How to "smear" a logical mask without looping

I would like to "smear" a logical mask - fast. There may be a proper term and even standard operation for this but I haven't been able to find them. The following code and image help describe the requirement:

r=20;
c=25;
a=false(r,c);
a(10,3)=true;
a(4,15)=true;
a(18,18)=true;
a2=a;
n=5; %length to "smear"
for j=1:c
  i=1;
  while i<=r  
    if a2(i,j);
        a2(i:(i+n),j)=true;
        i=i+n;
    end       
  i=i+1;
  end
end
a2=a2(1:r,:);

figure(1)
colormap(flipud(gray))
subplot(1,2,1)
imagesc(a)
title('Input')

subplot(1,2,2)
imagesc(a2)
title('Desired Output')

This is easy in a loop but very slow for large arrays. I've managed a few approaches without loops, some are faster but still messy and I'm sure there is a better way! Hence posting it here for the Gurus :)

Answer 1

[Greg Dionne]

If you have a recent copy (R2016a) try:

a2 = movmax(a, [n 0]);

[Guillaume]

Ooh yes, nice!. The minimum required version is R2016a, not R2017a.

[Sonomatic Australia]

Wow how good is that!? Time to update my Matlab then!

[Greg Dionne]

Thanks Guillaume,

I've updated the answer accordingly.

-G

Answer 2

[Stephen23]

I have no idea how fast this is, but it is relatively compact:

>> idx = cumsum(cumsum(a,1),1);
>> out = 0<idx & idx<=n
out =
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   1   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   1   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   1   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   1   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   1   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   1   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   1   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   1   0   0   0   0   0   0   0

Note that this method will not work if there are more than one non-zero value in a column. It could be adapted for that situation though.

[Sonomatic Australia]

Thank you Stephen, that's pretty neat! I do need to handle columns with more than one non-zero value. Will keep trying!

Answer 3

[Guillaume]

This would work regardless of the numbers of non-zero values in each column. The loop is only over the length of the smear, so should be fairly fast:

a2 = a;
smearlength = 5;
for s = 1 : smearlength
   a2 = a2 | [zeros(s, size(a, 2)); a(1:end-s, :)];
end

[Sonomatic Australia]

Nice one Guillaume, thank you for that! I'll have to run some comparisons with larger arrays and post results.

Answer 4

[Guillaume]

And here is a one liner that also works regardless of the numbers of ones in each column:

%a: logical matrix
%n: number of 1s to add to each smear

a2 = any(a(permute(toeplitz(1:size(a, 1), ones(1, n+1)), [1 3 2]) + (0:size(a, 1):numel(a)-1)), 3);

Requires R2016b or later (otherwise use bsxfun for the +) and is probably not faster than my loop answer.

edit: actually, it is faster than the loop on my machine. But not as fast as Stephen's answer.

[Sonomatic Australia]

Thank you Guillaume, I need some time to understand that one! For interest, I setup the following and did a quick test:

r=50000;
c=10000;
a=false(r,c);
ntrue=round(r*c*0.3);
b=round(rand(20,1)*numel(a));
a(b)=true;
n=30; %length to "smear"

Your method which loops for "n" took 183.267894 s. The method in the original question took 6.552720 s which surprised me.