How to order a matrix?

2019 4 23
2 个回答
更新时间：2019 4 23
23 次查看（30 天）

0 个投票

Hello,
I have a matrix,
A=[ 100 5 0 ; 200 0 -5 ; 300 0 0 ; 300 5 0 ; -100 0 5 ; 300 0 0; 300 0 5];
I want to get,
A= [100 5 0; 200 0 -5 ; 300 0 0; 300 5 0 ; 300 0 5 ; -100 0 5];
I am writing code:
B= unique(A(:, 2:3),'rows');
then it's removing the first row as 5 0 is also repeted for the 4th row 300 5 0 but I need to keep the 1st row.
Is it possible to do?

4 个评论
显示 2更早的评论 隐藏 2更早的评论

Adam
Adam 2019-4-23
What is wrong with just using
B = unique( A, 'rows' )
or
B = unique( A, 'rows', 'stable' )
though your chosen output ordering does not match either unique's sorted output nor its stable output which retains the original ordering.
Your version using 2:3 won't do this though either.
You can sort the output to your desired ordering afterwards though.
Sky Scrapper
Sky Scrapper 2019-4-23
My original matrix is 1000x20
what i have given in the question is the 1st four columns.I need to keep the corresponding whole columns. so what I did,
[~,idy,idx] = unique(A(:,2:3),'rows','stable');
B =A(idy,:);
then it's showing 1st row but it's also showing the repeted rows for 300 0 0 and so on.
Could you please clarify where is the mistake?
Adam
Adam 2019-4-23
It's not really obvious from your example what you are trying to achieve in a more general case. Your example just removes one row (which can be removed by using just a simple unique as mentioned) and then orders the remaining ones in a fairly unintuitive manner with respect to how they started.
What are you actually trying to keep? If you are trying to keep all unique rows then what I wrote works fine, if you are really trying to only run a unique on the 2nd and 3rd columns then your version does that, but from what you say, that isn't what you want to do.
Sky Scrapper
Sky Scrapper 2019-4-23
I could solve my problem. thanks for helping me!

请先登录，再进行评论。

请先登录，再回答此问题。

回答（2 个）

Alex Sune
Alex Sune 2019-4-23
编辑：Alex Sune 2019-4-23

0 个投票

Try this:
[~,ia,~] = unique(A(:, 2:3),'rows','stable');
B = A(ia,:);

7 个评论
显示 5更早的评论 隐藏 5更早的评论

Sky Scrapper
Sky Scrapper 2019-4-23
Yes, i have tried but it is not working for my original matrix.
Sky Scrapper
Sky Scrapper 2019-4-23
I think this should work but it's showing the same result!
Sky Scrapper
Sky Scrapper 2019-4-23
If I am doing,
[~,ia,~] = unique(A(:, 2:3),'rows','stable');
B = A(ia,:);
again it's removing 1st column. if i am doing,
[~,ia,~] = unique(A(:, 1:3),'rows','stable');
B = A(ia,:);
then it's showing the repeated column also!
Alex Sune
Alex Sune 2019-4-23
A = [ 100 5 0 ; 200 0 -5 ; 300 0 0 ; 300 5 0 ; -100 0 5 ; 300 0 0; 300 0 5];
[~,ia,~] = unique(A(:, 2:3),'rows','stable');
B = A(ia,:)
B =
100 5 0
200 0 -5
300 0 0
-100 0 5
It keeps the first column! And remove the repeated rows
Sky Scrapper
Sky Scrapper 2019-4-23
Yes, I got the solution of my problem. i need to take the tolerance of my values because i have some fractional values.
Thank you very much for your help.
Alex Sune
Alex Sune 2019-4-23
Using round
https://es.mathworks.com/help/matlab/ref/round.html
If you want to round the first column:
A = [ 100 5 0 ; 200 0 -5 ; 300 0 0 ; 300 5 0 ; -100 0 5 ; 300 0 0; 300 0 5];
A(:,1) = round(A(:,1),-2);
[~,ia,~] = unique(A(:, 2:3),'rows','stable');
B = A(ia,:);
Sky Scrapper
Sky Scrapper 2019-4-23
I have already done this. Again thanks a lot!

请先登录，再进行评论。

Guillaume
Guillaume 2019-4-23

0 个投票

how can i do that?
Use uniquetol instead of unique. Note that the 'rows' option of unique is 'ByRows', true with uniquetol
[~, irows] = uniquetol(A(:, [2 3]), 'ByRows', true); %if you are happy with the default tolerance
B = A(irows, :);

1 个评论
显示 -1更早的评论 隐藏 -1更早的评论

Sky Scrapper
Sky Scrapper 2019-4-23
Yes, i have already done that.
I had to take tolerance=0.02
Thank you very much for your kind support.

请先登录，再进行评论。

类别

在 帮助中心 和 File Exchange 中查找有关 Operating on Diagonal Matrices 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by