The size function does not return the right number

%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!

[Stephen23]

After measuring its size your code removes rows of m:

if m(j,2) > 4095
    m(j,:) = [];
..

After that code has run, it is entirely expected that m will have <=1000001 rows. Which apparently it does.

So far everything is behaving exactly as expected, it is unclear what the problem is.

[Xin]

Hi Stephen,

Thank you for the comments.

The problem is when I try to run the codes, this error shows up. I guess the index means len. What would you recommend to do to fix this error? Thank you!

[Stephen23]

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

[Xin]

It works!!!

Thank you so much for your comments!

Answer 1

[Jan]

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".

[Xin]

Thank you for the detailed explanation. Very useful!