please help generate a square correlation matrix
?s 20:47 de 25/07/2024, Yuan Chun Ding escreveu:
Hi Rui,
You are always very helpful!! Thank you,
I just modified your R codes to remove a row with zero values in both column pair as below for my real data.
Ding
dat<-gene22mut.coded
r <- P <- matrix(NA, nrow = 22L, ncol = 22L,
dimnames = list(names(dat), names(dat)))
for(i in 1:22) {
#i=1
x <- dat[[i]]
for(j in (1:22)) {
#j=2
if(i == j) {
# there's nothing to test, assign correlation 1
r[i, j] <- 1
} else {
tmp <-cbind(x,dat[[j]])
row0 <-rowSums(tmp)
tem2 <-tmp[row0!=0,]
tmp3 <- cor.test(tem2[,1],tem2[,2])
r[i, j] <- tmp3$estimate
P[i, j] <- tmp3$p.value
}
}
}
r<-as.data.frame(r)
P<-as.data.frame(P)
From: R-help <r-help-bounces at r-project.org> On Behalf Of Yuan Chun Ding via R-help
Sent: Thursday, July 25, 2024 11:26 AM
To: Rui Barradas <ruipbarradas at sapo.pt>; r-help at r-project.org
Subject: Re: [R] please help generate a square correlation matrix
HI Rui, Thank you for the help! You did not remove a row if zero values exist in both column pair, right? Ding From: Rui Barradas <ruipbarradas@?sapo.?pt> Sent: Thursday, July 25, 2024 11:?15 AM To: Yuan Chun Ding <ycding@?coh.?org>;
HI Rui,
Thank you for the help!
You did not remove a row if zero values exist in both column pair, right?
Ding
From: Rui Barradas <ruipbarradas at sapo.pt<mailto:ruipbarradas at sapo.pt>>
Sent: Thursday, July 25, 2024 11:15 AM
To: Yuan Chun Ding <ycding at coh.org<mailto:ycding at coh.org>>; r-help at r-project.org<mailto:r-help at r-project.org>
Subject: Re: [R] please help generate a square correlation matrix
?s 17:?39 de 25/07/2024, Yuan Chun Ding via R-help escreveu: > Hi R users, > > I generated a square correlation matrix for the dat dataframe below; > dat<-data.?frame(g1=c(1,0,0,1,1,1,0,0,0), > g2=c(0,1,0,1,0,1,1,0,0), > g3=c(1,1,0,0,0,1,0,0,0),
?s 17:39 de 25/07/2024, Yuan Chun Ding via R-help escreveu:
Hi R users,
I generated a square correlation matrix for the dat dataframe below;
dat<-data.frame(g1=c(1,0,0,1,1,1,0,0,0),
g2=c(0,1,0,1,0,1,1,0,0),
g3=c(1,1,0,0,0,1,0,0,0),
g4=c(0,1,0,1,1,1,1,1,0))
library("Hmisc")
dat.rcorr = rcorr(as.matrix(dat))
dat.r <-round(dat.rcorr$r,2)
however, I want to modify this correlation calculation;
my dat has more than 1000 rows and 22 columns;
in each column, less than 10% values are 1, most of them are 0;
so I want to remove a row with value of zero in both columns when calculate correlation between two columns.
I just want to check whether those values of 1 are correlated between two columns.
Please look at my code in the following;
cor.4gene <-matrix(0,nrow=4*4, ncol=4)
for (i in 1:4){
#i=1
for (j in 1:4) {
#j=1
d <-dat[,c(i,j)]%>%
filter(eval(as.symbol(colnames(dat)[i]))!=0 |
eval(as.symbol(colnames(dat)[j]))!=0)
c <-cor.test(d[,1],d[,2])
cor.4gene[i*j,]<-c(colnames(dat)[i],colnames(dat)[j],
c$estimate,c$p.value)
}
}
cor.4gene<-as.data.frame(cor.4gene)%>%filter(V1 !=0)
colnames(cor.4gene)<-c("gene1","gene2","cor","P")
Can you tell me what mistakes I made?
first, why cor is NA when calculation of correlation for g1 and g1, I though it should be 1.
cor.4gene$cor[is.na(cor.4gene$cor)]<-1
cor.4gene$cor[is.na(cor.4gene$P)]<-0
cor.4gene.sq <-pivot_wider(cor.4gene, names_from = gene1, values_from = cor)
Then this line of code above did not generate a square matrix as what the HMisc library did.
