Â 160.2Â 61.8Â -362.7Â 516.38452Â 1963.36394 5845.3 5754.580
> Jul 2002Â 53.2 -107.0Â -168.8Â -375.27765Â -891.66217 6005.5 6102.667
> Aug 2002Â 283.6Â 230.4Â 337.4Â 293.86703Â 669.14468 6058.7 6075.477
> Sep 2002 -111.6 -395.2Â -625.6Â 417.84519Â 123.97817 6342.3 6342.153
> Oct 2002Â 127.8Â 239.4Â 634.6Â 278.19881Â -139.64639 6230.7 7026.675
> Nov 2002Â -17.3 -145.1Â -384.5Â -968.59203 -1246.79084 6358.5 7989.396
> Dec 2002Â 286.3Â 303.6Â 448.7Â -314.19599Â 654.39605 6341.2 7983.524
> Jan 2003Â Â NAÂ Â NAÂ Â Â NAÂ Â Â Â Â NAÂ Â Â Â Â NA 6627.5 7663.457
>
> Thanks kindly. Lexi
> Â Â Â [[alternative HTML version deleted]]
>
> ______________________________________________
> [hidden email] mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html> and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
[[alternative HTML version deleted]]
______________________________________________
[hidden email] mailing list
https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
|
|
|
Thanks very much A.K. I have to admit that my problem was not clearly stated, with the structure of my data provided. Now all is well.
Â
Cheers
Lexi
________________________________
Cc: R help < [hidden email]>
Sent: Monday, July 2, 2012 4:40 PM
Subject: Re: [R] Adjusting length of series
Hello,
The class of your data is not dataframe.
Suppose I call your data as ydat1
str(ydat1)
 mts [1:24, 1:7] 68.1 -34.8 90.4 54.6 -172.3 ...
 - attr(*, "dimnames")=List of 2
 ..$ : NULL
 ..$ : chr [1:7] "DCred1" "DCred2" "DCred3" "DBoBC2" ...
 - attr(*, "tsp")= num [1:3] 2001 2003 12
 - attr(*, "class")= chr [1:2] "mts" "ts"
ydat2<-data.frame(ydat1)
str(ydat2)
'data.frame':   24 obs. of 7 variables:
 $ DCred1: num 68.1 -34.8 90.4 54.6 -172.3 ...
 $ DCred2: num NA -102.9 125.2 -35.8 -226.9 ...
 $ DCred3: num NA NA 228 -161 -191 ...
 $ DBoBC2: num NA -164.5 17.1 96 680.2 ...
 $ DBoBC3: num NA NA 181.5 78.9 584.3 ...
 $ CredL1: num 4937 5005 4970 5061 5115 ...
 $ BoBCL1: num 4188 4296 4240 4201 4258 ...
#Since you wanted only to do lm for these columns, I guess it doesn't really matter whether you have month and year in the dataset.
 #With NAs
 regCred<-lm(DCred1~DCred2+DCred3+DBoBC2+DBoBC3+CredL1+BoBCL1,data=ydat2)
> summary(regCred)
Call:
lm(formula = DCred1 ~ DCred2 + DCred3 + DBoBC2 + DBoBC3 + CredL1 +
   BoBCL1, data = ydat2)
Residuals:
       Min         1Q     Median         3Q        Max
-124.988463Â -33.133975Â Â Â 7.971083Â Â 23.607953Â Â 76.813601
Coefficients:
                Estimate   Std. Error t value  Pr(>|t|)  Â
