how can i go through a matrix line by line?
I have this matrix and would like to have the value of a spinner depending on the value.
x = [1:1:5; 6:1:10]

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Walter Roberson
Walter Roberson 2021년 4월 20일

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x = [1:1:5; 6:1:10; 11:15; 16:20; 21:25; 26:30; 31:35; [36, 0, -1, nan, nan]];
nrow = size(x,1);
for K = 1 : 5
spin = randi(nrow);
randrow = x(spin,:)
end
randrow = 1×5
11 12 13 14 15
randrow = 1×5
26 27 28 29 30
randrow = 1×5
26 27 28 29 30
randrow = 1×5
11 12 13 14 15
randrow = 1×5
31 32 33 34 35

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TheDice
TheDice 2021년 4월 20일
ok maybe i didn't write that correctly. when i enter 6 in my spinner i want the value from the second line in the first column.
if I query the so x(6) I get yes the value from the second row third column
It sounds like you want linear indexing, but row by row
x = [1:1:5; 6:1:10; 11:15; 16:20; 21:25; 26:30; 31:35; [36, 0, -1, nan, nan]];
nent = numel(x);
xt = x.';
for K = 1 : 5
spin = randi(nent);
randval = xt(spin);
[spin, randval]
end
ans = 1×2
1 1
ans = 1×2
37 0
ans = 1×2
19 19
ans = 1×2
35 35
ans = 1×2
30 30
The entries in the array happen to all be equal to their index when operating with that scheme, except for the last few, and you can see in my example that indeed it happened to pull out the stored value (0) at index 37
TheDice
TheDice 2021년 4월 20일
Thank you. Is it possible to fill the matrix from top to bottom not from left to right?

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도움말 센터File Exchange에서 Creating and Concatenating Matrices에 대해 자세히 알아보기

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