# 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 2023 年 2 月 26 日
コメント済み: Image Analyst 2023 年 2 月 27 日
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
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Image Analyst 2023 年 2 月 26 日
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.
If you have any more questions, then attach your data and code to read it in with the paperclip icon after you read this:
the cyclist 2023 年 2 月 26 日
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 件)

Jan 2023 年 2 月 26 日

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 2023 年 2 月 26 日
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 2023 年 2 月 27 日
@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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