[R] spearman correlation and p-value as a matrix
arun
smartpink111 at yahoo.com
Thu Feb 14 06:16:40 CET 2013
HI,
#ag data was created
bg<- as.matrix(read.table(text="
Otu00022 Otu00029 Otu00039 Otu00042 Otu00101 Otu00105 Otu00125 Otu00131 Otu00137 Otu00155 Otu00158 Otu00172 Otu00181 Otu00185 Otu00190 Otu00209 Otu00218
Gi20Jun11 0.001217 0 0.001217 0 0.000000 0 0 0 0.001217 0 0 0 0 0 0.001217 0 0.001217
Gi40Jun11 0.000000 0 0.000000 0 0.000000 0 0 0 0.000000 0 0 0 0 0 0.000000 0 0.000000
Gi425Jun11 0.000000 0 0.000000 0 0.000000 0 0 0 0.000000 0 0 0 0 0 0.000000 0 0.000000
Gi45Jun11 0.000000 0 0.000000 0 0.001513 0 0 0 0.000000 0 0 0 0 0 0.000000 0 0.000000
Gi475Jun11 0.000000 0 0.000000 0 0.000000 0 0 0 0.000000 0 0 0 0 0 0.000000 0 0.000000
Gi50Jun11 0.000000 0 0.000000 0 0.000000 0 0 0 0.000000 0 0 0 0 0 0.000000 0 0.000000
",sep="",header=TRUE,stringsAsFactors=F))
set.seed(128)
ag<- matrix(rnorm(30),nrow=6)
colnames(ag)<- paste("ag",1:5,sep="")
bg_ag<-expand.grid(colnames(bg),colnames(ag),stringsAsFactors=FALSE)
attr(bg_ag,"out.attrs")<- NULL
library(Hmisc)
#correlation
resr<-do.call(rbind,lapply(split(bg_ag,1:nrow(bg_ag)),function(x) {res<-rcorr(cbind(bg[,x[,1]],ag[,x[,2]]),type="spearman")$r; row.names(res)<- rep(paste(x[1],x[2],sep="_"),2);res}))
head(resr)
# [,1] [,2]
#Otu00022_ag1 1.0000000 0.1309307
#Otu00022_ag1 0.1309307 1.0000000
#Otu00029_ag1 1.0000000 NaN
#Otu00029_ag1 NaN 1.0000000
#Otu00039_ag1 1.0000000 0.1309307
#Otu00039_ag1 0.1309307 1.0000000
#p-values
resP<-do.call(rbind,lapply(split(bg_ag,1:nrow(bg_ag)),function(x) {res<-rcorr(cbind(bg[,x[,1]],ag[,x[,2]]),type="spearman")$P; row.names(res)<- rep(paste(x[1],x[2],sep="_"),2);res}))
head(resP)
# [,1] [,2]
#Otu00022_ag1 NA 0.8047262
#Otu00022_ag1 0.8047262 NA
#Otu00029_ag1 NA NaN
#Otu00029_ag1 NaN NA
#Otu00039_ag1 NA 0.8047262
#Otu00039_ag1 0.8047262 NA
#If you need only the values
indx<-row(resr)%%2!=1
resPnew<-as.matrix(resP[indx[,1],1])
resrnew<-as.matrix(resr[indx[,1],1])
head(resPnew)
# [,1]
#Otu00022_ag1 0.8047262
#Otu00029_ag1 NaN
#Otu00039_ag1 0.8047262
#Otu00042_ag1 NaN
#Otu00101_ag1 0.1583024
#Otu00105_ag1 NaN
head(resrnew)
# [,1]
#Otu00022_ag1 0.1309307
#Otu00029_ag1 NaN
#Otu00039_ag1 0.1309307
#Otu00042_ag1 NaN
#Otu00101_ag1 -0.6546537
#Otu00105_ag1 NaN
A.K.
----- Original Message -----
From: Ozgul Inceoglu <Ozgul.Inceoglu at ulb.ac.be>
To: r-help at r-project.org
Cc:
Sent: Wednesday, February 13, 2013 4:48 AM
Subject: [R] spearman correlation and p-value as a matrix
I have two data matrices that I want to make the correlation between each column from data1 and each column from data 2 and also calculate the p-value Matrices dont have the same size and I tried such a script.
> bg <- read.table (file.choose(), header=T, row.names)
> bg
> Otu00022 Otu00029 Otu00039 Otu00042 Otu00101 Otu00105 Otu00125 Otu00131 Otu00137 Otu00155 Otu00158 Otu00172 Otu00181 Otu00185 Otu00190 Otu00209 Otu00218
> Gi20Jun11 0.001217 0 0.001217 0 0.000000 0 0 0 0.001217 0 0 0 0 0 0.001217 0 0.001217
> Gi40Jun11 0.000000 0 0.000000 0 0.000000 0 0 0 0.000000 0 0 0 0 0 0.000000 0 0.000000
> Gi425Jun11 0.000000 0 0.000000 0 0.000000 0 0 0 0.000000 0 0 0 0 0 0.000000 0 0.000000
> Gi45Jun11 0.000000 0 0.000000 0 0.001513 0 0 0 0.000000 0 0 0 0 0 0.000000 0 0.000000
> Gi475Jun11 0.000000 0 0.000000 0 0.000000 0 0 0 0.000000 0 0 0 0 0 0.000000 0 0.000000
> Gi50Jun11 0.000000 0 0.000000 0 0.000000 0 0 0 0.000000 0 0 0 0 0 0.000000 0 0.000000
ag <- read.table (file.choose(), header=T, row.names)
for (i in 1:(ncol(bg)))
for (j in 1:(ncol(ag)))
print(c(i,j))
final_matrix <- matrix(rep("0",ncol(bg)*ncol(ag)),ncol=ncol(bg),nrow=ncol(ag))
cor <- cor.test(as.vector(as.matrix(bg[,i])),as.vector(as.matrix(ag[,j])), method="spearman")
#but the output is not matrice with all the values but a single correlation value
data: bg[, i] and ag[, j]
t = 2.2992, df = 26, p-value = 0.02978
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
0.04485289 0.67986803
sample estimates:
cor
0.4110515
# How I can creat an outfile with all the correlations and p-values?
Thank you very much!
Özgül
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