correlation - How to inform R that the first column of my dataset is row names? And how should change the class of data frame to vector or matrice? -

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   apd<- read.csv("apd.csv",header=false)   rar<- read.csv("rar.csv",header=false)  ## making column name labels  tree<-rep("tree",100)  tree_labels<-c(1:100)  colnames(apd)<-c(paste(tree,tree_labels, sep=""))   colnames(rar)<-c(paste(tree,tree_labels, sep=""))    correlation.csv<- cor(x=apd, y=rar, method = "spearman")

the above script suppose 2 calculated correlation between columns of 2 data sets . there 2 problem starts labeling output first column (which row name) last score gets na label. i'm not sure if i'm thinking correctly or not maybe same reason r thinks apd data frame not calculate last line.

cheers

subset of csv file

   v1              v2                v3    v4 t1  9.368703877 9.693286792 12.44129352 13.06908296 t10 8.128921254 8.940227268 11.40226603 12.17704779 t11 7.87062995  8.697508965 11.39250803 12.17704779

after read in it's below

    v2           v3            v4            v5     v1          v2               v3         v4  t1 9.368703877 9.693286792 12.44129352 13.06908296 t1  08.128921254    8.940227268 11.40226603 12.17704779 t11 7.87062995  8.697508965 11.39250803 12.17704779

for first problem, can use row.names:

apd <- read.csv("apd.csv", header = false, row.names = 1) rar <- read.csv("rar.csv", header = false, row.names = 1)

from documentation:

row.names: vector of row names. can vector giving actual row names, or single number giving column of table contains row names, or character string giving name of table column containing row names.

for second problem, can use mapply:

mapply(cor, apd, rar, moreargs = list(method = "spearman"))

assuming want correlation between 1st column of each table, 2nd column, , on.

using example:

> str(a) 'data.frame':   3 obs. of  4 variables:  $ v1: num  9.37 8.13 7.87  $ v2: num  9.69 8.94 8.7  $ v3: num  12.4 11.4 11.4  $ v4: num  13.1 12.2 12.2 > mapply(cor, a, -a, moreargs = list(method = "spearman")) v1 v2 v3 v4  -1 -1 -1 -1

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