(Intercept) -538.61375718Â 205.91179535 -2.61575Â Â 0.020344 *Â
DCred2Â Â Â Â Â Â Â Â 0.96401908Â Â Â 0.15623660Â 6.17025 2.4337e-05 ***
DCred3Â Â Â Â Â Â Â -0.25720355Â Â Â 0.08983607 -2.86303Â Â 0.012524 *Â
DBoBC2Â Â Â Â Â Â Â -0.11222347Â Â Â 0.07828182 -1.43358Â Â 0.173646Â Â Â
DBoBC3Â Â Â Â Â Â Â Â 0.04564621Â Â Â 0.03825169Â 1.19331Â Â 0.252578Â Â Â
CredL1Â Â Â Â Â Â Â Â 0.18499925Â Â Â 0.06565456Â 2.81777Â Â 0.013693 *Â
BoBCL1Â Â Â Â Â Â Â -0.07682710Â Â Â 0.03406916 -2.25503Â Â 0.040666 *Â
---
Signif. codes:Â 0 â***â 0.001 â**â 0.01 â*â 0.05 â.â 0.1 â â 1
Residual standard error: 54.44479 on 14 degrees of freedom
 (3 observations deleted due to missingness)
Multiple R-squared: 0.9324472,   Adjusted R-squared: 0.903496
F-statistic: 32.20757 on 6 and 14 DF, p-value: 2.046024e-07
Without NAs
> ydat3<-na.omit(ydat2)
> regCred<-lm(DCred1~DCred2+DCred3+DBoBC2+DBoBC3+CredL1+BoBCL1,data=ydat3)
> summary(regCred)
Call:
lm(formula = DCred1 ~ DCred2 + DCred3 + DBoBC2 + DBoBC3 + CredL1 +
   BoBCL1, data = ydat3)
Residuals:
       Min         1Q     Median         3Q        Max
-124.988463Â -33.133975Â Â Â 7.971083Â Â 23.607953Â Â 76.813601
Coefficients:
                Estimate   Std. Error t value  Pr(>|t|)  Â
(Intercept) -538.61375718Â 205.91179535 -2.61575Â Â 0.020344 *Â
DCred2Â Â Â Â Â Â Â Â 0.96401908Â Â Â 0.15623660Â 6.17025 2.4337e-05 ***
DCred3Â Â Â Â Â Â Â -0.25720355Â Â Â 0.08983607 -2.86303Â Â 0.012524 *Â
DBoBC2Â Â Â Â Â Â Â -0.11222347Â Â Â 0.07828182 -1.43358Â Â 0.173646Â Â Â
DBoBC3Â Â Â Â Â Â Â Â 0.04564621Â Â Â 0.03825169Â 1.19331Â Â 0.252578Â Â Â
CredL1Â Â Â Â Â Â Â Â 0.18499925Â Â Â 0.06565456Â 2.81777Â Â 0.013693 *Â
BoBCL1Â Â Â Â Â Â Â -0.07682710Â Â Â 0.03406916 -2.25503Â Â 0.040666 *Â
---
Signif. codes:Â 0 â***â 0.001 â**â 0.01 â*â 0.05 â.â 0.1 â â 1
Residual standard error: 54.44479 on 14 degrees of freedom
Multiple R-squared: 0.9324472,   Adjusted R-squared: 0.903496
F-statistic: 32.20757 on 6 and 14 DF, p-value: 2.046024e-
#Same result
Not sure what you meant by ("This is good, but couldn't I code the process for my 15 variable model?")
A.K.
________________________________
Cc: R help < [hidden email]>
Sent: Monday, July 2, 2012 5:13 AM
Subject: Re: [R]Â Â Adjusting length of series
Hi David and AK,
I have been trying to implement your suggestions since yesterday, but I encountered some challenges.
As for David's suggestions, I could only implement it after some modifications. Using an abridged version of my data, I dpud my dataset and then show my steps below.
