On 12-07-31 02:38 PM, Claudia Penaloza wrote:
Thank you very much Rui (and Eloi for your input)... this is (the very
simplified version of) what I will end up using:
Could we get the extended version ? Because right now, I don't know why
you need such complicated code that can be done in 1 line.
J <- 10
N <- 10
y <- matrix(0,N,J)
cols <- rep(c(TRUE, TRUE, FALSE, FALSE), 3)[seq_len(J)]
for(j in which(cols)){
for (q in 1:N){
if(j %% 2){
y[q,j]=1
}else{y[q,j]=2}
}
}
for(j in which(!cols)){
for (q in 1:N){
if(j %% 2){
y[q,j]="A"
}else{y[q,j]="B"}
}
}
There is no need for a double loop (on N) :
J <- 10
N <- 10
y <- matrix(0,N,J)
cols <- rep(c(TRUE, TRUE, FALSE, FALSE), 3)[seq_len(J)]
for(j in which(cols)){
if(j %% 2){
y[,j]=1
}else{y[,j]=2}
}
for(j in which(!cols)){
if(j %% 2){
y[,j]="A"
}else{y[,j]="B"}
}
if you really wants to use this code.
Cheers,
Eloi
Cheers,
Claudia
On Mon, Jul 30, 2012 at 5:38 PM, Mercier Eloi <emercier at chibi.ubc.ca
<mailto:emercier at chibi.ubc.ca>> wrote:
Or, assuming you only have 4 different elements :
mat<- matrix(rep(c(1,2,"A", "B"),each=10),10,10, byrow=F)
mat2 <- as.data.frame(mat)
mat
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] "1" "2" "A" "B" "1" "2" "A" "B" "1" "2"
[2,] "1" "2" "A" "B" "1" "2" "A" "B" "1" "2"
[3,] "1" "2" "A" "B" "1" "2" "A" "B" "1" "2"
[4,] "1" "2" "A" "B" "1" "2" "A" "B" "1" "2"
[5,] "1" "2" "A" "B" "1" "2" "A" "B" "1" "2"
[6,] "1" "2" "A" "B" "1" "2" "A" "B" "1" "2"
[7,] "1" "2" "A" "B" "1" "2" "A" "B" "1" "2"
[8,] "1" "2" "A" "B" "1" "2" "A" "B" "1" "2"
[9,] "1" "2" "A" "B" "1" "2" "A" "B" "1" "2"
[10,] "1" "2" "A" "B" "1" "2" "A" "B" "1" "2"
mat2
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
1 1 2 A B 1 2 A B 1 2
2 1 2 A B 1 2 A B 1 2
3 1 2 A B 1 2 A B 1 2
4 1 2 A B 1 2 A B 1 2
5 1 2 A B 1 2 A B 1 2
6 1 2 A B 1 2 A B 1 2
7 1 2 A B 1 2 A B 1 2
8 1 2 A B 1 2 A B 1 2
9 1 2 A B 1 2 A B 1 2
10 1 2 A B 1 2 A B 1 2
Cheers,
Eloi
On 12-07-30 04:28 PM, Rui Barradas wrote:
Hello,
Maybe something along the lines of
J <- 10
cols <- rep(c(TRUE, TRUE, FALSE, FALSE), 3)[seq_len(J)]
for(i in which(cols)) { do something }
for(i in which(!cols)) { do something else }
Hope this helps,
Rui Barradas
Em 31-07-2012 00 <tel:31-07-2012%2000>:18, Claudia Penaloza
escreveu:
Dear All,
I would like to apply two different "for loops" to each
set of four columns
of a matrix (the loops here are simplifications of the
actual loops I will
be running which involve multiple if/else statements).
I don't know how to "alternate" between the loops
depending on which column
is "running through the loop" at the time.
## Set up matrix
J <- 10
N <- 10
y <- matrix(0,N,J)
## Apply this loop to the first two of every four columns
([,1:2],
[,5:6],[,9:10], etc.)
for (q in 1:N){
for(j in 1:J){
if(j %% 2){
y[q,j]=1
}else{y[q,j]=2}
}
}
## Apply this loop to the next two of every four columns
([,3:4],[,7:8],[,11:12], etc.)
for (q in 1:N){
for(j in 1:J){
if(j %% 2){
y[q,j]="A"
}else{y[q,j]="B"}
}
}
I want to get something like this (preferably without the
quotes):
y
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] "1" "2" "A" "B" "1" "2" "A" "B" "1" "2"
[2,] "1" "2" "A" "B" "1" "2" "A" "B" "1" "2"
[3,] "1" "2" "A" "B" "1" "2" "A" "B" "1" "2"
[4,] "1" "2" "A" "B" "1" "2" "A" "B" "1" "2"
[5,] "1" "2" "A" "B" "1" "2" "A" "B" "1" "2"
[6,] "1" "2" "A" "B" "1" "2" "A" "B" "1" "2"
[7,] "1" "2" "A" "B" "1" "2" "A" "B" "1" "2"
[8,] "1" "2" "A" "B" "1" "2" "A" "B" "1" "2"
[9,] "1" "2" "A" "B" "1" "2" "A" "B" "1" "2"
[10,] "1" "2" "A" "B" "1" "2" "A" "B" "1" "2"
Any help greatly appreciated!
Claudia
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