Using find in a 4D matrix
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I have multiple 9 x 96 x 14 x 1356 matrixes. I want to find when one of them meets a certain criteria and then select the same indexes on the other matrixes.
I did this and looked liked it worked...
idx = find(neutral<24.5);
[r, c, v, l] = ind2sub(size(neutral),idx);
With this, I get a r, c, v, l with 4411652 x 1 each, instead of a 4D logical matrix, which was what I was expecting (like what happens with a 2D matrix when using find).
I tried using those indexes to cut my other variables, but I could't do it.
temp_AT = temp(r,c,v,l);
Thanks!
6 个评论
Rik
2019-8-5
Do you really need it to be 4D logical? If not, you can directly use the indices.
Thaís Lobato Sarmento
2019-8-5
d = other4Darray(neutral < 24.5) %assuming both arrays are same size
% d will be a vector
idx = find(neutral<24.5);
% d(i) comes from the index idx(i)
Thaís Lobato Sarmento
2019-8-6
Adam Danz
2019-8-6
When you remove or isolate elements from an array, they no longer have the same shape unless you're indexing all of the data. Here's a simple example.
d = [2, 3, 2;
5, 2, 2];
idx = d == 2; %same shape as d
d(idx) % a vector with 4 elements, not 6!
If you're trying to get rid of data and maintain the shape of the data, you can replace unwanted data with NaNs (or any other value).
d = [2, 3, 2;
5, 2, 2];
idx = d == 2; %same shape as d
d(~idx) = NaN; % same shape as d!
Thaís Lobato Sarmento
2019-8-6
回答（1 个）
Walter Roberson
2019-8-5
neutral<24.5
is the 4d logical matrix. No need for find()
2 个评论
Adam Danz
2019-8-5
+1 (moving my inferior answer here)
idx = find(neutral < 24.5);
logIdx = false(size(neutral)); %Logical index (default: all false)
logIdx(idx) = true; %Set 'idx' indices as true
Thaís Lobato Sarmento
2019-8-6
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