How to "smear" a logical mask without looping
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Sonomatic Australia
el 21 de Mzo. de 2017
Comentada: Greg Dionne
el 22 de Mzo. de 2017
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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Respuesta aceptada
Greg Dionne
el 22 de Mzo. de 2017
Editada: Greg Dionne
el 22 de Mzo. de 2017
If you have a recent copy (R2016a) try:
a2 = movmax(a, [n 0]);
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Más respuestas (3)
Stephen23
el 21 de Mzo. de 2017
Editada: Stephen23
el 21 de Mzo. de 2017
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.
Guillaume
el 21 de Mzo. de 2017
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
Guillaume
el 22 de Mzo. de 2017
Editada: Guillaume
el 22 de Mzo. de 2017
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.
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