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

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Len Jacob
Len Jacob on 3 Aug 2021
Commented: Jonas on 4 Aug 2021
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.
  2 Comments
Len Jacob
Len Jacob on 3 Aug 2021
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.

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Accepted Answer

Jonas
Jonas on 4 Aug 2021

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

  2 Comments
Jonas
Jonas on 4 Aug 2021

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)=[]

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