> dput(ydata)
structure(c(68.1000000000004, -34.8000000000002, 90.3999999999996,
54.6000000000004, -172.3, 51.8000000000002, 175, 79.8000000000002,
-35.7000000000007, 130.5, 116.8, -67.5, 164.5, 514.8, -326.1,
98.4000000000005, 160.2, 53.1999999999998, 283.6, -111.6, 127.8,
-17.3000000000002, 286.3, NA, NA, -102.900000000001, 125.2, -35.7999999999993,
-226.900000000001, 224.1, 123.2,
-95.1999999999998, -115.500000000001,
166.200000000001, -13.6999999999998, -184.3, 232, 350.3, -840.900000000001,
424.500000000001, 61.7999999999993, -107, 230.400000000001, -395.200000000001,
239.400000000001, -145.1, 303.6, NA, NA, NA, 228.1, -160.999999999999,
-191.100000000001, 451.000000000001, -100.900000000001, -218.4,
-20.3000000000011, 281.700000000002, -179.900000000001, -170.6,
416.3, 118.3, -1191.2, 1265.4, -362.700000000002, -168.799999999999,
337.400000000001, -625.600000000001, 634.600000000001, -384.500000000001,
448.700000000001, NA, NA, -164.457840999999, 17.0793539999995,
95.9767880000009, 680.238166999999, -491.348690999999, -274.694009,
-256.332907, 469.62296, -146.431891, -41.0772019999995, -106.970104,
757.688263999999, -1689.214533, 2320.098952, -1446.97942, 516.384521,
-375.277650999999, 293.867029999999, 417.845195, 278.198807,
-968.592033999999, -314.195986, NA, NA, NA,
181.537194999999,
78.8974340000013, 584.261378999998, -1171.586858, 216.654681999999,
18.3611019999998, 725.955867, -616.054851, 105.354689000001,
-65.8929020000005, 864.658367999999, -2446.902797, 4009.313485,
-3767.078372, 1963.363941, -891.662171999999, 669.144680999999,
123.978165, -139.646388, -1246.790841, 654.396048, NA, 4937,
5005.1, 4970.3, 5060.7, 5115.3, 4943, 4994.8, 5169.8, 5249.6,
5213.9, 5344.4, 5461.2, 5393.7, 5558.2, 6073, 5746.9, 5845.3,
6005.5, 6058.7, 6342.3, 6230.7, 6358.5, 6341.2, 6627.5, 4187.5,
4296.004835, 4240.051829, 4201.178177, 4258.281313, 4995.622616,
5241.615228, 5212.913831, 4927.879527, 5112.468183, 5150.624948,
5147.704511, 5037.81397, 5685.611693, 4644.194883, 5922.877025,
5754.579747, 6102.66699, 6075.476582, 6342.153204, 7026.675021,
7989.395645, 7983.524235, 7663.456839), .Dim = c(24L, 7L), .Dimnames = list(
   NULL, c("DCred1", "DCred2",
"DCred3", "DBoBC2", "DBoBC3",
   "CredL1", "BoBCL1")), .Tsp = c(2001.08333333333, 2003, 12
), class = c("mts", "ts"))
NB: the NAs in the dataset emanated from lagging or differencing the series
David's suggestion
 df<-data.frame(DCred1,DCred2,DCred3,DBoBC2,DBoBC3,CredL1,BoBCL1)
Error in data.frame(DCred1, DCred2, DCred3, DBoBC2, DBoBC3, CredL1, BoBCL1) :
 arguments imply differing number of rows: 23, 22, 21, 24
So I modified as follows:
length(DCred3)Â # finding the minimum length of various series
[1] 21
# Then dataframe construction
dframe<- data.frame(Dcre1=DCred1[1:21],Dcre2=DCred2[1:21],Dcre3=DCred3[1:21],
+ Dbobc2=DBoBC2[1:21],Dbobc3=DBoBC3[1:21],CredL=CredL1[1:21],BoBCL=BoBCL1[1:21])
# Then estimated regression
> regCred<- lm(Dcre1~Dcre2+Dcre3+Dbobc2+Dbobc3+CredL+BoBCL, data=dframe)
> summary(regCred)
# Worked well as shown by results below
Call:
lm(formula = Dcre1 ~ Dcre2 + Dcre3 + Dbobc2 + Dbobc3 + CredL +
   BoBCL, data = dframe)
Residuals:
   Min     1Q Median     3Q    Max
-69.516 -27.695Â -8.085Â 13.851 107.276
Coefficients:
            Estimate Std. Error t value Pr(>|t|)  Â
(Intercept) 159.32304Â 157.15209Â Â 1.014 0.327873Â Â Â
Dcre2Â Â Â Â Â Â Â -0.75527Â Â Â 0.17262Â -4.375 0.000634 ***
Dcre3Â Â Â Â Â Â Â -0.21006Â Â Â 0.08656Â -2.427 0.029329 *Â
Dbobc2Â Â Â Â Â Â Â 0.05111Â Â Â 0.06565Â Â 0.779 0.449197Â Â Â
Dbobc3Â Â Â Â Â Â Â 0.03106Â Â Â 0.03510Â Â 0.885 0.391108Â Â Â
CredLÂ Â Â Â Â Â Â -0.10967Â Â Â 0.04933Â -2.223 0.043177 *Â
BoBCLÂ Â Â Â Â Â Â Â 0.09756Â Â Â 0.03097Â Â 3.150 0.007087 **
---
Signif. codes:Â 0 â***â 0.001 â**â 0.01 â*â 0.05 â.â 0.1 â â 1
Residual standard error: 52.3 on 14 degrees of freedom
Multiple R-squared: 0.9331,    Adjusted R-squared: 0.9044
F-statistic: 32.55 on 6 and 14 DF, p-value: 1.911e-07
This is good, but couldn't I code the process for my 15 variable model?
