Richard O'Keefe
2024-Jul-27 11:07 UTC
[R] please help generate a square correlation matrix
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 at 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 at sapo.pt> wrote: > > > > ?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$P)]<-0 > > > > > > > > > > > >> cor.4gene.sq <-pivot_wider(cor.4gene, names_from = gene1, values_from = cor) > > > > > > > > > > > >> > > > > > > > > > > > >> Then this line of code above did not generate a square matrix as what the HMisc library did. > > > > > > > > > > > >> How to fix my code? > > > > > > > > > > > >> > > > > > > > > > > > >> Thank you, > > > > > > > > > > > >> > > > > > > > > > > > >> Ding > > > > > > > > > > > >> > > > > > > > > > > > >> > > > > > > > > > > > >> ---------------------------------------------------------------------- > > > > > > > > > > > >> ------------------------------------------------------------ > > > > > > > > > > > >> -SECURITY/CONFIDENTIALITY WARNING- > > > > > > > > > > > >> > > > > > > > > > > > >> This message and any attachments are intended solely for the individual or entity to which they are addressed. 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create the two results matrices beforehand > > > > > > > > > > > > r <- P <- matrix(NA, nrow = 4L, ncol = 4L, dimnames = list(names(dat), > > > > > > > > > > > > names(dat))) > > > > > > > > > > > > > > > > > > > > > > > > for(i in 1:4) { > > > > > > > > > > > > x <- dat[[i]] > > > > > > > > > > > > for(j in (1:4)) { > > > > > > > > > > > > if(i == j) { > > > > > > > > > > > > # there's nothing to test, assign correlation 1 > > > > > > > > > > > > r[i, j] <- 1 > > > > > > > > > > > > } else { > > > > > > > > > > > > tmp <- cor.test(x, dat[[j]]) > > > > > > > > > > > > r[i, j] <- tmp$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$P > > > > > > > > > > > > #> g1 g2 g3 g4 > > > > > > > > > > > > #> g1 NA 0.79797170 0.4070838 0.68452834 > > > > > > > > > > > > #> g2 0.7979717 NA 0.4070838 0.06758329 > > > > > > > > > > > > #> g3 0.4070838 0.40708382 NA 1.00000000 > > > > > > > > > > > > #> g4 0.6845283 0.06758329 1.0000000 NA > > > > > > > > > > > > P > > > > > > > > > > > > #> g1 g2 g3 g4 > > > > > > > > > > > > #> g1 NA 0.79797170 0.4070838 0.68452834 > > > > > > > > > > > > #> g2 0.7979717 NA 0.4070838 0.06758329 > > > > > > > > > > > > #> g3 0.4070838 0.40708382 NA 1.00000000 > > > > > > > > > > > > #> g4 0.6845283 0.06758329 1.0000000 NA > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You can put these two results in a list, like Hmisc::rcorr does. > > > > > > > > > > > > > > > > > > > > > > > > lst_rcorr <- list(r = r, P = P) > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > Hope this helps, > > > > > > > > > > > > > > > > > > > > > > > > Rui Barradas > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > -- > > > > > > > > > > > > Este e-mail foi analisado pelo software antiv?rus AVG para verificar a presen?a de v?rus. > > > > > > > > > > > > https://urldefense.com/v3/__http://www.avg.com__;!!Fou38LsQmgU!tyykZkQmOKcwoWXEpV2ohbnr02thhHMabAcYLL_-7dteKHAabK-eo4rGDnwgSFjniAy8SO00L6HbloMCQMI$<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$%3e%09%5b%5balternative>HTML version deleted]] > > > > > > > > > > > > ______________________________________________ > > > > > > R-help at r-project.org<mailto:R-help at r-project.org> mailing list -- To UNSUBSCRIBE and more, see > > > > > > https://urldefense.com/v3/__https://stat.ethz.ch/mailman/listinfo/r-help__;!!Fou38LsQmgU!s54ahZtZNAIyaIGV3C2p8lhXpYlHksC6XvFKZltf6g3ElJHOO3I1MYFecLQ4QeMO3MpP3qXz-YrhdUE$<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 at 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 at 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.
