- the 30 highest values,
- their indices,
- the size of the original matrix.
Build a matrix of zeros with the size of the initial matrix according to indices
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Hi,
I have an initial matrix M(466,504) where I extracted the 30 highest values in each row with the command:
[val,idx] = maxk(M,30,2)
I would like to get now a matrix of zeros of same size but with those 30 values at their initial position.
How could I build such a matrix ?
2 个评论
Stephen23
2019-1-8
You would need:
Do you have that information?
采纳的回答
Guillaume
2019-1-8
newM = zeros(size(M));
newM(sub2ind(size(M), repmat((1:size(idx, 1))', 1, size(idx, 2)), idx)) = val
3 个评论
Guillaume
2019-1-9
"It doesn't give the right dimensions"
Could you be a bit clearer about the problem you're having? "the right dimensions" of what? The code I've provided does exactly what you asked: e.g. with the 3 highest values of each row:
>> M = magic(10)
M =
92 99 1 8 15 67 74 51 58 40
98 80 7 14 16 73 55 57 64 41
4 81 88 20 22 54 56 63 70 47
85 87 19 21 3 60 62 69 71 28
86 93 25 2 9 61 68 75 52 34
17 24 76 83 90 42 49 26 33 65
23 5 82 89 91 48 30 32 39 66
79 6 13 95 97 29 31 38 45 72
10 12 94 96 78 35 37 44 46 53
11 18 100 77 84 36 43 50 27 59
>> [val, idx] = maxk(M, 3, 2);
>> newM = zeros(size(M));
newM(sub2ind(size(M), repmat((1:size(idx, 1))', 1, size(idx, 2)), idx)) = val
newM =
92 99 0 0 0 0 74 0 0 0
98 80 0 0 0 73 0 0 0 0
0 81 88 0 0 0 0 0 70 0
85 87 0 0 0 0 0 0 71 0
86 93 0 0 0 0 0 75 0 0
0 0 76 83 90 0 0 0 0 0
0 0 82 89 91 0 0 0 0 0
79 0 0 95 97 0 0 0 0 0
0 0 94 96 78 0 0 0 0 0
0 0 100 77 84 0 0 0 0 0
As you can see you get the highest 3 values of each row in exactly the same position as they were originally, in a matrix the same size as the original one.
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