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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