Richard O'Keefe
2024-Jul-27 11:46 UTC
[R] please help generate a square correlation matrix
Curses, my laptop is hallucinating again. Hope I can get through this. 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. (also D = length(x)-A-B-C) All the information is summarised in the 2-by-2 contingency table. Some years ago, Nathan Rountree and I supervised Yung-Sing Koh's data-mining PhD. She surveyed the data mining literature and found some 37 different "interestingness measures" for two-variable associations -- if I remember correctly; there were a lot of them. They fell into a much smaller number of qualitatively similar groups. At any rate, the Pearson correlation between x and y is (A*D - B*C)/sqrt((A+B)*(C+D)*(A+C)*(B+D)) So what happens when we delete the rows where x = 0 and y = 0? Right, it forces D to 0, leaving A B C unchanged. And looking at the numerator, If you delete rows with x = 0 y = 0 you MUST get a negative correlation. Quite a modest "true" correlation (based on all the data) like -0.2 can masquerade as quite a strong "zero-suppressed" correlation like -0.6. Even +0.2 can turn into -0.4. (These figures are from a particular simulation run and may not apply in your case.) Now one of the reasons why Yun-Sing Koh, Nathan Rountree, and I were interested in interestingness measures is perhaps coincidentally related to the file drawer/underreporting problem: it's quite common for rows where x = 0 and y = 0 never to have been reported to you, so we were hoping there were measures immune to that. I have argued for years that "till record analysis" for supermarkets &c is badly flawed by two facts: (a) it is hard to measure how much of a product people WOULD have bought if only you had offered it for sale (although you can make educated guesses) and (b) till records provide no evidence on what the people who walked out without buying anything wanted (was the price too high? could they not find it?). Problem (a) leads to a commercial variant of the Signor-Lipps effect: "when x and/or y were available for purchase" is not the same as "the period for which data were recorded", thus inflating D, perhaps massively. Methods developed for handling the Signor-Lipps effect in paleontology can be used to estimate when x and y were available helping you to recover a more realistic N=A+B+C+D. I really should have published that. All of which is a long-winded way of saying that - Pearson correlations on binary columns can be computed very efficiently - the rows with x=0 and y=0 may be very informative, even essential for analysis - delete them at your peril. - really, delete them at your peril. On Sat, 27 Jul 2024 at 23:07, Richard O'Keefe <raoknz at gmail.com> wrote:> > 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 at 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 at sapo.pt> wrote: > > > > > > ?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$P)]<-0 > > > > > > > > > > > > > > > >> cor.4gene.sq <-pivot_wider(cor.4gene, names_from = gene1, values_from = cor) > > > > > > > > > > > > > > > >> > > > > > > > > > > > > > > > >> Then this line of code above did not generate a square matrix as what the HMisc library did. > > > > > > > > > > > > > > > >> How to fix my code? > > > > > > > > > > > > > > > >> > > > > > > > > > > > > > > > >> Thank you, > > > > > > > > > > > > > > > >> > > > > > > > > > > > > > > > >> Ding > > > > > > > > > > > > > > > >> > > > > > > > > > > > > > > > >> > > > > > > > > > > > > > > > >> ---------------------------------------------------------------------- > > > > > > > > > > > > > > > >> ------------------------------------------------------------ > > > > > > > > > > > > > > > >> -SECURITY/CONFIDENTIALITY WARNING- > > > > > > > > > > > > > > > >> > > > > > > > > > > > > > > > >> This message and any attachments are intended solely for the individual or entity to which they are addressed. 