how to reduce the size of array as small as the smallest array to have them in one matrix

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arash rad
arash rad 2023-2-26
评论： Image Analyst 2023-2-27
Hello everyone
I have three arrays and size of each is x 1*104 , y is 1*100 and z is 1*95 and I Have them in a matrix like : T = [x ; y ;z]
How I reduce the size of y and x and make them as large as z to not have inconsistent error
Thanks in advance
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Image Analyst
Image Analyst 2023-2-26
What is in the extra 9 elements of the x row vector? What is in the extra 5 elements of the y row vector? Zeros? Nans? How are they all aligned? What criteria do you want to use to throw out 10 values from x and 5 values from y? What does "inconsistent" mean to you? Give an example with shorter arrays to explain the logic you want to use.
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the cyclist
the cyclist 2023-2-26
Considering a specific, smaller version of your problem, suppose your inputs are
x = [2 3 5 7 11]; % length 5
y = [13 17 19] % length 3
z = [23 29]; % length 2
What would you want the output to be?

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回答（1 个）

Jan
Jan 2023-2-26
编辑：Jan 2023-2-26
Ran in:
There are several possibilities:
  • Fill the shorter arrays with zeros or NaNs on the top, bottom or both.
  • Crop the longer arrays at the start or end.
  • Interpolate two vectors to have the same size as the 3rd one.
  • Interpolate all vectors to a greater or smaller number of elements.
With the shorter example of the cyclist:
x = [2 3 5 7 11]; % length 5
y = [13 17 19]; % length 3
z = [23 29]; % length 2
a = zeros(3, 5); % Or nan(3, 5)
a(1, :) = x;
a(2, 1:numel(y)) = y;
a(3, 1:numel(z)) = z
a = 3×5
2 3 5 7 11 13 17 19 0 0 23 29 0 0 0
nz = numel(z);
b1 = [x(1:nz); ...
y(1:nz);
z]
b1 = 3×2
2 3 13 17 23 29
b2 = [x(numel(x) - nz + 1:numel(x)); ...
y(numel(y) - nz + 1:numel(y)); ...
z]
b2 = 3×2
7 11 17 19 23 29
c1 = [x; ...
interp1(1:numel(y), y, linspace(1, numel(y), numel(x))); ...
interp1(1:numel(z), z, linspace(1, numel(z), numel(x)))]
c1 = 3×5
2.0000 3.0000 5.0000 7.0000 11.0000 13.0000 15.0000 17.0000 18.0000 19.0000 23.0000 24.5000 26.0000 27.5000 29.0000
c2 = [interp1(1:numel(x), x, linspace(1, numel(x), nz)); ...
interp1(1:numel(y), y, linspace(1, numel(y), nz)); ...
z]
c2 = 3×2
2 11 13 19 23 29
c3 = [interp1(1:numel(x), x, linspace(1, numel(x), 10)); ...
interp1(1:numel(y), y, linspace(1, numel(y), 10)); ...
interp1(1:numel(z), z, linspace(1, numel(z), 10))]
c3 = 3×10
2.0000 2.4444 2.8889 3.6667 4.5556 5.4444 6.3333 7.4444 9.2222 11.0000 13.0000 13.8889 14.7778 15.6667 16.5556 17.2222 17.6667 18.1111 18.5556 19.0000 23.0000 23.6667 24.3333 25.0000 25.6667 26.3333 27.0000 27.6667 28.3333 29.0000
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arash rad
arash rad 2023-2-26
Thank you very much for your answers
I find a better way
L = min([length(X), length(Y), length(Z)]);
T = [x(:,1:L) , y(:,1:L), z(:,1:L)]
Image Analyst
Image Analyst 2023-2-27
@arash rad OK, so you just wanted to crop off any part of the vectors that are beyond the length of Z. It would have eliminated a lot of confusion if you had just explained that in the very initial post.

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