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.
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Ameer Hamza
2020년 10월 16일
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(:));
댓글 수: 6
Rik
2020년 10월 16일
Ah, you must have posted this just before I refreshed the page. Interesting to see a different strategy.
Ameer Hamza
2020년 10월 16일
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
2020년 10월 16일
Rik
2020년 10월 16일
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
2020년 10월 17일
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
2020년 10월 17일
편집: Jack Gallahan
2020년 10월 17일
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