Set logical array values to true if value n values ahead is true

I'm curious if anyone can find a smart/clean way (other than a for loop of doing the following). Given a logical array x:

x = logical( [0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0] );

For n = 1 I'd like: y = [0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0];

For n = 2 I'd like: y = [0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0];

etc.

[Matt Fig]

Your examples conflict with your description. Do you want to place n values in the array ahead of a 1 as your examples show, or do you want to place a single 1 ahead of n 1s, as the title implies? If the latter, do you need to place a 1 only in front of exactly n 1s, or at least n 1s?

[Chris]

Sorry, it is indeed the latter. I want to place n values in the array ahead of a 1. Either exactly or at least are fine b/c x will always have "isolated" 1s.

Answer 1

[Andrei Bobrov]

x = [0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0] > 0;
n = 2;

x(bsxfun(@minus,find(x),(n:-1:1)')) = true;

Answer 2

[Matt J]

 n=2;

 y=regexprep(char(x+'0'),[repmat('.',1,n),'1'],   [repmat('1',1,n),'1'])-'0'

Answer 3

[Matt J]

y=logical(tril(triu(ones(length(x))),n)*x(:)).'