Hi, you could use sapply() on the list having each component=1 data.frame:
That is:
1) Let data be the list where>is.data.frame(data[[15]]) #say
[1] TRUE
and where the variables of interest are in the same position: 1,2 and 3 for
the response and the two explanatory variables.
2)Build some function to be applied to the list
fn.lm<-function(x,...){lm(x[,1]~x[,2]+x[,3], ...)}
3)Each component of out is a lm model.
out<-sapply(data, fn.lm)
4)Extract the p-values
fn.extract<-function(x,...){coef(summary(x))[,4]}
pvalue<-sapply(out, fn.extract)
#pvalue should be a vector of the p-values from every estimated model.
In my (small) experience using sapply() instead of for() is a bit more
efficient.
However for large list and/or model you could increase the memory limit:
memory.limit(500) #say
Also splitting the list "data" into several sub-lists should speed the
process.
Hope this help you,
best,
vito
----- Original Message -----
From: <yiping.fan at syngenta.com>
To: <r-help at stat.math.ethz.ch>
Sent: Tuesday, April 30, 2002 6:23 PM
Subject: [R] regression, for loop
> Hello, all
> Here are the questions I have:
> data samples:
> Data style x y
> data1 r 2 3
> data1 r 3 4
> data1 r 4 5
> data1 g 4 5
> data1 g 6 7
> data1 g 6 7
> data2 r 8 9
> data2 r 8 9
> data2 r 8 9
> data2 g 8 9
> data2 g 8 9
> data2 g 8 9
> ...
> datan
>
>
> we have n data (data1 - datan). For each data, I want to do a simple
> regression:
> y ~ x+style
> I am using a for loop now
> for(i in 1:n){
> data.name=Data[i];
> res<-lm(y~x+style, data, subset=(Data == data.name) )
> }
>
> I am wondering whether it is possible to do the same without using for
loop?> Also, after regression, I am only interested in getting the P value for
> style, how to extract it?
>
> Thank you very much!
>
> Best regards,
>
> Y. Fan
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