A = [5 6];
EDU>> B = [3;4];
EDU>> C = 3;
EDU>> B(2) = 2;
EDU>> B(:,2) = A'
B =
3 5
2 6
EDU>> B = B - C
B =
0 2
-1 3
EDU>> t = A*B
t =
-6 28
EDU>> t = [t; B(2,:)];
EDU>> [i,j] = find(t>0)
i =
1
2
j =
2
2
I was wondering, why do i and j come out as column arrays, and not row arrays??

 채택된 답변

Geoff Hayes
Geoff Hayes 2014년 8월 2일

0 개 추천

According to the documentation for this function at find,
ind = find(X) locates all nonzero elements of array X, and returns the linear indices of those elements in vector ind. If X is a row vector, then ind is a row vector; otherwise, ind is a column vector. If X contains no nonzero elements or is an empty array, then ind is an empty array.
The text doesn't really say what will happen if the input is a matrix. I tried a couple of examples for square matrices, ones with more rows than columns, ones with more columns that rows. The result each time included two column vectors. So this may just be the default behaviour for when the input is a matrix.
Why were you expecting row vectors/arrays?

추가 답변 (2개)

Image Analyst
Image Analyst 2014년 8월 2일

1 개 추천

Because MATLAB stores arrays in memory in that order, column major order. It's just the way they chose to design the language.
Azzi Abdelmalek
Azzi Abdelmalek 2014년 8월 2일
편집: Azzi Abdelmalek 2014년 8월 2일

0 개 추천

Your question should be like this
t = [-6 28; -1 3]
[i,j] = find(t>0)
i =
1
2
j =
2
2
I was wondering, why do i and j come out as column arrays, and not row arrays??
Because the previous calculations have nothing to do with your question. And i are j are displayed as columns, because Matlab has to display them as columns or rows, it's not important if they are displayed as column or as rows, because the most important thing is to know the position (i,j) of the elements in t that match the conditions.

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질문:

2014년 8월 2일

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2014년 8월 2일

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