Reshape Nx1 struct with field of Mx1 elements to N*Mx1 vector

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I have a struct A of [1000x1] size. There is a field "bar" which has [100x1] elements.
I want to have 1 variable, let's say B, which will be [1000*100x1] of "bar" elements.
How can I do it without iterating a struct and appending on some prepared zeroed vector? Can reshape do it?
  1 Comment
Stephen23
Stephen23 on 22 May 2022
"How can I do it without iterating a struct and appending on some prepared zeroed vector? Can reshape do it?!
No, but a comma-separated list can do it:

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Accepted Answer

Bruno Luong
Bruno Luong on 22 May 2022
Edited: Bruno Luong on 22 May 2022
Something like
B = cat(1,A.bar)

More Answers (1)

MJFcoNaN
MJFcoNaN on 22 May 2022
This is an example:
bar=rand(100,1)
bar = 100×1
0.4600 0.1687 0.4194 0.2287 0.7427 0.7976 0.8690 0.4868 0.9914 0.4464
A=struct;
A.bar=bar;
% copy A.bar to all 1000 elements
A(2:1000,1)=A(1)
A = 1000×1 struct array with fields:
bar
% table is a good intermediate form
tbl=struct2table(A);
B=cell2mat(cellfun(@transpose, tbl.bar, 'UniformOutput', false))
B = 1000×100
0.4600 0.1687 0.4194 0.2287 0.7427 0.7976 0.8690 0.4868 0.9914 0.4464 0.1298 0.0337 0.5265 0.2551 0.7338 0.6966 0.2969 0.3447 0.0830 0.7853 0.9779 0.2714 0.3507 0.4244 0.3516 0.3386 0.8395 0.6153 0.2840 0.6013 0.4600 0.1687 0.4194 0.2287 0.7427 0.7976 0.8690 0.4868 0.9914 0.4464 0.1298 0.0337 0.5265 0.2551 0.7338 0.6966 0.2969 0.3447 0.0830 0.7853 0.9779 0.2714 0.3507 0.4244 0.3516 0.3386 0.8395 0.6153 0.2840 0.6013 0.4600 0.1687 0.4194 0.2287 0.7427 0.7976 0.8690 0.4868 0.9914 0.4464 0.1298 0.0337 0.5265 0.2551 0.7338 0.6966 0.2969 0.3447 0.0830 0.7853 0.9779 0.2714 0.3507 0.4244 0.3516 0.3386 0.8395 0.6153 0.2840 0.6013 0.4600 0.1687 0.4194 0.2287 0.7427 0.7976 0.8690 0.4868 0.9914 0.4464 0.1298 0.0337 0.5265 0.2551 0.7338 0.6966 0.2969 0.3447 0.0830 0.7853 0.9779 0.2714 0.3507 0.4244 0.3516 0.3386 0.8395 0.6153 0.2840 0.6013 0.4600 0.1687 0.4194 0.2287 0.7427 0.7976 0.8690 0.4868 0.9914 0.4464 0.1298 0.0337 0.5265 0.2551 0.7338 0.6966 0.2969 0.3447 0.0830 0.7853 0.9779 0.2714 0.3507 0.4244 0.3516 0.3386 0.8395 0.6153 0.2840 0.6013 0.4600 0.1687 0.4194 0.2287 0.7427 0.7976 0.8690 0.4868 0.9914 0.4464 0.1298 0.0337 0.5265 0.2551 0.7338 0.6966 0.2969 0.3447 0.0830 0.7853 0.9779 0.2714 0.3507 0.4244 0.3516 0.3386 0.8395 0.6153 0.2840 0.6013 0.4600 0.1687 0.4194 0.2287 0.7427 0.7976 0.8690 0.4868 0.9914 0.4464 0.1298 0.0337 0.5265 0.2551 0.7338 0.6966 0.2969 0.3447 0.0830 0.7853 0.9779 0.2714 0.3507 0.4244 0.3516 0.3386 0.8395 0.6153 0.2840 0.6013 0.4600 0.1687 0.4194 0.2287 0.7427 0.7976 0.8690 0.4868 0.9914 0.4464 0.1298 0.0337 0.5265 0.2551 0.7338 0.6966 0.2969 0.3447 0.0830 0.7853 0.9779 0.2714 0.3507 0.4244 0.3516 0.3386 0.8395 0.6153 0.2840 0.6013 0.4600 0.1687 0.4194 0.2287 0.7427 0.7976 0.8690 0.4868 0.9914 0.4464 0.1298 0.0337 0.5265 0.2551 0.7338 0.6966 0.2969 0.3447 0.0830 0.7853 0.9779 0.2714 0.3507 0.4244 0.3516 0.3386 0.8395 0.6153 0.2840 0.6013 0.4600 0.1687 0.4194 0.2287 0.7427 0.7976 0.8690 0.4868 0.9914 0.4464 0.1298 0.0337 0.5265 0.2551 0.7338 0.6966 0.2969 0.3447 0.0830 0.7853 0.9779 0.2714 0.3507 0.4244 0.3516 0.3386 0.8395 0.6153 0.2840 0.6013
  1 Comment
Sabyrzhan Tasbolatov
Sabyrzhan Tasbolatov on 22 May 2022
Thanks, I'm going to use cat as it's faster
Elapsed time is 0.001430 seconds.
I've also tried your option, it takes:
Elapsed time is 0.020119 seconds.

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