[R] please help generate a square correlation matrix

Richard O'Keefe r@oknz @end|ng |rom gm@||@com
Sat Jul 27 13:07:46 CEST 2024 

Let's go back to the original posting.

> >
> >> 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.
> >

So we're talking about correlations between binary variables.
Suppose we have two 0-1-valued variables, x and y.
Let A <- sum(x*y)  # number of cases where x and y are both 1.
Let B <- sum(x)-a  # number of cases where x is 1 and y is 0
Let C <- sum(y)-a # number of cases where y is 1 and x is 0
Let D <- sum(!x * !y) # number of cases where x and y are both 0.

N

On Fri, 26 Jul 2024 at 12:07, Bert Gunter <bgunter.4567 using gmail.com> wrote:
>
> If I have understood the request, I'm not sure that omitting all 0
> pairs for each pair of columns makes much sense, but be that as it
> may, here's another way to do it by using the 'FUN' argument of combn
> to encapsulate any calculations that you do. I just use cor() as the
> calculation -- you can use anything you like that takes two vectors of
> 0's and 1's and produces fixed length numeric results (or fromm which
> you can extract such).
>
> I encapsulated it all in a little function. Note that I first
> converted the data frame to a matrix. Because of their generality,
> data frames carry a lot of extra baggage that can slow purely numeric
> manipulations down.
>
> Anyway, here's the function, 'somecors' (I'm a bad name picker :(  ! )
>
>    somecors <- function(dat, func = cor){
>       dat <- as.matrix(dat)
>       indx <- seq_len(ncol(dat))
>          combn(indx, 2, FUN = \(z) {
>             i <- z[1]; j <- z[2]
>             k <- dat[, i ] | dat[, j ]
>             c(z,func(dat[k,i ], dat[k,j ]))
>          })
>    }
>
> Results come out as a matrix with combn(ncol(dat),2) columns, the
> first 2 rows giving the pair of column numbers for each column,and
> then 1 or more rows (possibly extracted) from whatever func you use.
> Here's the results for your data formatted to 2 decimal places:
>
> > round(somecors(dat),2)
>      [,1]  [,2]  [,3]  [,4] [,5]  [,6]
> [1,]  1.0  1.00  1.00  2.00    2  3.00
> [2,]  2.0  3.00  4.00  3.00    4  4.00
> [3,] -0.5 -0.41 -0.35 -0.41   NA -0.47
> Warning message:
> In func(dat[k, i], dat[k, j]) : the standard deviation is zero
>
> The NA and warning comes in the 2,4 pair of columns because after
> removing all zero rows in the pair, dat[,4] is all 1's, giving a zero
> in the denominator of the cor() calculation -- again, assuming I have
> correctly understood your request. If so, this might be something you
> need to worry about.
>
> Again, feel free to ignore if  I have misinterpreterd or this does not suit.
>
> Cheers,
> Bert
>
>
> On Thu, Jul 25, 2024 at 2:01 PM Rui Barradas <ruipbarradas using sapo.pt> wrote:
> >
> > 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-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$P
> > >
> > >
> > >
> > > #>           g1         g2        g3         g4
> > >
> > >
> > >
> > > #> g1        NA 0.79797170 0.4070838 0.68452834
> > >
> > >
> > >
> > > #> g2 0.7979717         NA 0.4070838 0.06758329
> > >
> > >
> > >
> > > #> g3 0.4070838 0.40708382        NA 1.00000000
> > >
> > >
> > >
> > > #> g4 0.6845283 0.06758329 1.0000000         NA
> > >
> > >
> > >
> > > P
> > >
> > >
> > >
> > > #>           g1         g2        g3         g4
> > >
> > >
> > >
> > > #> g1        NA 0.79797170 0.4070838 0.68452834
