# [R] correlation by factor

Bert Gunter bgunter@4567 @end|ng |rom gm@||@com
Wed Jul 28 04:30:12 CEST 2021

```Well, first of all, your example is messed up. You missed the "c" in front
of the ( in Freq <-; and all of the Freq entries need to be enclosed in
quotes for proper syntax. A simpler way to do it is just to use paste() and
rep():

Freq <- paste0("a", rep(1:5,3))
(If you are not familiar with such "utility" functions, you should consider
spending time with a basic R tutorial or two.)

Ordinarily, your individual vectors, R, Day and Freq, would be in a data
frame or similar (e.g. a tibble or data.table) structure and you would use
functions like by() in base R; or "tidyverse" or "data.table" package
equivalents/elaborations of these.

Here is a base R version (you must have version 4.1.x for the anonymous
function shortcut, \(x)  ) using by, but you may prefer tidyverse or
data.table versions that others may  provide:

> out <- by(cbind(R,Day), factor(Freq), FUN = \(x)cor(x)[1,2]) ## to just
get the off-diagonal of the 2x2 cor matrix
> as.list(out)
\$a1
[1] 1

\$a2
[1] -0.7559289

\$a3
[1] 0

\$a4
[1] 0.1889822

\$a5
[1] -0.8660254

See ?by and ?cor for details as needed.

Bert Gunter

"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )

On Tue, Jul 27, 2021 at 5:30 PM Marlin Keith Cox <marlinkcox using gmail.com>
wrote:

> I am having problems making a correlation/association between two variables
> by a factor.
>
> In the case below, I need to know the correlation between R and Day at each
> frequency (a1-a5). Each frequency would have a corresponding correlation
> between R and day.
>
> I have found a lm function that is similar to what I need.
> lm(R~Day*Freq), but this wont apply to the cor function.
>
> Mind you, I have hundreds of these to with these same three columns, so if
> there is an association package, I would be interested in those too.  I did
> research it, but it quickly went over my head, so I thought I would
> approach my problem this way.
>
> Data is below.
>
> Keith
>
> R<-c(1,8,3,6,7,2,3,7,2,3,3,4,3,7,3)
> Day<-c(1,1,1,1,1,2,2,2,2,2,3,3,3,3,3)
> Freq<-(a1,a2,a3,a4,a5,a1,a2,a3,a4,a5,a1,a2,a3,a4,a5,)
>
>
>
> M. Keith Cox, Ph.D.
> Principal
> MKConsulting
> 17415 Christine Ave.
> Juneau, AK 99801
> U.S. 907.957.4606
>
>         [[alternative HTML version deleted]]
>
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