find sequences of 1's
in [Matlab]
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From: mb on 14 Apr 2010 04:13 I have a matrix of 1's and 0's. I want to go through the matrix column for column and find sequences of 1's and also find the length of these sequences. An example: Matrix A: [1 0 0 1 1 1 0 0 1 1 0 0 1 0 0 1] I then want an array like this: B = [2 1 0 1], so for every time a sequence of length n is found I will add 1 at position n in B. Anyone who know how I can solve this?
From: us on 14 Apr 2010 04:37 "mb " <marit.berger(a)student.lu.se> wrote in message <hq3teh$qal$1(a)fred.mathworks.com>... > I have a matrix of 1's and 0's. > I want to go through the matrix column for column and find sequences of 1's and also find the length of these sequences. > > An example: > > Matrix A: [1 0 0 1 > 1 1 0 0 > 1 1 0 0 > 1 0 0 1] > > I then want an array like this: B = [2 1 0 1], so for every time a sequence of length n is found I will add 1 at position n in B. > > Anyone who know how I can solve this? i don't see how you get from A to B given your task... otherwise, one of the solutions m=[ 1 0 0 1 1 1 0 0 1 1 0 0 1 0 0 1 ]; n=histc(sum(m),1:4) % n = 0 2 0 1 us
From: mb on 14 Apr 2010 04:51 > > i don't see how you get from A to B given your task... > otherwise, > > one of the solutions > > m=[ > 1 0 0 1 > 1 1 0 0 > 1 1 0 0 > 1 0 0 1 > ]; > n=histc(sum(m),1:4) > % n = 0 2 0 1 > > us I don't want to take the sum of the columns, cause if it is a sequence of zeros between the ones I want to count the ones seperatly. So in matrix M I have 2 "sequences" with 1 ones, 1 sequence with 2 ones and 1 sequence with 4 ones. mb
From: Bruno Luong on 14 Apr 2010 04:59 % One among solutions: [m n] = size(A); B = zeros(m+2,n); B(2:end-1,:) = A; B = reshape(B,1,[]); lgt = findstr(B, [1 0])-findstr(B, [0 1]); B = accumarray(lgt(:),1,[m 1])' % Bruno
From: us on 14 Apr 2010 05:09 "mb " <marit.berger(a)student.lu.se> wrote in message <hq3vlo$snc$1(a)fred.mathworks.com>... > > > > > i don't see how you get from A to B given your task... > > otherwise, > > > > one of the solutions > > > > m=[ > > 1 0 0 1 > > 1 1 0 0 > > 1 1 0 0 > > 1 0 0 1 > > ]; > > n=histc(sum(m),1:4) > > % n = 0 2 0 1 > > > > us > > I don't want to take the sum of the columns, cause if it is a sequence of zeros between the ones I want to count the ones seperatly. So in matrix M I have 2 "sequences" with 1 ones, 1 sequence with 2 ones and 1 sequence with 4 ones. > > mb well... should have been said at the beginning... anyhow, one of the many solutions m=[ 1 0 0 1 1 1 0 0 1 1 0 0 1 0 0 1 ]; r=regexp(char([0,m(:).',0]+'0'),'0','split'); n=histc(cellfun(@numel,r),1:4) % n = 2 1 0 1 us
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