Distribution of binary values

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..

Answer 1

[David Fletcher]

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

[Amitrajit Mukherjee]

what is test+48 ??

[David Fletcher]

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]

Can you please tell me what is the use of "char(test+48)" ?

[Amitrajit Mukherjee]

Thank you so much..It works... :)