use of find command for 3d array
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How to use find command for a 3d array
eg. A = rand(5,5,3)
I want to get i,j,k location of the array A for value [0.5 ,0.2,0.3]
psuedo code
[i,j,k] = find(A(:,:,1) == .5 & A(:,:,2)== 0.2 & A(:,:,3) ==0.3)
but this is not working any help is highly appreciated
采纳的回答
For arrays with more than two dimensions you have to get the linear index (1 output version of find) and then transform that into ND indices using ind2sub:
idx = find(yourlogicalarray);
[row, col, page] = ind2sub(size(yourlogicalarray), idx);
However, in your example the array you're passing to find is a 2D array, not 3D. So, in your case:
[row, col] = find(A(:,:,1) == .5 & A(:,:,2)== 0.2 & A(:,:,3) ==0.3);
is all that is required. And to get the 3 values that match these coordinates:
A(row, col, :)
2 个评论
+1. A marginal note: Some years ago I asked the support if expressions like ".5" and "5." are guaranteed to work and if this is documented. The answer was: "We don't know. Everything, which can be created by sprintf, is supported. Stay at this." All interpreters I know accept ".5" so I think this is just a question of taste - but not documented. :-)
Guillaume
2017-3-15
To be honest, I just copy/pasted the original code for that bit. Didn't even notice the .5 (and the inconsistent use of blank spaces).
That the answer is "We don't know" is alarming! I would expect the grammar of the parser to be well defined, not left to random chance. As far as I know, both are perfectly fine.
更多回答（1 个）
Adam
2017-3-15
doc ind2sub
Use the standard form of find that returns linear indices and convert them using ind2sub to 3d.
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