Perhaps that is where the use of
Dcr<- lapply(..., function(x) ...)
comes in?
AK, if you spare some minutes, please use my dput data to illustrate the suggestion you made, I searched the lapply function (using ??lapply) but could not get a handle of how to use it in my case. My dput data is as shown below.
        DCred1 DCred2 DCred3     DBoBC2     DBoBC3 CredL1  BoBCL1
Feb 2001Â Â 68.1Â Â Â Â NAÂ Â Â Â Â NAÂ Â Â Â Â Â Â Â Â NAÂ Â Â Â Â Â Â Â Â NA 4937.0 4187.500
Mar 2001Â -34.8 -102.9Â Â Â Â Â NAÂ -164.45784Â Â Â Â Â Â Â Â Â NA 5005.1 4296.005
Apr 2001Â Â 90.4Â 125.2Â Â
228.1Â Â Â 17.07935Â Â 181.53719 4970.3 4240.052
May 2001Â Â 54.6Â -35.8Â -161.0Â Â Â 95.97679Â Â Â 78.89743 5060.7 4201.178
Jun 2001 -172.3 -226.9Â -191.1Â Â 680.23817Â Â 584.26138 5115.3 4258.281
Jul 2001Â Â 51.8Â 224.1Â Â 451.0Â -491.34869 -1171.58686 4943.0 4995.623
Aug 2001Â 175.0Â 123.2Â -100.9Â -274.69401Â Â 216.65468 4994.8 5241.615
Sep 2001Â Â 79.8Â -95.2Â -218.4Â -256.33291Â Â Â 18.36110 5169.8 5212.914
Oct 2001Â -35.7 -115.5Â Â -20.3Â Â 469.62296Â Â 725.95587 5249.6 4927.880
Nov 2001Â 130.5Â 166.2Â Â 281.7Â -146.43189Â -616.05485 5213.9 5112.468
Dec 2001Â 116.8Â -13.7Â -179.9Â Â -41.07720Â Â 105.35469 5344.4 5150.625
Jan 2002Â -67.5
-184.3Â -170.6Â -106.97010Â Â -65.89290 5461.2 5147.705
Feb 2002Â 164.5Â 232.0Â Â 416.3Â Â 757.68826Â Â 864.65837 5393.7 5037.814
Mar 2002Â 514.8Â 350.3Â Â 118.3 -1689.21453 -2446.90280 5558.2 5685.612
Apr 2002 -326.1 -840.9 -1191.2Â 2320.09895Â 4009.31348 6073.0 4644.195
May 2002Â Â 98.4Â 424.5Â 1265.4 -1446.97942 -3767.07837 5746.9 5922.877
Jun 2002Â 160.2Â Â 61.8Â -362.7Â Â 516.38452Â 1963.36394 5845.3 5754.580
Jul 2002Â Â 53.2 -107.0Â -168.8Â -375.27765Â -891.66217 6005.5 6102.667
Aug 2002Â 283.6Â 230.4Â Â 337.4Â Â 293.86703Â Â 669.14468 6058.7 6075.477
Sep 2002 -111.6 -395.2Â -625.6Â Â 417.84519Â Â 123.97817 6342.3 6342.153
Oct 2002Â 127.8Â 239.4Â Â 634.6Â Â 278.19881Â
-139.64639 6230.7 7026.675
Nov 2002Â -17.3 -145.1Â -384.5Â -968.59203 -1246.79084 6358.5 7989.396
Dec 2002Â 286.3Â 303.6Â Â 448.7Â -314.19599Â Â 654.39605 6341.2 7983.524
Jan 2003Â Â Â Â NAÂ Â Â Â NAÂ Â Â Â Â NAÂ Â Â Â Â Â Â Â Â NAÂ Â Â Â Â Â Â Â Â NA 6627.5 7663.457
Thanks kindly. Lexi      Â
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https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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Hi,
One more thing,
ydat1: original dataset
ydat2<-data.frame(ydat1)
#Not sure ,how you did this step on original data because::
dframe<- data.frame(Dcre1=DCred1[1:21],Dcre2=DCred2[1:21],Dcre3=DCred3[1:21],
Dbobc2=DBoBC2[1:21],Dbobc3=DBoBC3[1:21],CredL=CredL1[1:21],BoBCL=BoBCL1[1:21])
I am getting errors for that step, when I used ydat1.
