Sort rows of a matrix based on a specific array

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Amirhossein Moosavi
Amirhossein Moosavi 2020 年 7 月 13 日
コメント済み: madhan ravi 2020 年 7 月 13 日
Hi everyone,
Let us assum Matrix A and array B as follows:
A = [1 0 1
2 0 0
4 0 0
3 1 1
6 0 0
5 0 0]
B = [3 4 1]
Array B always include values stored in the first column of Matrix A. So, I want to sort rows of Matrix A based on Array B. Whatever comes first in Array B must come first in Matrix A. Moreover, I do not want to change the position of any other rows in Matrix A that their values (first column) do not exist in Array B.
Having said this, sorted version of Matrix A based on Array B will be as follows:
An = [3 1 1
2 0 0
4 0 0
1 0 1
6 0 0
5 0 0]
Would you please guide me what is the fastest way to do this?
Regards,
Amir

採用された回答

Stephen23
Stephen23 2020 年 7 月 13 日
編集済み: Stephen23 2020 年 7 月 13 日
>> [X,Y] = ismember(A(:,1),B);
>> [~,Z] = sort(Y(X));
>> T = A(X,:);
>> A(X,:) = T(Z,:)
A =
3 1 1
2 0 0
4 0 0
1 0 1
6 0 0
5 0 0
  1 件のコメント
Amirhossein Moosavi
Amirhossein Moosavi 2020 年 7 月 13 日
Thanks Stephen, your code works!

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その他の回答 (1 件)

madhan ravi
madhan ravi 2020 年 7 月 13 日
v = sort(B);
A([1, v(end)], :) = A([v(end), 1], :)
  2 件のコメント
Amirhossein Moosavi
Amirhossein Moosavi 2020 年 7 月 13 日
Thank you, but it does not work for some cases. Let us say Array B is as follows:
B = [3 4]
Then, Matrix A becomes like this (using your code):
A = [3 1 1
2 0 0
4 0 0
1 0 1
6 0 0
5 0 0]
While, Matrix A must have become as follows:
A = [1 0 1
2 0 0
3 1 1
4 0 0
6 0 0
5 0 0]
madhan ravi
madhan ravi 2020 年 7 月 13 日
I misunderstood the question.

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