Re: [R] please help generate a square correlation matrix

Rui Barradas Thu, 25 Jul 2024 14:01:34 -0700

Às 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-boun...@r-project.org> On Behalf Of Yuan Chun Ding via 
R-help
Sent: Thursday, July 25, 2024 11:26 AM
To: Rui Barradas <ruipbarra...@sapo.pt>; r-help@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 <ruipbarra...@sapo.pt<mailto:ruipbarra...@sapo.pt>>

Sent: Thursday, July 25, 2024 11:15 AM

To: Yuan Chun Ding <ycd...@coh.org<mailto:ycd...@coh.org>>; 
r-help@r-project.org<mailto:r-help@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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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 seethe results, cor is symmetric (as it should) so the calculations arerepeated. If you only run the cor.tests on the combinations ofnames(dat) by groups of 2, it will save a lot of work. But the output isa 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


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Re: [R] please help generate a square correlation matrix

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