Reshaping a Char Array
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Hi,
I am currently working on a Project where I need to order a Char Array like shown below
A = ['12';'12';'12';'12';'12';'12';'12';'12';'34';'34';'34';'34';'34';'34';'34';'34';'56';'56';'56';'56';'56';'56';'56';'56';'78';'78';'78';'78';'78';'78';'78';'78';'AA';'AA';'AA';'AA';'AA';'AA';'AA';'AA';'BB';'BB';'BB';'BB';'BB';'BB';'BB';'BB';'CC';'CC';'CC';'CC';'CC';'CC';'CC';'CC';'DD';'DD';'DD';'DD';'DD';'DD';'DD';'DD'];
I want to re-arrange the char array like:
'1234'
'1234'
'1234'
'1234'
'1234'
'1234'
'1234'
'1234'
'5678'
'5678'
'5678'
'5678'
'5678'
'5678'
'5678'
'5678'
'AABB'
'AABB'
'AABB'
'AABB'
'AABB'
'AABB'
'AABB'
'AABB'
'CCDD'
'CCDD'
'CCDD'
'CCDD'
'CCDD'
'CCDD'
'CCDD'
'CCDD'
What is the best way to do that? I tried using reshape function but it doesn't seem to work for this case?
Thanks in advance.
采纳的回答
Rik
2020-10-16
This should do it:
A = ['12';'12';'12';'12';'12';'12';'12';'12';'34';'34';'34';'34';'34';'34';'34';'34';'56';'56';'56';'56';'56';'56';'56';'56';'78';'78';'78';'78';'78';'78';'78';'78';'AA';'AA';'AA';'AA';'AA';'AA';'AA';'AA';'BB';'BB';'BB';'BB';'BB';'BB';'BB';'BB';'CC';'CC';'CC';'CC';'CC';'CC';'CC';'CC';'DD';'DD';'DD';'DD';'DD';'DD';'DD';'DD'];
B=mat2cell(A,ones(size(A,1),1),2);
B=reshape(B,8,[]);
part1=B(:,1:2:end);part1=part1(:);
part2=B(:,2:2:end);part2=part2(:);
B=cell2mat([part1 part2]);
更多回答（1 个）
Ameer Hamza
2020-10-16
Something like this
A = ['12';'12';'12';'12';'12';'12';'12';'12';'34';'34';'34';'34';'34';'34';'34';'34';'56';'56';'56';'56';'56';'56';'56';'56';'78';'78';'78';'78';'78';'78';'78';'78';'AA';'AA';'AA';'AA';'AA';'AA';'AA';'AA';'BB';'BB';'BB';'BB';'BB';'BB';'BB';'BB';'CC';'CC';'CC';'CC';'CC';'CC';'CC';'CC';'DD';'DD';'DD';'DD';'DD';'DD';'DD';'DD'];
idx = find([1; diff(A(:,1))]);
B = reshape(string(A), diff(idx(1:2)), []);
B = B(:,1:2:end) + B(:,2:2:end);
B = char(B(:));
6 个评论
Rik
2020-10-16
Ah, you must have posted this just before I refreshed the page. Interesting to see a different strategy.
Ameer Hamza
2020-10-16
Interestingly another question today had a somewhat common structure: https://www.mathworks.com/matlabcentral/answers/615958-how-to-create-arrays-from-repeated-matrix-raws#answer_515683?s_tid=prof_contriblnk, so writing this answer just required a bit of modification.
Jack Gallahan
2020-10-16
Rik
2020-10-16
Have you seen my answer below? This answer can also be modified for such cases, but mine would have an obvious parameter to edit.
Ameer Hamza
2020-10-17
It appears to be working in that case too
A = [repmat('12', 256, 1); repmat('34', 256, 1); repmat('56', 256, 1); repmat('78', 256, 1); repmat('AA', 256, 1); repmat('BB', 256, 1); repmat('CC', 256, 1); repmat('DD', 256, 1)];
idx = find([1; diff(A(:,1))]);
B = reshape(string(A), diff(idx(1:2)), []);
B = B(:,1:2:end) + B(:,2:2:end);
B = char(B(:));
The only case it will fail is if the second letter changes and the first letter remain same, for example
A = ['12';'12';'14';'14';'56';'56';'78';'78';'AA';'AA';'BB';'BB';'CC';'CC';'DD';'DD'];
In that case, following modification
idx = find([1; any(diff(A),2)]);
works, which is a more general form of the line in my answer and work for the original input too.
Jack Gallahan
2020-10-17
编辑：Jack Gallahan
2020-10-17
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