Why replacement has length zero? And How can I fix it?

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Why replacement has length zero? And How can I fix it?

Jiaqi.Zhang
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Hi, all,

I am working on a project to run a simulation. I am now working on the main body of the simulation body, I generate the data and all work well before I identify a variable "IAP", there is always an error message as "replacement has length zero", and I do not know how to fix it. Can anyone help?

Here is the code:

library (mvtnorm)

options(digits=2)

# Set variable name #
namelist1 <- c("Age", "Gender", "SES", "ISS", "IQTe", "PreTe")

# Generate Data #
Sample.Size <- 10 # Variable Sample Size #
CorMat <- matrix(c(1.0 ,0.0 ,0.0 , 0.5, 0.2, 0.5,
                                        0.0, 1.0 ,0.0, 0.0, 0.0, 0.0,
                                        0.0 ,0.0 ,1.0, 0.5, 0.4, 0.5,
                                        0.5, 0.0, 0.5, 1.0, 0.4, 0.7,
                                        0.2, 0.0, 0.4, 0.4, 1.0, 0.9,
                                        0.5, 0.0, 0.5, 0.7, 0.9, 1.0),ncol=6) # Correlation Matrix #

DataSet <- rmvnorm(Sample.Size, mean=c(0,0,0,0,0,0), sigma=CorMat) # Draw Sample.Size covariate set all mean 0 #
DataSet <- as.data.frame(DataSet)
names(DataSet) <- namelist1
DataSet

# For Treatment IAP #
IAP <- matrix(0, nrow=Sample.Size)
for (i in 1:Sample.Size){
        IAP[i] <- if (DataSet$SES>0)
        {
                ifelse(runif[1]>0.75, 1, 0) # High SES, higher chance to be in Treatment #
        }
        IAP[i] <- if (DataSet$SES<=0)
        {
                ifelse(runif[1]<=0.25, 1, 0) # Low SES, lower chance to be in Treatment #
        }
} # End loop #
IAP

Error in IAP[i] <- if (DataSet$SES > 0) { : replacement has length zero
In addition: Warning message:
In if (DataSet$SES > 0) { :
  the condition has length > 1 and only the first element will be used

Thanks!
Sincerely yours,

Jiaqi Zhang
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Re: Why replacement has length zero? And How can I fix it?

soon yi
Hi

for the loop section runif needs curved brackets

Try

IAP <-NA
for (i in 1:Sample.Size){
        if (DataSet$SES[i]>0)   {
                IAP[i] <-  ifelse(runif(1)>0.75, 1, 0) # High SES, higher chance to be in Treatment #
  }
        else  {
  IAP[i] <- ifelse(runif(1)<=0.25, 1, 0) # Low SES, lower chance to be in Treatment #
                 }
        } # End loop #
IAP




IAP
zjiaqi19880219 wrote
Hi, all,

I am working on a project to run a simulation. I am now working on the main body of the simulation body, I generate the data and all work well before I identify a variable "IAP", there is always an error message as "replacement has length zero", and I do not know how to fix it. Can anyone help?

Here is the code:

library (mvtnorm)

options(digits=2)

# Set variable name #
namelist1 <- c("Age", "Gender", "SES", "ISS", "IQTe", "PreTe")

# Generate Data #
Sample.Size <- 10 # Variable Sample Size #
CorMat <- matrix(c(1.0 ,0.0 ,0.0 , 0.5, 0.2, 0.5,
                                        0.0, 1.0 ,0.0, 0.0, 0.0, 0.0,
                                        0.0 ,0.0 ,1.0, 0.5, 0.4, 0.5,
                                        0.5, 0.0, 0.5, 1.0, 0.4, 0.7,
                                        0.2, 0.0, 0.4, 0.4, 1.0, 0.9,
                                        0.5, 0.0, 0.5, 0.7, 0.9, 1.0),ncol=6) # Correlation Matrix #

DataSet <- rmvnorm(Sample.Size, mean=c(0,0,0,0,0,0), sigma=CorMat) # Draw Sample.Size covariate set all mean 0 #
DataSet <- as.data.frame(DataSet)
names(DataSet) <- namelist1
DataSet

# For Treatment IAP #
IAP <- matrix(0, nrow=Sample.Size)
for (i in 1:Sample.Size){
        IAP[i] <- if (DataSet$SES>0)
        {
                ifelse(runif[1]>0.75, 1, 0) # High SES, higher chance to be in Treatment #
        }
        IAP[i] <- if (DataSet$SES<=0)
        {
                ifelse(runif[1]<=0.25, 1, 0) # Low SES, lower chance to be in Treatment #
        }
} # End loop #
IAP

Error in IAP[i] <- if (DataSet$SES > 0) { : replacement has length zero
In addition: Warning message:
In if (DataSet$SES > 0) { :
  the condition has length > 1 and only the first element will be used

Thanks!
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Re: Why replacement has length zero? And How can I fix it?