head(ydat1)
[1] 68.1 -34.8 90.4 54.6 -172.3 51.8
head(ydat2)
DCred1 DCred2 DCred3 DBoBC2 DBoBC3 CredL1 BoBCL1
1 68.1 NA NA NA NA 4937.0 4187.500
2 -34.8 -102.9 NA -164.45784 NA 5005.1 4296.005
3 90.4 125.2 228.1 17.07935 181.53719 4970.3 4240.052
4 54.6 -35.8 -161.0 95.97679 78.89743 5060.7 4201.178
5 -172.3 -226.9 -191.1 680.23817 584.26138 5115.3 4258.281
6 51.8 224.1 451.0 -491.34869 -1171.58686 4943.0 4995.623
#I analyzed [1:21] again in ydat2.
dframe<-data.frame(Dcre1=ydat2$DCred1[1:21],Dcre2=ydat2$DCred2[1:21],Dcre3=ydat2$DCred3[1:21],Dbobc2=ydat2$DBoBC2[1:21],Dbobc3=ydat2$DBoBC3[1:21],CredL=ydat2$CredL1[1:21],BoBCL=ydat2$BoBCL1[1:21])
But, the results are bit different than in my earlier post, because, here the NAs are still present in different rows. So, the observations in those rows will be deleted while it is analyzed.
regCred<- lm(Dcre1~Dcre2+Dcre3+Dbobc2+Dbobc3+CredL+BoBCL, data=dframe)
> summary(regCred)
Call:
lm(formula = Dcre1 ~ Dcre2 + Dcre3 + Dbobc2 + Dbobc3 + CredL +
BoBCL, data = dframe)
Residuals:
Min 1Q Median 3Q Max
-118.687 -25.568 -5.334 35.035 69.992
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -485.42427 209.47952 -2.317 0.038958 *
Dcre2 0.95097 0.18156 5.238 0.000209 ***
Dcre3 -0.28676 0.10787 -2.658 0.020852 *
Dbobc2 -0.09512 0.09334 -1.019 0.328278
Dbobc3 0.03199 0.04933 0.648 0.528936
CredL 0.14825 0.07193 2.061 0.061645 .
BoBCL -0.04844 0.04333 -1.118 0.285540
---
A.K.
________________________________
From: "Lekgatlhamang, lexi Setlhare" < [hidden email]>
To: arun < [hidden email]>
Cc: R help < [hidden email]>
Sent: Monday, July 2, 2012 11:43 AM
Subject: Re: [R] Adjusting length of series
Thanks very much A.K. I have to admit that my problem was not clearly stated, with the structure of my data provided. Now all is well.
Cheers
Lexi
From: arun < [hidden email]>
To: "Lekgatlhamang, lexi Setlhare" < [hidden email]>
Cc: R help < [hidden email]>
Sent: Monday, July 2, 2012 4:40 PM
Subject: Re: [R] Adjusting length of series
Hello,
The class of your data is not dataframe.
