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coefficients lm of data.frame

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Alfredo Alessandrini

coefficients lm of data.frame

Hi,

I've a data frame like this:

> as.data.frame(cbind(rnorm(1:12),rnorm(1:12)))
V1 V2
1 -1.30849402 -0.52094136
2 0.96157302 0.76217871
3 -0.44223351 -1.72630871
4 -0.10432438 -1.04732942
5 -1.38748914 0.95877311
6 -0.63965975 0.65494811
7 -0.24058318 0.19496830
8 -0.11172988 1.01680655
9 0.08065333 0.22168589
10 0.25196536 0.84619914
11 -0.59536986 -0.08243074
12 1.09115054 0.49822977

I need to add two columns as result of the fitting of linear model
based on a preset numbers of row.

For example if I need to compute a lm each 4 rows, I get the
data.frame below, where intercept1 and coeff1 is obtained from V1 and
V2 of first 4 rows lm(V2 ~ V1), and so on...

V1 V2 "intercept" "coeff"
1 0.6931694 0.05797771 intercept1 coeff1
2 -1.4069786 0.23983307 intercept1 coeff1
3 -1.4901708 0.45079601 intercept1 coeff1
4 0.2215696 1.87888983 intercept1 coeff1
5 -0.5828106 0.90376622 intercept2 coeff2
6 -0.7607985 0.71419938 intercept2 coeff2
7 0.1273495 0.06199312 intercept2 coeff2
8 -0.5612245 1.02223971 intercept2 coeff2
9 -0.1439178 0.92135354 intercept3 coeff3
10 -1.1011662 0.02894731 intercept3 coeff3
11 -0.4098710 -0.01231322 intercept3 coeff3
12 1.1511811 -0.63923140 intercept3 coeff3

Thanks in advance,

Alfredo

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PIKAL Petr

Odp: coefficients lm of data.frame

Hi

divide and conquer

Split your data frame to required portions

dat<-as.data.frame(cbind(rnorm(1:12),rnorm(1:12)))
dat.s<-split(dat, rep(1:4, each=3))

apply lm to each portion

sapply(dat.s, function(x) coef(lm(V2~V1, x)))
1 2 3 4
(Intercept) 1.21745328 -0.7783777 -0.1736094 0.1042114
V1 0.03639838 2.6203724 1.3167433 -0.2350888

Add those values to your data frame.

hint: use rep(..., each=4)

Regards
Petr

[hidden email] napsal dne 07.07.2011 16:09:51:

> Alfredo Alessandrini <[hidden email]>
> Odeslal: [hidden email]
>
> 07.07.2011 16:09
>
> Komu
>
> [hidden email]
>
> Kopie
>
> Předmět
>
> [R] coefficients lm of data.frame
>
> Hi,
>
> I've a data frame like this:
>
> > as.data.frame(cbind(rnorm(1:12),rnorm(1:12)))
> V1 V2
> 1 -1.30849402 -0.52094136
> 2 0.96157302 0.76217871
> 3 -0.44223351 -1.72630871
> 4 -0.10432438 -1.04732942
> 5 -1.38748914 0.95877311
> 6 -0.63965975 0.65494811
> 7 -0.24058318 0.19496830
> 8 -0.11172988 1.01680655
> 9 0.08065333 0.22168589
> 10 0.25196536 0.84619914
> 11 -0.59536986 -0.08243074
> 12 1.09115054 0.49822977
>
> I need to add two columns as result of the fitting of linear model
> based on a preset numbers of row.
>
> For example if I need to compute a lm each 4 rows, I get the
> data.frame below, where intercept1 and coeff1 is obtained from V1 and
> V2 of first 4 rows lm(V2 ~ V1), and so on...
>
>
> V1 V2 "intercept" "coeff"
> 1 0.6931694 0.05797771 intercept1 coeff1
> 2 -1.4069786 0.23983307 intercept1 coeff1
> 3 -1.4901708 0.45079601 intercept1 coeff1
> 4 0.2215696 1.87888983 intercept1 coeff1
> 5 -0.5828106 0.90376622 intercept2 coeff2
> 6 -0.7607985 0.71419938 intercept2 coeff2
> 7 0.1273495 0.06199312 intercept2 coeff2
> 8 -0.5612245 1.02223971 intercept2 coeff2
> 9 -0.1439178 0.92135354 intercept3 coeff3
> 10 -1.1011662 0.02894731 intercept3 coeff3
> 11 -0.4098710 -0.01231322 intercept3 coeff3
> 12 1.1511811 -0.63923140 intercept3 coeff3
>
>
> Thanks in advance,
>
> Alfredo
>
> ______________________________________________
> [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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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.

Filipe Leme Botelho

RES: coefficients lm of data.frame

In reply to this post by Alfredo Alessandrini

For sure some smart cookie has a much better solution but this can do
the trick

dt <- as.data.frame(cbind(rnorm(1:12),rnorm(1:12)))

regs <- sapply(seq(1,9,by=4), function(x) coef(lm(dt[(x:(x+3)),1] ~
dt[(x:(x+3)),2])))
tmp <- c()
for (i in 1:ncol(regs)) {
tmp <- cbind(tmp, sapply(1:4, function(y) regs[,i]))
}

cbind(dt, t(tmp))

Cheers,
Filipe

-----Mensagem original-----
De: [hidden email] [mailto:[hidden email]]
Em nome de Alfredo Alessandrini
Enviada em: quinta-feira, 7 de julho de 2011 11:10
Para: [hidden email]
Assunto: [R] coefficients lm of data.frame

Hi,

I've a data frame like this:

> as.data.frame(cbind(rnorm(1:12),rnorm(1:12)))
V1 V2
1 -1.30849402 -0.52094136
2 0.96157302 0.76217871
3 -0.44223351 -1.72630871
4 -0.10432438 -1.04732942
5 -1.38748914 0.95877311
6 -0.63965975 0.65494811
7 -0.24058318 0.19496830
8 -0.11172988 1.01680655
9 0.08065333 0.22168589
10 0.25196536 0.84619914
11 -0.59536986 -0.08243074
12 1.09115054 0.49822977

I need to add two columns as result of the fitting of linear model
based on a preset numbers of row.

