Hello,
And here are three more ways. I will put the data, corrected in Bert's
post, in a data.frame.
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 <- paste0("a", rep(1:5,3))
df1 <- data.frame(R, Day, Freq)
# Base R, as for the function, see Bert's post
sapply(split(df1[-3], df1$Freq), \(x) cor(x)[1,2])
# tidyverse
library(dplyr)
df1 %>%
group_by(Freq) %>%
summarise(Cor = cor(R, Day))
# data.table
library(data.table)
setDT(df1)[, .(Cor = cor(R, Day)), by = Freq]
Hope this helps,
Rui Barradas
Às 03:30 de 28/07/21, Bert Gunter escreveu:
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 <marlink...@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]]
______________________________________________
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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.
[[alternative HTML version deleted]]
______________________________________________
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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.
______________________________________________
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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.