Suppose I call your data as ydat1
str(ydat1)
mts [1:24, 1:7] 68.1 -34.8 90.4 54.6 -172.3 ...
- attr(*, "dimnames")=List of 2
..$ : NULL
..$ : chr [1:7] "DCred1" "DCred2" "DCred3" "DBoBC2" ...
- attr(*, "tsp")= num [1:3] 2001 2003 12
- attr(*, "class")= chr [1:2] "mts" "ts"
ydat2<-data.frame(ydat1)
str(ydat2)
'data.frame': 24 obs. of 7 variables:
$ DCred1: num 68.1 -34.8 90.4 54.6 -172.3 ...
$ DCred2: num NA -102.9 125.2 -35.8 -226.9 ...
$ DCred3: num NA NA 228 -161 -191 ...
$ DBoBC2: num NA -164.5 17.1 96 680.2 ...
$ DBoBC3: num NA NA 181.5 78.9 584.3 ...
$ CredL1: num 4937 5005 4970 5061 5115 ...
$ BoBCL1: num 4188 4296 4240 4201 4258 ...
#Since you wanted only to do lm
for these columns, I guess it doesn't really matter whether you have month and year in the dataset.
#With NAs
regCred<-lm(DCred1~DCred2+DCred3+DBoBC2+DBoBC3+CredL1+BoBCL1,data=ydat2)
> summary(regCred)
Call:
lm(formula = DCred1 ~ DCred2 + DCred3 + DBoBC2 + DBoBC3 + CredL1 +
BoBCL1, data = ydat2)
Residuals:
Min 1Q Median 3Q Max
-124.988463 -33.133975 7.971083 23.607953 76.813601
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
-538.61375718 205.91179535 -2.61575 0.020344 *
DCred2 0.96401908 0.15623660 6.17025 2.4337e-05 ***
DCred3 -0.25720355 0.08983607 -2.86303 0.012524 *
DBoBC2 -0.11222347 0.07828182 -1.43358 0.173646
DBoBC3 0.04564621 0.03825169 1.19331 0.252578
CredL1 0.18499925 0.06565456 2.81777 0.013693 *
BoBCL1 -0.07682710 0.03406916 -2.25503 0.040666 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01
‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 54.44479 on 14 degrees of freedom
(3 observations deleted due to missingness)
Multiple R-squared: 0.9324472, Adjusted R-squared: 0.903496
F-statistic: 32.20757 on 6 and 14 DF, p-value: 2.046024e-07
Without NAs
> ydat3<-na.omit(ydat2)
> regCred<-lm(DCred1~DCred2+DCred3+DBoBC2+DBoBC3+CredL1+BoBCL1,data=ydat3)
> summary(regCred)
Call:
lm(formula = DCred1 ~ DCred2 + DCred3 + DBoBC2 + DBoBC3 + CredL1 +
BoBCL1, data = ydat3)
Residuals:
Min 1Q Median 3Q Max
-124.988463 -33.133975 7.971083 23.607953 76.813601
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -538.61375718 205.91179535 -2.61575 0.020344 *
DCred2 0.96401908 0.15623660 6.17025 2.4337e-05 ***
DCred3 -0.25720355 0.08983607 -2.86303 0.012524 *
DBoBC2 -0.11222347 0.07828182 -1.43358 0.173646
DBoBC3 0.04564621 0.03825169 1.19331 0.252578
CredL1 0.18499925 0.06565456
2.81777 0.013693 *
BoBCL1 -0.07682710 0.03406916 -2.25503 0.040666 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 54.44479 on 14 degrees of freedom
Multiple R-squared: 0.9324472, Adjusted R-squared: 0.903496
F-statistic: 32.20757 on 6 and 14 DF, p-value: 2.046024e-
#Same result
Not sure what you meant by ("This is good, but couldn't I code the process for my 15 variable model?")
A.K.
________________________________
From: "Lekgatlhamang, lexi Setlhare" < [hidden email]>
To: arun < [hidden email]>
Cc: R help
< [hidden email]>
Sent: Monday, July 2, 2012 5:13 AM
Subject: Re: [R] Adjusting length of series
Hi David and AK,
I have been trying to implement your suggestions since yesterday, but I encountered some challenges.
