Detecting length and number of occurrences in a logical array

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Jason MacNaughton
Jason MacNaughton 2019-11-25
评论： Image Analyst 2022-6-8
EX:
array1 = [0 0 0 0 0 1 1 1 1 0 0 1 1 0 0 0 0 1 1 1 1 1 1 0 0 0 1]
I want to extract the number of times of consecutive 1's in a separate array eg: array2 = [4 2 6 1], where the length of array 2 is equal to the amount of groups of consecutive 1s and the values is the length of the chain of 1s. I also want to ensure that at 1's at the beginning and end of the arrays are captured.

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采纳的回答

the cyclist
the cyclist 2019-11-25
编辑：the cyclist 2019-11-25
Download Jan's RunLength utility from the File Exchange. It will do exactly what you want.
array1 = [0 0 0 0 0 1 1 1 1 0 0 1 1 0 0 0 0 1 1 1 1 1 1 0 0 0 1];
[b n] = RunLength(array1);
array2 = n(b==1)
array2 =
4 2 6 1

更多回答（2 个）

Image Analyst
Image Analyst 2019-11-25
Jason, if you want a simple way to do it using built-in Mathworks functions and without using some third party File Exchange submission, and if you have the Image Processing Toolbox, you can simply use regionprops(). Here's an example:
array1 = [0 0 0 0 0 1 1 1 1 0 0 1 1 0 0 0 0 1 1 1 1 1 1 0 0 0 1]
props = regionprops(logical(array1), 'Area')
allLengths = [props.Area]
regionprops() labels each contiguous run of 1's and then measures each run and puts the results into a structure array.
props =
4×1 struct array with fields:
Area
Using the bracket trick in the third line concatenates all Area fields from the structure into one single vector.
allLengths =
4 2 6 1
Andrei Bobrov
Andrei Bobrov 2019-11-25
编辑：Andrei Bobrov 2019-11-26
Without Toolboxes and Fileexchanges
a = accumarray(cumsum([0;diff(array1(:))] == 1).*array1(:)+1,1);
out = a(2:end);
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Amanda Beatty
Amanda Beatty 2022-6-8
@Image Analyst I don't have the image processing toolbox so I couldn't use regionprops. My array was array1 = [1 1 1 1 0].
%original
a = accumarray(cumsum([0;diff(array1(:))] == 1).*array1(:)+1,1);
out = a(2:end)
out =
1×0 empty double row vector
%my edit
a = accumarray(cumsum(diff([0;array1(:)]) == 1).*array1(:)+1,1);
out = a(2:end)
out =
4
I don't know, maybe I just had a typo or something that haven't noticed yet, but in the end it did the job, so that's what matters :)

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