How to "smear" a logical mask without looping

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Sonomatic Australia
评论： Greg Dionne 2017-3-22
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 :)

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采纳的回答

Greg Dionne
Greg Dionne 2017-3-22
编辑：Greg Dionne 2017-3-22
If you have a recent copy (R2016a) try:
a2 = movmax(a, [n 0]);
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Sonomatic Australia
Sonomatic Australia 2017-3-22
Wow how good is that!? Time to update my Matlab then!
Greg Dionne
Greg Dionne 2017-3-22
Thanks Guillaume,
I've updated the answer accordingly.
-G

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更多回答（3 个）

Stephen23
Stephen23 2017-3-21
编辑：Stephen23 2017-3-21
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.
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Sonomatic Australia
Sonomatic Australia 2017-3-21
Thank you Stephen, that's pretty neat! I do need to handle columns with more than one non-zero value. Will keep trying!

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Guillaume
Guillaume 2017-3-21
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
  1 个评论
Sonomatic Australia
Sonomatic Australia 2017-3-22
Nice one Guillaume, thank you for that! I'll have to run some comparisons with larger arrays and post results.

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Guillaume
Guillaume 2017-3-22
编辑：Guillaume 2017-3-22
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.
  1 个评论
Sonomatic Australia
Sonomatic Australia 2017-3-22
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.

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