As for David's suggestions, I could only implement it after some modifications. Using an abridged version of my data, I dpud my dataset and then show my steps below.
> dput(ydata)
structure(c(68.1000000000004, -34.8000000000002, 90.3999999999996,
54.6000000000004, -172.3, 51.8000000000002, 175, 79.8000000000002,
-35.7000000000007, 130.5, 116.8, -67.5, 164.5, 514.8, -326.1,
98.4000000000005, 160.2, 53.1999999999998, 283.6, -111.6, 127.8,
-17.3000000000002, 286.3, NA, NA, -102.900000000001, 125.2, -35.7999999999993,
-226.900000000001, 224.1, 123.2,
-95.1999999999998,
-115.500000000001,
166.200000000001, -13.6999999999998, -184.3, 232, 350.3, -840.900000000001,
424.500000000001, 61.7999999999993, -107, 230.400000000001, -395.200000000001,
239.400000000001, -145.1, 303.6, NA, NA, NA, 228.1, -160.999999999999,
-191.100000000001, 451.000000000001, -100.900000000001, -218.4,
-20.3000000000011, 281.700000000002, -179.900000000001, -170.6,
416.3, 118.3, -1191.2, 1265.4, -362.700000000002, -168.799999999999,
337.400000000001, -625.600000000001, 634.600000000001, -384.500000000001,
448.700000000001, NA, NA, -164.457840999999, 17.0793539999995,
95.9767880000009, 680.238166999999, -491.348690999999, -274.694009,
-256.332907, 469.62296, -146.431891, -41.0772019999995, -106.970104,
757.688263999999, -1689.214533, 2320.098952, -1446.97942, 516.384521,
-375.277650999999, 293.867029999999, 417.845195, 278.198807,
-968.592033999999, -314.195986, NA, NA, NA,
181.537194999999,
78.8974340000013, 584.261378999998, -1171.586858, 216.654681999999,
18.3611019999998, 725.955867, -616.054851, 105.354689000001,
-65.8929020000005, 864.658367999999, -2446.902797, 4009.313485,
-3767.078372, 1963.363941, -891.662171999999, 669.144680999999,
123.978165, -139.646388, -1246.790841, 654.396048, NA, 4937,
5005.1, 4970.3, 5060.7, 5115.3, 4943, 4994.8, 5169.8, 5249.6,
5213.9, 5344.4, 5461.2, 5393.7, 5558.2, 6073, 5746.9, 5845.3,
6005.5, 6058.7, 6342.3, 6230.7, 6358.5, 6341.2, 6627.5, 4187.5,
4296.004835, 4240.051829, 4201.178177, 4258.281313, 4995.622616,
5241.615228, 5212.913831, 4927.879527, 5112.468183, 5150.624948,
5147.704511, 5037.81397, 5685.611693, 4644.194883, 5922.877025,
5754.579747, 6102.66699, 6075.476582, 6342.153204, 7026.675021,
7989.395645, 7983.524235, 7663.456839), .Dim = c(24L, 7L), .Dimnames = list(
NULL, c("DCred1", "DCred2",
"DCred3", "DBoBC2",
"DBoBC3",
"CredL1", "BoBCL1")), .Tsp = c(2001.08333333333, 2003, 12
), class = c("mts", "ts"))
NB: the NAs in the dataset emanated from lagging or differencing the series
David's suggestion
df<-data.frame(DCred1,DCred2,DCred3,DBoBC2,DBoBC3,CredL1,BoBCL1)
Error in data.frame(DCred1, DCred2, DCred3, DBoBC2, DBoBC3, CredL1, BoBCL1) :
arguments imply differing number of rows: 23, 22, 21, 24
So I modified as follows:
length(DCred3) # finding the minimum length of various series
[1] 21
# Then dataframe construction
dframe<- data.frame(Dcre1=DCred1[1:21],Dcre2=DCred2[1:21],Dcre3=DCred3[1:21],
