Using find in a 4D matrix

조회 수: 9 (최근 30일)
Thaís Lobato Sarmento
Thaís Lobato Sarmento 2019년 8월 5일
댓글: Thaís Lobato Sarmento 2019년 8월 6일
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!
  댓글 수: 6
Adam Danz
Adam Danz 2019년 8월 6일
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
Thaís Lobato Sarmento 2019년 8월 6일
Thanks, this worked!

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답변 (1개)

Walter Roberson
Walter Roberson 2019년 8월 5일
neutral<24.5
is the 4d logical matrix. No need for find()
  댓글 수: 2
Adam Danz
Adam Danz 2019년 8월 5일
+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
Thaís Lobato Sarmento 2019년 8월 6일
Thank you! This also worked!

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