How to fix my code?
Thank you,
Ding
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Hello,
You are complicating the code, there's no need for as.symbol/eval, the
column numbers do exactly the same.
# create the two results matrices beforehand
r <- P <- matrix(NA, nrow = 4L, ncol = 4L, dimnames = list(names(dat),
names(dat)))
for(i in 1:4) {
x <- dat[[i]]
for(j in (1:4)) {
if(i == j) {
# there's nothing to test, assign correlation 1
r[i, j] <- 1
} else {
tmp <- cor.test(x, dat[[j]])
r[i, j] <- tmp$estimate
P[i, j] <- tmp$p.value
}
}
}
# these two results are equal up to floating-point precision
dat.rcorr$r
#> g1 g2 g3 g4
#> g1 1.0000000 0.1000000 0.3162278 0.1581139
#> g2 0.1000000 1.0000000 0.3162278 0.6324555
#> g3 0.3162278 0.3162278 1.0000000 0.0000000
#> g4 0.1581139 0.6324555 0.0000000 1.0000000
r
#> g1 g2 g3 g4
#> g1 1.0000000 0.1000000 3.162278e-01 1.581139e-01
#> g2 0.1000000 1.0000000 3.162278e-01 6.324555e-01
#> g3 0.3162278 0.3162278 1.000000e+00 1.355253e-20
#> g4 0.1581139 0.6324555 1.355253e-20 1.000000e+00
# these two results are equal up to floating-point precision
dat.rcorr$P
#> g1 g2 g3 g4
#> g1 NA 0.79797170 0.4070838 0.68452834
#> g2 0.7979717 NA 0.4070838 0.06758329
#> g3 0.4070838 0.40708382 NA 1.00000000
#> g4 0.6845283 0.06758329 1.0000000 NA
P
#> g1 g2 g3 g4
#> g1 NA 0.79797170 0.4070838 0.68452834
#> g2 0.7979717 NA 0.4070838 0.06758329
#> g3 0.4070838 0.40708382 NA 1.00000000
#> g4 0.6845283 0.06758329 1.0000000 NA
You can put these two results in a list, like Hmisc::rcorr does.
lst_rcorr <- list(r = r, P = P)
Hope this helps,
Rui Barradas
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Hello,
Here are two other ways.
The first is equivalent to your long format attempt.
library(tidyverse)
dat %>%
names() %>%
expand.grid(., .) %>%
apply(1L, \(x) {
tmp <- dat[rowSums(dat[x]) > 0, ]
tmp2 <- cor.test(tmp[[ x[1L] ]], tmp[[ x[2L] ]])
c(tmp2$estimate, P = tmp2$p.value)
}) %>%
t() %>%
as.data.frame() %>%
cbind(tmp_df, .) %>%
na.omit()
The second is, in my opinion the one that makes more sense. If you see
the results, cor is symmetric (as it should) so the calculations are
repeated. If you only run the cor.tests on the combinations of
names(dat) by groups of 2, it will save a lot of work. But the output is
a much smaller a data.frame.
cbind(
combn(names(dat), 2L) %>%
t() %>%
as.data.frame(),
combn(dat, 2L, FUN = \(d) {
d2 <- d[rowSums(d) > 0, ]
tmp2 <- cor.test(d2[[1L]], d2[[2L]])
c(tmp2$estimate, P = tmp2$p.value)
}) %>% t()
) %>% na.omit()
Hope this helps,
Rui Barradas