how to reduce the size of array as small as the smallest array to have them in one matrix
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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
3 个评论
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.
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
2023-2-26
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?
回答(1 个)
Jan
2023-2-26
编辑:Jan
2023-2-26
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
nz = numel(z);
b1 = [x(1:nz); ...
y(1:nz);
z]
b2 = [x(numel(x) - nz + 1:numel(x)); ...
y(numel(y) - nz + 1:numel(y)); ...
z]
c1 = [x; ...
interp1(1:numel(y), y, linspace(1, numel(y), numel(x))); ...
interp1(1:numel(z), z, linspace(1, numel(z), numel(x)))]
c2 = [interp1(1:numel(x), x, linspace(1, numel(x), nz)); ...
interp1(1:numel(y), y, linspace(1, numel(y), nz)); ...
z]
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))]
4 个评论
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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