# How do I assign 3D variables when third dimension has size one?

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Thomas on 10 Nov 2014
Commented: Thomas on 11 Nov 2014
Is this a bug? If not, how am I supposed to code this so dd1 has shape [2 3 1]?
>> sd1 = reshape(1:12,[4,3,1])
sd1 =
1 5 9
2 6 10
3 7 11
4 8 12
>> sd2 = reshape(1:24,[4,3,2])
sd2(:,:,1) =
1 5 9
2 6 10
3 7 11
4 8 12
sd2(:,:,2) =
13 17 21
14 18 22
15 19 23
16 20 24
>> for k = 1:2, dd1(k,:,:) = sd1(2*k,:,:); end
>> for k = 1:2, dd2(k,:,:) = sd2(2*k,:,:); end
>> size(dd1)
ans =
2 1 3
>> size(dd2)
ans =
2 3 2
Matt J on 10 Nov 2014
I'm assuming dd1 was not pre-allocated prior to your for-loop.

Matt J on 10 Nov 2014
dd1=sd1(2:2:end,:,:)
##### 3 CommentsShow 1 older commentHide 1 older comment
Matt J on 11 Nov 2014
Edited: Matt J on 11 Nov 2014
I doubt it's a bug. If dd1 and dd2 are not pre-defined, then note that your for-loop results are consistent with the "shiftdim rule" that I describe in your other thread.
The bottom line - it's just one more reason why its dangerous to define or modify the size/shape of an array through assignment.
Thomas on 11 Nov 2014
>> it113 = rand(1,1,3)
it113(:,:,1) =
0.9572
it113(:,:,2) =
0.4854
it113(:,:,3) =
0.8003
>> vom113(1,:,:) = it
vom113 =
0.2785 0.5469 0.9575
>> it131 = rand(1,3,1)
it131 =
0.1419 0.4218 0.9157
>> vom131(1,:,:) = it131
vom131(:,:,1) =
0.1419
vom131(:,:,2) =
0.4218
vom131(:,:,3) =
0.9157

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