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

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]

Answer 1

[Image Analyst]

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

[Image Analyst]

Abbas, are you still there?

[Abbas Husain]

YOU ARE MY HERO <3

[Abbas Husain]

Thank you so much, one more question:

Is there a function that can take [ 0 0 0 0 1 1 1 1 0 1 1 1 0 0 0] say and return [ 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0], i.e. bridge the gap in the sequence of ones?

[Abbas Husain]

found brmorph 'bridge' to do a good job of this, for the example:

[ 0 0 0 0 1 1 1 1 0 1 1 1 0 0 0] ------> [ 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0]

but is there a way to specify say if I wanted this to work when there were more 0s?

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

[Image Analyst]

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]

Brilliant, worked a charm, thank you!!!

Answer 2

[Jan]

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];

[Abbas Husain]

Thanks Jan :)