Removing zeros from a binary array if length of zeros is lesser than a value

Hello,

I have an array:

input = [0;0;0;0;0;0;0;1;0;1;1;1;0;0;0;0;0;0;1;1;0;0;0;0]

I would like to remove only the 0's between the 1's if the stretch of 0's is lesser than 5, between the 1's. So, for this instance, the output would look something like:

output = [0;0;0;0;0;0;0;1;1;1;1;0;0;0;0;0;0;1;1;0;0;0;0].

Any help would be appreciated.

Thanks!

Answer 1

[KSSV]

input = [0;0;0;0;0;0;0;1;0;1;1;1;0;0;0;0;0;0;1;1;0;0;0;0] ; 

idx = input==1;  % get the locations of 1 
% split zeros and one's into cell 
idr = diff(find([1;diff(idx);1]));
D = mat2cell(input,idr(:),size(input,2));
% get the lengths of each cell 
L = cellfun(@length,D) ; 
L(2:2:end) = NaN ; % make length of 1's NaN as it is not needed 
% get the length of zeros which are less then 4 
id = L<4 ; 
% remove those zeros 
D(id) = [] ; 
% convert to matrix 
output = cell2mat(D) ; 

[Meghana Balasubramanian]

This works perfectly!

What if I were to replace the removed zero values with 1's ?

[KSSV]

Simple...replace

D(id) = []  ; 

with

D(id) = 1 ; 

[Meghana Balasubramanian]

I tried that. However, I get the error, "Conversion to cell from double is not possible."

[Stephen23]

The RHS would need to be a scalar cell array:

D(id) = {1};

[Meghana Balasubramanian]

Stephen,

Some of the cells have zeros of varying lengths. Replacing it with {1} would simply reduce the entire cell content to 1. Should I run a for loop to find and replace the zero's between the 1's using values stored in "L = cellfun(@length,D);" ?

Answer 2

[Raymond MacNeil]

k = [0;0;0;0;0;0;0;1;0;1;1;1;0;0;0;0;0;0;1;1;0;0;0;0];
y = regexp(num2str(k)', '(?<=[0]{5,})0*', 'split');
z = strjoin(cellfun(@(x) replace(x, '', ' '), y, 'UniformOutput', false));
out = str2num(z)';

[Raymond MacNeil]

Could also use regexprep. I realize I added more steps than was necessary. Also, other ways exist that don't require regexp. This just happened to be the first solution that came to mind.