?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. This communication may contain information
that is privileged, confidential, or exempt from disclosure under applicable law
(e.g., personal health information, research data, financial information).
Because this e-mail has been sent without encryption, individuals other than the
intended recipient may be able to view the information, forward it to others or
tamper with the information without the knowledge or consent of the sender. If
you are not the intended recipient, or the employee or person responsible for
delivering the message to the intended recipient, any dissemination,
distribution or copying of the communication is strictly prohibited. If you
received the communication in error, please notify the sender immediately by
replying to this message and deleting the message and any accompanying files
from your system. If, due to the security risks, you do not wish to rec
> eive further communications via e-mail, please reply to this message and
inform the sender that you do not wish to receive further e-mail from the
sender. (LCP301)
> ------------------------------------------------------------
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> 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.
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.
www.avg.com