Cutting a certain pattern from an array

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beatlewalrus
beatlewalrus el 23 de Sept. de 2015
Comentada: Stephen23 el 24 de Sept. de 2015
Hello everyone!
I have a simple question regarding deleting certain pattern from array. For example I have a random array
A=[1 2 3 4 5 6 0 0 0 0 1 2 3 4 5 6 7 8 9 10 0 0 0 0 ...]
I want to delete all fragments, built of four zeroes. So cut away all B=[0 0 0 0] from a 1d array.
How can I implement it?

Respuesta aceptada

C.J. Harris
C.J. Harris el 23 de Sept. de 2015
A = [1 2 3 4 5 6 0 0 0 0 1 2 3 4 5 6 7 8 9 10 0 0 0 0];
B = [0 0 0 0];
C = A(setxor(cell2mat(arrayfun(@(x)(x:x+length(B)-1),strfind(A,B), 'UniformOutput', false)), 1:length(A)));
  2 comentarios
Stephen23
Stephen23 el 23 de Sept. de 2015
Editada: Stephen23 el 23 de Sept. de 2015
See my answer for a much simpler and 100x faster solution.
beatlewalrus
beatlewalrus el 24 de Sept. de 2015
Thank you, but it turned out to be not applicable in my case. I had to be more precise: i'm using MathScript node in LabView. So for example C=setxor(A,B) works ok in MatLab but doesn't work in LabVIEW

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Más respuestas (1)

Stephen23
Stephen23 el 23 de Sept. de 2015
Editada: Stephen23 el 23 de Sept. de 2015
>> A = [1,2,3,4,5,6,0,0,0,0,1,2,3,4,5,6,7,8,9,10,0,0,0,0];
>> B = [0,0,0,0];
>> +strrep(char(A),char(B),'')
ans =
1 2 3 4 5 6 1 2 3 4 5 6 7 8 9 10
  2 comentarios
Walter Roberson
Walter Roberson el 24 de Sept. de 2015
Hee, cute.
Works fine for non-negative integers up to 65535 but not in general.
Stephen23
Stephen23 el 24 de Sept. de 2015
True. It is also more than 100x faster than the accepted solution :)

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