+ Dbobc2=DBoBC2[1:21],Dbobc3=DBoBC3[1:21],CredL=CredL1[1:21],BoBCL=BoBCL1[1:21])
# Then estimated regression
> regCred<- lm(Dcre1~Dcre2+Dcre3+Dbobc2+Dbobc3+CredL+BoBCL, data=dframe)
> summary(regCred)
# Worked well as shown by results
below
Call:
lm(formula = Dcre1 ~ Dcre2 + Dcre3 + Dbobc2 + Dbobc3 + CredL +
BoBCL, data = dframe)
Residuals:
Min 1Q Median 3Q Max
-69.516 -27.695 -8.085 13.851 107.276
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 159.32304 157.15209 1.014 0.327873
Dcre2 -0.75527 0.17262 -4.375 0.000634 ***
Dcre3 -0.21006 0.08656 -2.427 0.029329 *
Dbobc2 0.05111 0.06565 0.779 0.449197
Dbobc3 0.03106 0.03510 0.885 0.391108
CredL -0.10967 0.04933 -2.223 0.043177 *
BoBCL 0.09756 0.03097 3.150 0.007087 **
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 52.3 on 14 degrees of freedom
Multiple R-squared: 0.9331, Adjusted R-squared: 0.9044
F-statistic: 32.55 on 6 and 14 DF, p-value: 1.911e-07
This is good, but couldn't I code the process for my 15 variable model?
Perhaps that is where the use of
Dcr<- lapply(..., function(x) ...)
comes in?
AK, if you spare some minutes, please use my dput data to illustrate the suggestion you made, I searched the
lapply function (using ??lapply) but could not get a handle of how to use it in my case. My dput data is as shown below.
DCred1 DCred2 DCred3 DBoBC2 DBoBC3 CredL1 BoBCL1
Feb 2001 68.1 NA NA NA NA 4937.0 4187.500
Mar 2001 -34.8 -102.9 NA -164.45784 NA 5005.1 4296.005
Apr 2001 90.4 125.2
228.1 17.07935 181.53719 4970.3 4240.052
May 2001 54.6 -35.8 -161.0 95.97679 78.89743 5060.7 4201.178
Jun 2001 -172.3 -226.9
-191.1 680.23817 584.26138 5115.3 4258.281
Jul 2001 51.8 224.1 451.0 -491.34869 -1171.58686 4943.0 4995.623
Aug 2001 175.0 123.2 -100.9 -274.69401 216.65468 4994.8 5241.615
Sep 2001 79.8 -95.2 -218.4 -256.33291 18.36110 5169.8 5212.914
Oct 2001 -35.7 -115.5 -20.3 469.62296 725.95587 5249.6 4927.880
Nov 2001 130.5 166.2 281.7 -146.43189 -616.05485 5213.9 5112.468
Dec 2001 116.8 -13.7 -179.9 -41.07720 105.35469 5344.4 5150.625
Jan 2002 -67.5
-184.3 -170.6 -106.97010 -65.89290 5461.2 5147.705
Feb 2002 164.5 232.0 416.3 757.68826 864.65837 5393.7 5037.814
Mar 2002 514.8
350.3 118.3 -1689.21453 -2446.90280 5558.2 5685.612
Apr 2002 -326.1 -840.9 -1191.2 2320.09895 4009.31348 6073.0 4644.195
May 2002 98.4 424.5 1265.4 -1446.97942 -3767.07837 5746.9 5922.877
Jun 2002 160.2 61.8 -362.7 516.38452 1963.36394 5845.3 5754.580
Jul 2002 53.2 -107.0 -168.8 -375.27765 -891.66217 6005.5 6102.667
Aug 2002 283.6 230.4 337.4 293.86703 669.14468 6058.7 6075.477
Sep 2002 -111.6 -395.2 -625.6 417.84519 123.97817 6342.3 6342.153
Oct 2002 127.8 239.4 634.6 278.19881
-139.64639 6230.7 7026.675
Nov 2002 -17.3 -145.1 -384.5 -968.59203 -1246.79084 6358.5 7989.396
Dec 2002 286.3 303.6 448.7 -314.19599 654.39605 6341.2
7983.524
Jan 2003 NA NA NA NA NA 6627.5 7663.457
Thanks kindly. Lexi
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