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(LCP301) > > > > > > > > > > > > > > > >> ------------------------------------------------------------ > > > > > > > > > > > > > > > >> > > > > > > > > > > > > > > > >> [[alternative HTML version deleted]] > > > > > > > > > > > > > > > >> > > > > > > > > > > > > > > > >> ______________________________________________ > > > > > > > > > > > > > > > >> R-help at r-project.org<mailto:R-help at r-project.org<mailto:R-help at r-project.org%3cmailto:R-help at r-project.org>> mailing list -- To UNSUBSCRIBE and more, see > > > > > > > > > > > > > > > >> https://urldefense.com/v3/__https://stat.ethz.ch/mailman/listinfo/r-help__;!!Fou38LsQmgU!tyykZkQmOKcwoWXEpV2ohbnr02thhHMabAcYLL_-7dteKHAabK-eo4rGDnwgSFjniAy8SO00L6Hb8338TBM$<https://urldefense.com/v3/__https:/stat.ethz.ch/mailman/listinfo/r-help__;!!Fou38LsQmgU!tyykZkQmOKcwoWXEpV2ohbnr02thhHMabAcYLL_-7dteKHAabK-eo4rGDnwgSFjniAy8SO00L6Hb8338TBM$><https://urldefense.com/v3/__https:/stat.ethz.ch/mailman/listinfo/r-help__;!!Fou38LsQmgU!tyykZkQmOKcwoWXEpV2ohbnr02thhHMabAcYLL_-7dteKHAabK-eo4rGDnwgSFjniAy8SO00L6Hb8338TBM$%3chttps:/urldefense.com/v3/__https:/stat.ethz.ch/mailman/listinfo/r-help__;!!Fou38LsQmgU!tyykZkQmOKcwoWXEpV2ohbnr02thhHMabAcYLL_-7dteKHAabK-eo4rGDnwgSFjniAy8SO00L6Hb8338TBM$%3e%3e> > > > > > > > > <https://urldefense.com/v3/__https:/stat.ethz.ch/mailman/listinfo/r-help__;!!Fou38LsQmgU!tyykZkQmOKcwoWXEpV2ohbnr02thhHMabAcYLL_-7dteKHAabK-eo4rGDnwgSFjniAy8SO00L6Hb8338TBM$%3chttps:/urldefense.com/v3/__https:/stat.ethz.ch/mailman/listinfo/r-help__;!!Fou38LsQmgU!tyykZkQmOKcwoWXEpV2ohbnr02thhHMabAcYLL_-7dteKHAabK-eo4rGDnwgSFjniAy8SO00L6Hb8338TBM$%3e%3e> > > > > > > > >> <https://urldefense.com/v3/__https:/stat.ethz.ch/mailman/listinfo/r-help__;!!Fou38LsQmgU!tyykZkQmOKcwoWXEpV2ohbnr02thhHMabAcYLL_-7dteKHAabK-eo4rGDnwgSFjniAy8SO00L6Hb8338TBM$%3chttps:/urldefense.com/v3/__https:/stat.ethz.ch/mailman/listinfo/r-help__;!!Fou38LsQmgU!tyykZkQmOKcwoWXEpV2ohbnr02thhHMabAcYLL_-7dteKHAabK-eo4rGDnwgSFjniAy8SO00L6Hb8338TBM$%3e%3e>PLEASE do read the posting guide https://urldefense.com/v3/__http://www.R-project.org/posting-guide.html__;!!Fou38LsQmgU!tyykZkQmOKcwoWXEpV2ohbnr02thhHMabAcYLL_-7dteKHAabK-eo4rGDnwgSFjniAy8SO00L6Hb880tLw0$<https://urldefense.com/v3/__http:/www.R-project.org/posting-guide.html__;!!Fou38LsQmgU!tyykZkQmOKcwoWXEpV2ohbnr02thhHMabAcYLL_-7dteKHAabK-eo4rGDnwgSFjniAy8SO00L6Hb880tLw0$><https://urldefense.com/v3/__http:/www.R-project.org/posting-guide.html__;!!Fou38LsQmgU!tyykZkQmOKcwoWXEpV2ohbnr02thhHMabAcYLL_-7dteKHAabK-eo4rGDnwgSFjniAy8SO00L6Hb880tLw0$%3chttps:/urldefense.com/v3/__http:/www.R-project.org/posting-guide.html__;!!Fou38LsQmgU!tyykZkQmOKcwoWXEpV2ohbnr02thhHMabAcYLL_-7dteKHAabK-eo4rGDnwgSFjniAy8SO00L6Hb880tLw0$%3e%3e> > > > > > > > > <https://urldefense.com/v3/__http:/www.R-project.org/posting-guide.html__;!!Fou38LsQmgU!tyykZkQmOKcwoWXEpV2ohbnr02thhHMabAcYLL_-7dteKHAabK-eo4rGDnwgSFjniAy8SO00L6Hb880tLw0$%3chttps:/urldefense.com/v3/__http:/www.R-project.org/posting-guide.html__;!!Fou38LsQmgU!tyykZkQmOKcwoWXEpV2ohbnr02thhHMabAcYLL_-7dteKHAabK-eo4rGDnwgSFjniAy8SO00L6Hb880tLw0$%3e%3e> > > > > > > > >> <https://urldefense.com/v3/__http:/www.R-project.org/posting-guide.html__;!!Fou38LsQmgU!tyykZkQmOKcwoWXEpV2ohbnr02thhHMabAcYLL_-7dteKHAabK-eo4rGDnwgSFjniAy8SO00L6Hb880tLw0$%3chttps:/urldefense.com/v3/__http:/www.R-project.org/posting-guide.html__;!!Fou38LsQmgU!tyykZkQmOKcwoWXEpV2ohbnr02thhHMabAcYLL_-7dteKHAabK-eo4rGDnwgSFjniAy8SO00L6Hb880tLw0$%3e%3e>and