How to count the number of times that values changes?

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Mia Dier
Mia Dier 2021-1-9
评论： Mario Malic 2021-1-9
A=[1 1 1 2 1 3 3 1 1]' I want to get:
B=[NaN 0 0 1 2 3 3 4 4]'
The second and third values of B are 0 because A doesn't change value. Fourth value is 1 because A changes value. Fifth value is 2 because A changes value again. And so on...

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采纳的回答

Stephen23
Stephen23 2021-1-9
编辑：Stephen23 2021-1-9
Ran in:
A = [1;1;1;2;1;3;3;1;1]
A = 9×1
1 1 1 2 1 3 3 1 1
B = [NaN;cumsum(diff(A)~=0)]
B = 9×1
NaN 0 0 1 2 3 3 4 4
●

更多回答（1 个）

Mario Malic
Mario Malic 2021-1-9
编辑：Mario Malic 2021-1-9
Hello,
This should do the trick.
B = [NaN; cumsum(diff(A) ~= 0)]';
  2 个评论
Stephen23
Stephen23 2021-1-9
Ran in:
A = [1 1 1 2 1 3 3 1 1]';
B = [NaN, cumsum(diff(A) ~= 0)]'
Error using horzcat
Dimensions of arrays being concatenated are not consistent.

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