[R] please help generate a square correlation matrix
Yuan Chun Ding
ycd|ng @end|ng |rom coh@org
Thu Jul 25 21:47:13 CEST 2024
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 using r-project.org> On Behalf Of Yuan Chun Ding via R-help
Sent: Thursday, July 25, 2024 11:26 AM
To: Rui Barradas <ruipbarradas using sapo.pt>; r-help using 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 using sapo.pt<mailto:ruipbarradas using sapo.pt>>
Sent: Thursday, July 25, 2024 11:15 AM
To: Yuan Chun Ding <ycding using coh.org<mailto:ycding using coh.org>>; r-help using r-project.org<mailto:r-help using r-project.org>
Subject: Re: [R] please help generate a square correlation matrix
Às 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),
Às 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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