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

2 visualizaciones (últimos 30 días)
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!

Respuesta aceptada

KSSV
KSSV el 19 de Feb. de 2020
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) ;
  5 comentarios
Stephen23
Stephen23 el 19 de Feb. de 2020
The RHS would need to be a scalar cell array:
D(id) = {1};
Meghana Balasubramanian
Meghana Balasubramanian el 19 de Feb. de 2020
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);" ?

Iniciar sesión para comentar.

Más respuestas (1)

Raymond MacNeil
Raymond MacNeil el 19 de Feb. de 2020
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)';
  1 comentario
Raymond MacNeil
Raymond MacNeil el 19 de Feb. de 2020
Editada: Raymond MacNeil el 19 de Feb. de 2020
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.

Iniciar sesión para comentar.

Categorías

Más información sobre Logical en Help Center y File Exchange.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by