For example if I need to compute a lm each 4 rows, I get the
data.frame below, where intercept1 and coeff1 is obtained from V1 and
V2 of first 4 rows lm(V2 ~ V1), and so on...

V1 V2 "intercept" "coeff"
1 0.6931694 0.05797771 intercept1 coeff1
2 -1.4069786 0.23983307 intercept1 coeff1
3 -1.4901708 0.45079601 intercept1 coeff1
4 0.2215696 1.87888983 intercept1 coeff1
5 -0.5828106 0.90376622 intercept2 coeff2
6 -0.7607985 0.71419938 intercept2 coeff2
7 0.1273495 0.06199312 intercept2 coeff2
8 -0.5612245 1.02223971 intercept2 coeff2
9 -0.1439178 0.92135354 intercept3 coeff3
10 -1.1011662 0.02894731 intercept3 coeff3
11 -0.4098710 -0.01231322 intercept3 coeff3
12 1.1511811 -0.63923140 intercept3 coeff3

Thanks in advance,

Alfredo

______________________________________________
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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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______________________________________________
[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.

djmuseR

Re: coefficients lm of data.frame

In reply to this post by Alfredo Alessandrini

Hi:

Here's another approach using the plyr package:

library(plyr)
df <- data.frame(gp = factor(rep(1:3, each = 4)), x = rnorm(12), y = rnorm(12))
mylst <- split(df, df$gp)
mycoefs <- ldply(mylst, function(d) coef(lm(y ~ x, data = d)))
names(mycoefs) <- c('gp', 'intercept', 'slope')
merge(df, mycoefs, by = 'gp', all.x = TRUE)

One could write this more compactly as

dfr <- merge(df, ldply(split(df, df$gp), function(d) coef(lm(y ~ x, data = d))),
by.x = 'gp', by.y = '.id', all.x = TRUE)
names(dfr) <- c('x', 'y', 'intercept', 'slope')
dfr

HTH,
Dennis

On Thu, Jul 7, 2011 at 7:09 AM, Alfredo Alessandrini
<[hidden email]> wrote:

> Hi,
>
> I've a data frame like this:
>
>> as.data.frame(cbind(rnorm(1:12),rnorm(1:12)))
> V1 V2
> 1 -1.30849402 -0.52094136
> 2 0.96157302 0.76217871
> 3 -0.44223351 -1.72630871
> 4 -0.10432438 -1.04732942
> 5 -1.38748914 0.95877311
> 6 -0.63965975 0.65494811
> 7 -0.24058318 0.19496830
> 8 -0.11172988 1.01680655
> 9 0.08065333 0.22168589
> 10 0.25196536 0.84619914
> 11 -0.59536986 -0.08243074
> 12 1.09115054 0.49822977
>
> I need to add two columns as result of the fitting of linear model
> based on a preset numbers of row.
>
> For example if I need to compute a lm each 4 rows, I get the
> data.frame below, where intercept1 and coeff1 is obtained from V1 and
> V2 of first 4 rows lm(V2 ~ V1), and so on...
>
>
> V1 V2 "intercept" "coeff"
> 1 0.6931694 0.05797771 intercept1 coeff1
> 2 -1.4069786 0.23983307 intercept1 coeff1
> 3 -1.4901708 0.45079601 intercept1 coeff1
> 4 0.2215696 1.87888983 intercept1 coeff1
> 5 -0.5828106 0.90376622 intercept2 coeff2
> 6 -0.7607985 0.71419938 intercept2 coeff2
> 7 0.1273495 0.06199312 intercept2 coeff2
> 8 -0.5612245 1.02223971 intercept2 coeff2
> 9 -0.1439178 0.92135354 intercept3 coeff3
> 10 -1.1011662 0.02894731 intercept3 coeff3
> 11 -0.4098710 -0.01231322 intercept3 coeff3
> 12 1.1511811 -0.63923140 intercept3 coeff3
>
>
> Thanks in advance,
>
> Alfredo
>
> ______________________________________________
> [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.
>

______________________________________________
[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.

Hadley Wickham-2

Re: coefficients lm of data.frame

On Thu, Jul 7, 2011 at 5:24 PM, Dennis Murphy <[hidden email]> wrote:

> Hi:
>
> Here's another approach using the plyr package:
>
> library(plyr)
> df <- data.frame(gp = factor(rep(1:3, each = 4)), x = rnorm(12), y = rnorm(12))
> mylst <- split(df, df$gp)
> mycoefs <- ldply(mylst, function(d) coef(lm(y ~ x, data = d)))
> names(mycoefs) <- c('gp', 'intercept', 'slope')
> merge(df, mycoefs, by = 'gp', all.x = TRUE)
>
> One could write this more compactly as
>
> dfr <- merge(df, ldply(split(df, df$gp), function(d) coef(lm(y ~ x, data = d))),
> by.x = 'gp', by.y = '.id', all.x = TRUE)

Or even more compactly as

dfr <- merge(df, ddply(df, "gp", function(d) coef(lm(y ~ x, data = d))),
by.x = 'gp', by.y = '.id', all.x = TRUE)

Hadley

--
Assistant Professor / Dobelman Family Junior Chair
Department of Statistics / Rice University
http://had.co.nz/

______________________________________________
[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.