Jiaqi.Zhang
This post has NOT been accepted by the mailing list yet.
Thanks a lot, it works when I follow your suggestion. But if I change [] to () in my orignal code, it appears the same error message without using a it...else... code, I actually do not know why.
Sincerely yours,

Jiaqi Zhang
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Re: Why replacement has length zero? And How can I fix it?

djmuseR
In reply to this post by soon yi
Hi:

No offense, but this is code is inefficient on several levels.
Firstly, a loop is unnecessary (see below); secondly,  ifelse() is a
vectorized function and you're attempting to apply it to each row of
DataSet. Try this instead, given

 dput(DataSet)
structure(list(Age = c(-0.552450789929175, -2.06988485889524,
-1.1290497440272, -0.268777167923272, -0.012748181465016, 0.554761979685086,
0.179401201802416, 0.0973216900180107, -0.914782595075178, -1.53708818506571
), Gender = c(-0.0705367578252014, 0.969426004829301, 0.0556317051882303,
1.33716753655214, -0.638382962845772, 0.480939398350137, -0.755053441639217,
1.25393329162734, 1.45933385954704, 0.672871276697159), SES =
c(-1.88102131807556,
0.529181612783923, 0.271835979074789, -0.656415898217512, -0.517622943602359,
0.639094817790355, -1.08205962796926, -0.193116856300625, 0.707374091321319,
1.98180909069287), ISS = c(-1.74208565286036, -1.235721877648,
0.459828219284939, -0.87781916182417, -0.429262433047679, 0.379335629940134,
0.949088546564315, 0.689758637278236, -1.14832161185134, 1.40363651773545
), IQTe = c(-0.266523341317252, -0.32477349599139, -0.75723700635683,
0.74556355929622, -0.238794158234391, 0.0906173696194765, 0.331200954821995,
-0.373009866084321, 0.417237160703897, 0.755603172535694), PreTe =
c(-0.743400287678305,
-1.1501727014007, -0.668278397501659, -0.0766957410591086, -0.232360222985187,
0.320050787942651, 0.474114337341715, -0.0181639478062117, -0.0905594374421625,
0.517614989060291)), .Names = c("Age", "Gender", "SES", "ISS",
"IQTe", "PreTe"), row.names = c(NA, -10L), class = "data.frame")

u <- runif(10)
> u
 [1] 0.72 0.57 0.71 0.64 0.38 0.53 0.14 0.14 0.82 0.12

# Remember two things:
#  (i) ifelse() is a vectorized function;
#  (ii) logical statements return either TRUE (1) or FALSE (0).
#  As a result,

# less efficient:
IAP0 <- ifelse(DataSet$SES > 0, ifelse(u > 0.75, 1, 0),
                                 ifelse(u <= 0.25, 1, 0))
# more efficient:
IAP1 <- ifelse(DataSet$SES > 0, u > 0.75, u <= 0.25) * 1
IAP1
 [1] 0 0 0 0 0 0 1 1 1 0
> identical(IAP0, IAP1)
[1] TRUE

# Verification:
cbind(DataSet$SES, u, IAP1)

This result conforms to what one would expect with u as the realized
uniform random vector. Moreover, the code is transparent: if SES > 0,
then apply the test u > 0.75, else apply the test u <= 0.25). The
post-multiplication by 1 ensures that the result is numeric rather
than logical.

Dennis


On Sat, Feb 2, 2013 at 12:57 PM, soon yi <[hidden email]> wrote:

> Hi
>
> for the loop section runif needs curved brackets
>
> Try
>
> IAP <-NA
> for (i in 1:Sample.Size){
>         if (DataSet$SES[i]>0)   {
>                 IAP[i] <-  ifelse(runif(1)>0.75, 1, 0) # High SES, higher chance to be in
> Treatment #
>                 }
>         else  {
>                 IAP[i] <- ifelse(runif(1)<=0.25, 1, 0) # Low SES, lower chance to be in
> Treatment #
>                  }
>         } # End loop #
> IAP
>
>
>
>
> IAP
> zjiaqi19880219 wrote
>> Hi, all,
>>
>> I am working on a project to run a simulation. I am now working on the
>> main body of the simulation body, I generate the data and all work well
>> before I identify a variable "IAP", there is always an error message as
>> "replacement has length zero", and I do not know how to fix it. Can anyone
>> help?
>>
>> Here is the code:
>>
>> library (mvtnorm)
>>
>> options(digits=2)
>>
>> # Set variable name #
>> namelist1 <- c("Age", "Gender", "SES", "ISS", "IQTe", "PreTe")
>>
>> # Generate Data #
>> Sample.Size <- 10 # Variable Sample Size #
>> CorMat <- matrix(c(1.0 ,0.0 ,0.0 , 0.5, 0.2, 0.5,
>>                                       0.0, 1.0 ,0.0, 0.0, 0.0, 0.0,
>>                                       0.0 ,0.0 ,1.0, 0.5, 0.4, 0.5,
>>                                       0.5, 0.0, 0.5, 1.0, 0.4, 0.7,
>>                                       0.2, 0.0, 0.4, 0.4, 1.0, 0.9,
>>                                       0.5, 0.0, 0.5, 0.7, 0.9, 1.0),ncol=6) # Correlation Matrix #
>>
>> DataSet <- rmvnorm(Sample.Size, mean=c(0,0,0,0,0,0), sigma=CorMat) # Draw
>> Sample.Size covariate set all mean 0 #
>> DataSet <- as.data.frame(DataSet)
>> names(DataSet) <- namelist1
>> DataSet
>>
>> # For Treatment IAP #
>> IAP <- matrix(0, nrow=Sample.Size)
>> for (i in 1:Sample.Size){
>>       IAP[i] <- if (DataSet$SES>0)
>>       {
>>               ifelse(runif[1]>0.75, 1, 0) # High SES, higher chance to be in Treatment
>> #
>>       }
>>       IAP[i] <- if (DataSet$SES<=0)
>>       {
>>               ifelse(runif[1]<=0.25, 1, 0) # Low SES, lower chance to be in Treatment
>> #
>>       }
>> } # End loop #
>> IAP
>>
>> Error in IAP[i] <- if (DataSet$SES > 0) { : replacement has length zero
>> In addition: Warning message:
>> In if (DataSet$SES > 0) { :
>>   the condition has length > 1 and only the first element will be used
>>
>> Thanks!
>
>
>
>
>
> --
> View this message in context: http://r.789695.n4.nabble.com/Why-replacement-has-length-zero-And-How-can-I-fix-it-tp4657373p4657380.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> [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.

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