> > >
> > >
> > >
> > > #> g2 0.7979717         NA 0.4070838 0.06758329
> > >
> > >
> > >
> > > #> g3 0.4070838 0.40708382        NA 1.00000000
> > >
> > >
> > >
> > > #> g4 0.6845283 0.06758329 1.0000000         NA
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > > You can put these two results in a list, like Hmisc::rcorr does.
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > > lst_rcorr <- list(r = r, P = P)
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > > Hope this helps,
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > > Rui Barradas
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > > --
> > >
> > >
> > >
> > > Este e-mail foi analisado pelo software antivírus AVG para verificar a presença de vírus.
> > >
> > >
> > >
> > > https://urldefense.com/v3/__http://www.avg.com__;!!Fou38LsQmgU!tyykZkQmOKcwoWXEpV2ohbnr02thhHMabAcYLL_-7dteKHAabK-eo4rGDnwgSFjniAy8SO00L6HbloMCQMI$<https://urldefense.com/v3/__http:/www.avg.com__;!!Fou38LsQmgU!tyykZkQmOKcwoWXEpV2ohbnr02thhHMabAcYLL_-7dteKHAabK-eo4rGDnwgSFjniAy8SO00L6HbloMCQMI$><https://urldefense.com/v3/__http:/www.avg.com__;!!Fou38LsQmgU!tyykZkQmOKcwoWXEpV2ohbnr02thhHMabAcYLL_-7dteKHAabK-eo4rGDnwgSFjniAy8SO00L6HbloMCQMI$%3chttps:/urldefense.com/v3/__http:/www.avg.com__;!!Fou38LsQmgU!tyykZkQmOKcwoWXEpV2ohbnr02thhHMabAcYLL_-7dteKHAabK-eo4rGDnwgSFjniAy8SO00L6HbloMCQMI$%3e%09%5b%5balternative>
> > >
> > >   <https://urldefense.com/v3/__http:/www.avg.com__;!!Fou38LsQmgU!tyykZkQmOKcwoWXEpV2ohbnr02thhHMabAcYLL_-7dteKHAabK-eo4rGDnwgSFjniAy8SO00L6HbloMCQMI$%3chttps:/urldefense.com/v3/__http:/www.avg.com__;!!Fou38LsQmgU!tyykZkQmOKcwoWXEpV2ohbnr02thhHMabAcYLL_-7dteKHAabK-eo4rGDnwgSFjniAy8SO00L6HbloMCQMI$%3e%09%5b%5balternative>
> > >
> > >                 [[alternative<https://urldefense.com/v3/__http:/www.avg.com__;!!Fou38LsQmgU!tyykZkQmOKcwoWXEpV2ohbnr02thhHMabAcYLL_-7dteKHAabK-eo4rGDnwgSFjniAy8SO00L6HbloMCQMI$%3chttps:/urldefense.com/v3/__http:/www.avg.com__;!!Fou38LsQmgU!tyykZkQmOKcwoWXEpV2ohbnr02thhHMabAcYLL_-7dteKHAabK-eo4rGDnwgSFjniAy8SO00L6HbloMCQMI$%3e%09%5b%5balternative>HTML version deleted]]
> > >
> > >
> > >
> > > ______________________________________________
> > >
> > > R-help using r-project.org<mailto:R-help using r-project.org> mailing list -- To UNSUBSCRIBE and more, see
> > >
> > > https://urldefense.com/v3/__https://stat.ethz.ch/mailman/listinfo/r-help__;!!Fou38LsQmgU!s54ahZtZNAIyaIGV3C2p8lhXpYlHksC6XvFKZltf6g3ElJHOO3I1MYFecLQ4QeMO3MpP3qXz-YrhdUE$<https://urldefense.com/v3/__https:/stat.ethz.ch/mailman/listinfo/r-help__;!!Fou38LsQmgU!s54ahZtZNAIyaIGV3C2p8lhXpYlHksC6XvFKZltf6g3ElJHOO3I1MYFecLQ4QeMO3MpP3qXz-YrhdUE$>
> > >
> > > PLEASE do read the posting guide https://urldefense.com/v3/__http://www.R-project.org/posting-guide.html__;!!Fou38LsQmgU!s54ahZtZNAIyaIGV3C2p8lhXpYlHksC6XvFKZltf6g3ElJHOO3I1MYFecLQ4QeMO3MpP3qXzNRZxc6s$<https://urldefense.com/v3/__http:/www.R-project.org/posting-guide.html__;!!Fou38LsQmgU!s54ahZtZNAIyaIGV3C2p8lhXpYlHksC6XvFKZltf6g3ElJHOO3I1MYFecLQ4QeMO3MpP3qXzNRZxc6s$>
> > >
> > > and provide commented, minimal, self-contained, reproducible code.
> > >
> > 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
> >
> >
> > ______________________________________________
> > R-help using 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 using 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.

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