Hi,
I'm solve a problem where I want to select every 3rd row from a matrix. I do
this for all rows in a loop. This gives me "i" matrices with different
number of rows
I want to stack these matrices in an array and fill the blanks with zero.
Does anyone have any suggestions on how to go about this?
So i want to go from
[,1] [,2] [,3]
[1,] 1 1 1
[2,] 2 2 2
[3,] 3 3 3
[4,] 4 4 4
[5,] 5 5 5
[6,] 6 6 6
[7,] 7 7 7
[8,] 8 8 8
[9,] 9 9 9
to
, , 1
[,1] [,2] [,3]
[1,] 1 1 1
[2,] 4 4 4
[3,] 7 7 7
, , 2
[,1] [,2] [,3]
[1,] 2 2 2
[2,] 5 5 5
[3,] 8 8 8
, , 3
[,1] [,2] [,3]
[1,] 3 3 3
[2,] 6 6 6
[3,] 9 9 9
, , 4
[,1] [,2] [,3]
[1,] 4 4 4
[2,] 7 7 7
[3,] 0 0 0
...
, , 9
[,1] [,2] [,3]
[1,] 9 9 9
[2,] 0 0 0
[3,] 0 0 0
Here's my go at it
z <- array(c(1:9), c(9,3))
z1 <- array(0, c(3,3,9))
for(i in 1:9){
if(nrow(z[seq(i,9,3),]) == 2)
z1[,,i] <- rbind(z[seq(i,9,3),],c(0,0,0))
else
z1[,,i] <- z[seq(i,9,3),]
}
print(z1)
I get the following error: Error in if (nrow(z[seq(i, 9, 3), ]) == 2) z1[, ,
i] <- rbind(z[seq(i, :
argument is of length zero
This is because nrow(z[seq(i, 9, 3) becomes NULL.
Furthermore, this doesn't take care of the case where nrow(z[seq(i,9,3),])
==1
Sorry for the long email and thank you in advance for reading it ;)
Kristian
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