Loop step size issue
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Hoping there is a very simple solution to this but being new to matlab can't quite see where I'm going wrong:
for i = 10:5:100;
sums = conv(S, ones(1, i), 'valid');
Output(i) = nnz(sums>2 & sums<3);
end
where S is a column vector of values The code searches for instances where the sum of a block of data contained in window (i) is either >2 or <3 and counts them.
I was expecting the Output to return 19 values based on the step size of 5. Instead I get 100 values. I only want to return the counts for window sizes equal to i. Any thoughts would be much appreciated.
1 commentaire
Image Analyst
le 22 Jan 2013
That's because conv() moves the window over 1 element at a time. If you want to move in "jumps" of the window size, you need to use blockproc().
Réponses (3)
Walter Roberson
le 22 Jan 2013
Modifié(e) : Walter Roberson
le 22 Jan 2013
ivals = 10:5:100;
ni = length(ivals);
for K = 1 : ni
i = ivals(K);
sums = conv(S, ones(1, i), 'valid');
Output(K) = nnz(sums>2 & sums<3);
end
2 commentaires
Matt J
le 22 Jan 2013
Modifié(e) : Matt J
le 22 Jan 2013
I don't think there's a way to vectorize fully, but if the S(i) are all non-negative, you should be sure to bail out of the loop once any(sums<3)==0. Enlarging the window from that point on will not change the result.
ivals = 10:5:100;
ni = length(ivals);
Output=zeros(ni,1);
flag=false;
for K = 1 : ni
i = ivals(K);
sums = conv(S, ones(1, i), 'valid');
Usums=sums<3;
if ~any(Usums)
break
end
Lsums=sums>2;
Output(K) = sum(Usums&Lsums);
end
Thorsten
le 22 Jan 2013
This runs about 4 times faster on my machine:
cS = cumsum(S);
for K = 1 : ni
sums = cS(ivals(K):end) - [0 cS(1:end-ivals(K))];
Output(K) = nnz(sums>2 & sums<3);
end
2 commentaires
Matt J
le 22 Jan 2013
Modifié(e) : Matt J
le 22 Jan 2013
Not on mine.
S=rand(1,1e6);
ivals = 10:5:100;
ni = length(ivals);
Output=zeros(ni,1);
tic
for K = 1 : ni
i = ivals(K);
sums = conv(S, ones(1, i), 'valid');
Output(K) = nnz(sums>2 & sums<3);
end
toc
%Elapsed time is 0.313358 seconds.
tic;
cS = cumsum(S);
for K = 1 : ni
sums = cS(ivals(K):end) - [0 cS(1:end-ivals(K))];
Output(K) = sum(sums>2 & sums<3);
end
toc;
%Elapsed time is 0.437812 seconds.
Matt J
le 22 Jan 2013
I really do think the CUMSUM approach should be the better one, though. A MEX is probably needed to avoid repeated memory allocation operations in
sums = cS(ivals(K):end) - [0 cS(1:end-ivals(K))];
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