I have a function, which takes two vectors and computes a numeric value (like cor correlation would do). However, I have two datasets with around 6000 columns (the two datasets have the same dimensions), where the function should return one vector with the values of the correlation.
The code with a loop would look like this:
set.seed(123)
m=matrix(rnorm(9),ncol=3)
n=matrix(rnorm(9,10),ncol=3)
colNumber=dim(m)[2]
ReturnData=rep(NA,colNumber)
for (i in 1:colNumber){
ReturnData[i]=cor(m[,i],n[,i])
}
This works fine, but for efficiency reasons I want to use the apply-family, obviously, the mapply function.
However,mapply(cor,m,n) returns a vector with length 9 of NAs, where it should return:
> ReturnData
[1] 0.1247039 -0.9641188 0.5081204
EDIT/SOLUTION
The solution as given by @akrun was the usage of dataframes instead of matrices.
Furthermore, a speed test between the two proposed solutions showed, that the mapply-version is faster than sapply:
require(rbenchmark)
set.seed(123)
#initiate the two dataframes for the comparison
m=data.frame(matrix(rnorm(10^6),ncol=100))
n=data.frame(matrix(rnorm(10^6),ncol=100))
#indx is needed for the sapply function to get the column numbers
indx=seq_len(ncol(m))
benchmark(s1=mapply(cor, m,n), s2=sapply(indx, function(i) cor(m[,i], n[,i])), order="elapsed", replications=100)
# test replications elapsed relative user.self sys.self user.child sys.child
# 2 s2 100 4.16 1.000 4.15 0 NA NA
# 1 s1 100 4.33 1.041 4.32 0 NA NA
Because your dataset is matrix, the mapply would loop through each element instead of each column. To avoid that, convert to dataframe. I am not sure how efficient this would be for big datasets.
mapply(cor, as.data.frame(m), as.data.frame(n))
# V1 V2 V3
#0.1247039 -0.9641188 0.5081204
Another option is to use sapply without converting to data.frame
indx <- seq_len(ncol(m))
sapply(indx, function(i) cor(m[,i], n[,i]))
#[1] 0.1247039 -0.9641188 0.5081204