Apply the corr function to a matrix using levels of a factor?

Question

I'm trying using the corr() function to calculate weighted ponderations. The way it works is the first argument should be a matrix with two columns corresponding to the two variables whose correlation we wish to calculate and the second a vector of weights to be applied to each pair of observations.

Here is an example.

> head(d)
 Shade_tolerance htot
1            4.56 25.0
2            2.73 23.5
3            2.73 21.5
4            3.97 17.0
5            4.00 25.5
6            4.00 23.5

> head(poids)
[1] 5.200440e-07 5.200440e-07 1.445016e-06 1.445016e-06 1.445016e-06 1.445016e-06

> corr(d,poids)
[1] 0.1357279

So I got it and I'm able to use it on my matrix but I would like to compute different correlations according to the levels of a factor. Let's say as if I was using the tapply() function.

> head(d2)
  Shade_tolerance htot idp
1            4.56 25.0  19
2            2.73 23.5  19
3            2.73 21.5  19
4            3.97 17.0  18
5            4.00 25.5  18
6            4.00 23.5  18

So my dream would be to do something like this:

tapply(as.matrix(d2[,c(1,2)]), d2$idp, corr)

Except that as you know in tapply() the first element needs to be avector not a matrix.

Would someone have any solution for me?

Thanks a lot for your help.

EDIT: I just realized that I am missing the weights for the weighted correlation in the part of the data frame I showed you. So it would have some how to take both the matrix and the weights according to the levels of the factor.

> head(df)
  Shade_tolerance htot idp        poids
1            4.56 25.0  19 5.200440e-07
2            2.73 23.5  19 5.200440e-07
3            2.73 21.5  19 1.445016e-06
4            3.97 17.0  19 1.445016e-06
5            4.00 25.5  19 1.445016e-06
6            4.00 23.5  19 1.445016e-06

I hope it is clear.

Answer 1

If you've a "huge" data.frame, then using data.table might help:

require(data.table)
dt <- as.data.table(df)
setkey(dt, "idp")
dt[, list(corr = corr(cbind(Shade_tolerance, htot), poids)), by=idp]

#    idp      corr
# 1:  18 0.9743547
# 2:  19 0.8387363

Answer 2

Here is a solution using function ddply() from library plyr.

ddply(df,.(idp),
   summarise,kor=corr(cbind(Shade_tolerance, htot),poids))
  idp       kor
1  18 0.9743547
2  19 0.8387363

Answer 3

Using by and cbind,

 library(boot)
 by(dat,dat$idp,FUN=function(x)corr(cbind(x$Shade_tolerance,x$htot),x$poids))
dat$idp: 18
[1] 0.9743547
--------------------------------------------------------------------------------------- 
dat$idp: 19
[1] 0.7474093