Distribution of binary values
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Hi I have a random binary data set [1 1 1 0 0 1 1 1 1 1 0 0 0 0 1 1 0 1 0 1 1 0 0 0 0 1 1 0 1], I want to count how many duration of 1 and 0 are there. Answer will be [3 5 2 1 2 2 1] and [2 4 1 1 4 1] for 1 and 0 respectively..
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David Fletcher
2018-3-29
编辑：David Fletcher
2018-3-29
test=[1 1 1 0 0 1 1 1 1 1 0 0 0 0 1 1 0 1 0 1 1 0 0 0 0 1 1 0 1]
out=regexp(char(test+48),'[0]+|[1]+','match')
start=test(1)
sz=length(out)
if logical(start)
type1=cellfun(@length,out(1:2:sz))
type0=cellfun(@length,out(2:2:sz))
else
type0=cellfun(@length,out(1:2:sz))
type1=cellfun(@length,out(2:2:sz))
end
type1 =
3 5 2 1 2 2 1
type0 =
2 4 1 1 4 1
Not bothered to add any error handling, but I'm sure you can amend it according to your needs
4 个评论
Amitrajit Mukherjee
2018-3-29
David Fletcher
2018-3-29
It converts the numeric array to an array of chars for the purpose of using regexp. You could use mat2str instead if you wanted.
Amitrajit Mukherjee
2018-4-2
Amitrajit Mukherjee
2018-4-2
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2018-3-29
2018-4-2
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