Turning 1s to 0s in a logical vector when element distance between 1s is below threshold

Hello,

I have a logical vector of 1.5 million elements. I need to look through the elements in the vector, and whenever I find a 1, I need to turn any potential subsequent 1s in the next N elements into 0s. While this would be easy to implement with a for loop, I'm struggling to figure out a more efficient way to do it.

Example:

If [ 1 1 0 0 0 0 1 0 1 0 0 0 1 ] is my vector, and my N threshold is 5, I want to turn that vector into [ 1 0 0 0 0 0 1 0 0 0 0 0 1 ].

Thank you in advance.

[Chien-Han Su]

How about using find() to get indices of all trues, get distance from those indices by subtraction and turn true to false for those distance bellow threshold?

[Len Jacob]

This is a good starting point but I'm running into one issue--if I run something like this:

test = [ 1 1 0 0 0 0 1 0 1 0 0 0 1 ];
diff(find(test))

ans =

     1     5     2     4

I would end up removing the last 1, but in practice I actually want to keep it since the 1 before that will be getting removed first. To clarify, since I want to be moving through the vector from left to right (i.e. moving forward in time), the distance between 1s will change as I flip them to 0, and I need to accommodate that.

Answer 1

[Jonas]

you can try the follwing, but i don't know if it is faster

H = [ 1 1 0 0 0 0 1 0 1 0 0 0 1 ];
eraseNAfter=5;
Hstr=num2str(H);
out=regexprep(Hstr,['1' repmat('.',[1 3*eraseNAfter])],['1' repmat('  0',[1 eraseNAfter])]);
asDouble=str2mat(out)

i am also sure the regexp could be written nicer

[Len Jacob]

This was fast enough, thanks!

EDIT: Heads up that this fails to convert 1s at the end of your vector depending on vector size and threshold size. Not an issue for my application but something to keep in mind for others who are reading this and might want to use this implementation for something else.

[Jonas]

that's true, fast solition would be padding the array and removing the padded elements at the end

H = [ 1 1 0 0 0 0 1 0 1 0 0 0 1 1];
eraseNAfter=5;
H=[H repmat(0,[1 eraseNAfter])];
Hstr=num2str(H);
out=regexprep(Hstr,['1' repmat('.',[1 3*eraseNAfter])],['1' repmat(' 0',[1 eraseNAfter])]);
asDouble=str2mat(out);
asDouble((end-eraseNAfter+1):end)=[]