The size function does not return the right number
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%trim the calibrated data
m = unique([in_M, code_M],'sorted', 'rows');
[len, notImp] = size(m);
for j = 1:1:len
if m(j,2) > 4095
m(j,:) = [];
else
continue;
end
end
The size of m is 999961 x 2, but the returned value of the number of rows(len) is 1000001
Could you help me with this matter? Thank you in advance!
4 commentaires
Stephen23
le 10 Nov 2022
The problem is that inside the loop you are removing rows and then trying to access rows which no longer exist.
Consider this vector: [1,2,3,4]
we start running a loop over it and remove the 2nd element, so the vector now look like this: [1,3,4]
then the loop keeps running and then we try to remove the 4th element. But does the vector have a 4th element? (hint: no)
Thus the error: you are trying to remove rows which do not exist.
The usual solution is to loop over the rows from bottom to top:
for j = len:-1:1
Réponse acceptée
Jan
le 10 Nov 2022
Modifié(e) : Jan
le 10 Nov 2022
You do not even need a loop:
m = unique([in_M, code_M],'sorted', 'rows');
[len, notImp] = size(m);
m = m(m(j,2) <= 4095, :);
This is even faster. The iterative shrinking of arrays is as expensive as the growing. Example:
x = [];
for k = 1:1e6
x(k) = k * rand;
end
The final array has 8 MB only (8 byte per double). But in each iteration a new array is created an the contents of the old one is copied. This let Matlab allocate sum(1:1e6)*8 bytes : over 4 TB ! Although the memory is released in the next iteration, this is a lot of work. A small change can avoid this - pre-allocation:
x = zeros(1, 1e6); % Pre-allocation
for k = 1:1e6
x(k) = k * rand;
end
Now only the final array of 8 MB is allocated.
Using the logical indexing m(j,2)<=4095 avoid the iterative change of the array also.
Another hint: continue let the loop perform the next iteration. In your case this is useless, because this the obvious next step at all. So simply omit the "else continue".
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