provide commented, minimal, self-contained, reproducible code. > > > > > > > > > > > > > > > > Hello, > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You are complicating the code, there's no need for as.symbol/eval, the > > > > > > > > > > > > > > > > column numbers do exactly the same. > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > # create the two results matrices beforehand > > > > > > > > > > > > > > > > r <- P <- matrix(NA, nrow = 4L, ncol = 4L, dimnames = list(names(dat), > > > > > > > > > > > > > > > > names(dat))) > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > for(i in 1:4) { > > > > > > > > > > > > > > > > x <- dat[[i]] > > > > > > > > > > > > > > > > for(j in (1:4)) { > > > > > > > > > > > > > > > > if(i == j) { > > > > > > > > > > > > > > > > # there's nothing to test, assign correlation 1 > > > > > > > > > > > > > > > > r[i, j] <- 1 > > > > > > > > > > > > > > > > } else { > > > > > > > > > > > > > > > > tmp <- cor.test(x, dat[[j]]) > > > > > > > > > > > > > > > > r[i, j] <- tmp$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$P > > > > > > > > > > > > > > > > #> g1 g2 g3 g4 > > > > > > > > > > > > > > > > #> g1 NA 0.79797170 0.4070838 0.68452834 > > > > > > > > > > > > > > > > #> g2 0.7979717 NA 0.4070838 0.06758329 > > > > > > > > > > > > > > > > #> g3 0.4070838 0.40708382 NA 1.00000000 > > > > > > > > > > > > > > > > #> g4 0.6845283 0.06758329 1.0000000 NA > > > > > > > > > > > > > > > > P > > > > > > > > > > > > > > > > #> g1 g2 g3 g4 > > > > > > > > > > > > > > > > #> g1 NA 0.79797170 0.4070838 0.68452834 > > > > > > > > > > > > > > > > #> g2 0.7979717 NA 0.4070838 0.06758329 > > > > > > > > > > > > > > > > #> g3 0.4070838 0.40708382 NA 1.00000000 > > > > > > > > > > > > > > > > #> g4 0.6845283 0.06758329 1.0000000 NA > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You can put these two results in a list, like Hmisc::rcorr does. > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > lst_rcorr <- list(r = r, P = P) > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > Hope this helps, > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > Rui Barradas > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > -- > > > > > > > > > > > > > > > > Este e-mail foi analisado pelo software antiv?rus AVG para verificar a presen?a de v?rus. > > > > > > > > > > > > > > > > https://urldefense.com/v3/__http://www.avg.com__;!!Fou38LsQmgU!tyykZkQmOKcwoWXEpV2ohbnr02thhHMabAcYLL_-7dteKHAabK-eo4rGDnwgSFjniAy8SO00L6HbloMCQMI$<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$%3e%09%5b%5balternative>HTML version deleted]] > > > > > > > > > > > > > > > > ______________________________________________ > > > > > > > > R-help at r-project.org<mailto:R-help at r-project.org> mailing list -- To UNSUBSCRIBE and more, see > > > > > > > > https://urldefense.com/v3/__https://stat.ethz.ch/mailman/listinfo/r-help__;!!Fou38LsQmgU!s54ahZtZNAIyaIGV3C2p8lhXpYlHksC6XvFKZltf6g3ElJHOO3I1MYFecLQ4QeMO3MpP3qXz-YrhdUE$<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 at 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 at 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.