Subscripted assignment dimension mismatch. error in for loop
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for i = 1 : (length(locs_min)*2) + 1;
output2(:,i+1) = output(locs_min(i):locs_max(i));
i = i + 1;
n = n + 1;
end
gives error:
|Subscripted assignment dimension mismatch.
Error in Ralphuh (line 74) output2(:,i+1) = output(locs_min(i):locs_max(i));|
output2 is a new variable, output is a 53957x1 double array, and locs_min and locs_max are both 1x538 double arrays.
I have tried transposing the different matrices, changing around the format of my indices, etc...nothing seems to work, please help, thanks!
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Respuestas (3)
dpb
el 2 de Nov. de 2015
Unless the difference between the indices is the same for each iteration, you'll "go boom" the first time the length is different than the preceding.
Use a cell array in this case instead...Matlab's only support for "jagged" arrays.
2 comentarios
dpb
el 3 de Nov. de 2015
Precisely what you see as the error and go on to explain is similar to what I presumed; the lengths of the various subsections aren't the same from one step to the next. If, say the first is min-max -->[1 20] the array will have been allocated by the first assignment as 1,20. Now if the next two locations are [21 40], you're fine but as soon as one isn't precisely 20 elements, then the assignment to that row will be off in the second dimension.
for i=1:length(locs_min)
output2{i} = output(locs_min(i):locs_max(i));
end
output will now be a cell array of the number of elements in the two location arrays. The {} curlies in the indexing instead of () normal parens make it such.
Star Strider
el 2 de Nov. de 2015
What do you want to do?
5 comentarios
Star Strider
el 3 de Nov. de 2015
What you want to do will only work if the vectors are at least (or can be truncated to be exactly) 1000 elements long.
Guessing here since I don’t know how your data are organised, but something like this should work:
Vc = {1:1111}; % Original Cell Vector
V = Vc{1}(1:1000); % Original Cell Vector (Truncated)
V = V(:); % Convert To Column Vector
Vr = reshape(V(:), 100, 10); % Reshape To Desired Matrix
drVr = diff(Vr, [], 2); % Differences Along Columns
dcVr = diff(Vr, [], 1); % Differences Along Rows
I’m not quite certain what you want with respect to diff, so I present you with two options.
Ralph
el 3 de Nov. de 2015
1 comentario
Star Strider
el 3 de Nov. de 2015
My pleasure.
Note that the zeros are valid data. It might be best to pad them with NaN instead. However, I would see if the code in my comment to my Question does what you want before your start with padded matrices.
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