Setting lone 1s to 0s in a logical vector/matrix

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Abbas Husain 2021-4-2

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评论： Abbas Husain 2021-4-3

If I have this vector for example: [ 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0]

And I want to replace the lone 1 (the sixth element in this case) with a zero, is there an existing algorithm which can do this?

i.e. [ 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0] ----> [ 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0]

For context, this is an example of a row of pixels which has been colour thresholded to return 1s in areas of green and 0s otherwise.

The lone 1 represents an anomalous point, there is no green there and hence I wish to replace it with a 0.

For a further condition, I'd like to do the same for any section of 1s less than 3 in length.

i.e. [0 0 0 0 0 1 1 0 0 0 0 0 1 1 1 1 1 1 0 0 0 0] would go to [0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 0 0 0 0]

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Answer 1

Image Analyst 2021-4-2

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You can use bwareaopen() in the Image Processing Toolbox. You specify the smallest region length you want to keep, like to get rid of single 1's keep runs of 2 or longer

v =  [ 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0]
v2 = bwareaopen(v, 2) % Keep runs of 2 or longer

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Image Analyst 2021-4-3

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It's a little tricky but it can still be done with only bwareaopen(). I've done it in multiple steps so you can follow long each step of the way:

v = [ 0 0 0 0 1 1 1 1 0 0 1 1 0 0 0 1 1 0] % Need to remove one stretch of 2 zeros.
% Allow 3 or more 0's in a row (they remain).  2 or less will get filled.
minAllowableHoleSize = 3;  
% Pad with 1's on each end so we do not remove any zeros on the ends
% no matter how long or short they are because there is no one outside of them.
v2 = [ones(1, minAllowableHoleSize), ~v, ones(1, minAllowableHoleSize)]
v3 = bwareaopen(logical(v2), minAllowableHoleSize)
% Remove padding that we added
v4 = v3(minAllowableHoleSize+1:end-minAllowableHoleSize)
v5 = ~v4

v5 is the same as v except any runs of 2 or fewer 0s are filled in with 1s.

Abbas Husain 2021-4-3

Brilliant, worked a charm, thank you!!!

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Answer 2

Jan 2021-4-2

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https://ww2.mathworks.cn/matlabcentral/answers/790774-setting-lone-1s-to-0s-in-a-logical-vector-matrix#answer_665629

编辑：Jan 2021-4-2

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With FEX: RunLength :

a = [0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0];
[val, n] = RunLength(a);
del      = (val == 1) & (n == 1);  % Or:  n < limit
val(del) = 0;
b        = RunLength(val, n);
% b = [0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0];