Hello Nikhil and Wu,
Thank you very much for your reply!
What I want is to calculate the student's score column-wise by ignoring the
specific values such as zeros for example only using c(2,1) in column 8 in the
tmp1 and generate tmp2. I changed the zeros to NAs and modified my stud fun to
student as below. So I can ignore these specific values but can not put back
the NAs to the right position in the resulted matrix.
Any suggestions to put back NAs to their original positions in column 7 and 8
either to list tmp4 or matrix tmp5?
tmp1[tmp1==0]<-NA
student<- function(x){
x<-x[!is.na(x)]
x<-(x-mean(x))/sd(x)
return (x)
}
tmp4<-apply(tmp1, 2, student)
tmp4
[[1]]
[1] 1.1296201 0.1613743 -1.2909944 1.1296201 -0.3227486 -0.8068715
[[2]]
[1] 0.0000000 0.0000000 -1.4680505 -0.5872202 0.5872202 1.4680505
[[3]]
[1] -0.5207910 -0.4817316 -0.4036130 0.2994548 1.9008870 -0.7942062
[[4]]
[1] -0.8630035 -0.2380699 -0.7737273 -0.5951748 1.5474546 0.9225210
[[5]]
[1] -0.1913482 -0.6944435 -0.7202433 0.5826446 1.7565336 -0.7331431
[[6]]
[1] -0.1132277 0.5661385 -1.4719601 -0.7925939 0.5661385 1.2455047
[[7]]
[1] -0.9561829 0.2390457 1.4342743 0.2390457 -0.9561829
[[8]]
[1] 0.7071068 -0.7071068
tmp5<- matrix(unlist(tmp4),nrow=6)
tmp5
[,1] [,2] [,3] [,4] [,5] [,6]
[,7] [,8]
[1,] 1.1296201 0.0000000 -0.5207910 -0.8630035 -0.1913482 -0.1132277
-0.9561829 -0.7071068
[2,] 0.1613743 0.0000000 -0.4817316 -0.2380699 -0.6944435 0.5661385
0.2390457 1.1296201
[3,] -1.2909944 -1.4680505 -0.4036130 -0.7737273 -0.7202433 -1.4719601
1.4342743 0.1613743
[4,] 1.1296201 -0.5872202 0.2994548 -0.5951748 0.5826446 -0.7925939
0.2390457 -1.2909944
[5,] -0.3227486 0.5872202 1.9008870 1.5474546 1.7565336 0.5661385
-0.9561829 1.1296201
[6,] -0.8068715 1.4680505 -0.7942062 0.9225210 -0.7331431 1.2455047
0.7071068 -0.3227486
Joshua
----- Original Message -----
From: "Shuhua Zhan" <sz...@uoguelph.ca>
To: r-help@r-project.org
Sent: Thursday, July 15, 2010 11:08:34 PM GMT -05:00 US/Canada Eastern
Subject: [R] how to skip a specific value when using apply() function to a
matrix?
Hello R experts,
I'd like to studentize a matrix (tmp1) by column using apply() function and
skip some specific values such as zeros in the example below to tmp2 but not
tmp3. I used the script below and only can get a matrix tmp3. Could you please
help me to studentize the matrix (tmp1) without changing the zeros and generate
a new matrix tmp2?
Thanks,
Joshua
tmp1
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 15 7 9 49 60 3 2 0
[2,] 13 7 10 56 21 4 3 0
[3,] 10 2 12 50 19 1 0 2
[4,] 15 5 30 52 120 2 4 0
[5,] 12 9 71 76 211 4 3 1
[6,] 11 12 2 69 18 5 2 0
tmp2
[1,] 1.1296201 0.0000000 -0.5207910 -0.8630035 -0.1913482 -0.1132277
-0.9561829 0.0000000
[2,] 0.1613743 0.0000000 -0.4817316 -0.2380699 -0.6944435 0.5661385
0.2390457 0.0000000
[3,] -1.2909944 -1.4680505 -0.4036130 -0.7737273 -0.7202433 -1.4719601
0.0000000 0.7071068
[4,] 1.1296201 -0.5872202 0.2994548 -0.5951748 0.5826446 -0.7925939
1.4342743 0.0000000
[5,] -0.3227486 0.5872202 1.9008870 1.5474546 1.7565336 0.5661385
0.2390457 -0.7071068
[6,] -0.8068715 1.4680505 -0.7942062 0.9225210 -0.7331431 1.2455047
-0.9561829 0.0000000
tmp3
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[,8]
[1,] 1.1296201 0.0000000 -0.5207910 -0.8630035 -0.1913482 -0.1132277
-0.243975 -0.5976143
[2,] 0.1613743 0.0000000 -0.4817316 -0.2380699 -0.6944435 0.5661385
0.487950 -0.5976143
[3,] -1.2909944 -1.4680505 -0.4036130 -0.7737273 -0.7202433 -1.4719601
-1.707825 1.7928429
[4,] 1.1296201 -0.5872202 0.2994548 -0.5951748 0.5826446 -0.7925939
1.219875 -0.5976143
[5,] -0.3227486 0.5872202 1.9008870 1.5474546 1.7565336 0.5661385
0.487950 0.5976143
[6,] -0.8068715 1.4680505 -0.7942062 0.9225210 -0.7331431 1.2455047
-0.243975 -0.5976143
Here is my script:
stud<- function(x){
x<-(x-mean(x))/sd(x)
return (x)
}
tmp3<-apply(tmp1,2,stud)
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______________________________________________
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
