升序排序，但最后保留零 [英] Sort in ascending order, but keep zeros at last

问题描述

假设我有以下形式的矩阵A.

Suppose I have a matrix A, on the following form.

A =
    35     1     6
     3    32     0
     0     9     0
     0     0     0

我想按升序对其进行排序，但最后保留零.

I want to sort it in ascending order, but keep the zeros at last.

我知道我可以用inf替换所有零，对其进行排序，然后再次将inf替换为零，如

I know I can subsitute all zeros with inf, sort it, and replace the infs with zeros again, as proposed in this question.

我认为有一种更简单的方法.至少因为我的零已经在最下面的行中了.我可以单行吗?

I would think there was a simpler way. At least since my zeros are already in the bottom rows. Can I do this in a single line?

我想要什么:

A =
     3     1     6
     35    9     0
     0     32    0
     0     0     0

谢谢！

关于Eitan的答案的开销存在一个问题.结果如下(平均，以及预热后):

There was a question regarding the overhead of Eitan's answer. Here are the results (averaged, and after warm up):

B =  kron(A,ceil(rand(2000)*1000));  % 8000x6000 matrix
C = B;

%% Eitan's solution:
t1 = tic; B(B ~= 0) = nonzeros(sort(B)); toc(t1)
Elapsed time is  1.768782 seconds.

%% From question text:
B = C;
t1 = tic; B(B==0)=Inf; B = sort(B); B(B==Inf)=0; toc(t1) 
Elapsed time is 1.938374 seconds.

%% evading's solution (in the comments):
B = C;
t1 = tic; for i = 1:size(B,2)  index = B(:,i) ~= 0; B(index, i) = sort(B(index, i)); end
toc(t1)
Elapsed time is 1.954454 seconds.

%% Shai's solution (in the comments):
B = C;
t1 = tic; sel = B==0; B(sel)=inf;B=sort(B);B(sel)=0; toc(t1)
Elapsed time is 1.880054 seconds.

推荐答案

如果可以保证零仅位于每一列的底部，则可以执行以下操作:

If you can guarantee that the zeros are only at the bottom of each column, you can do:

A(A ~= 0) = nonzeros(sort(A));

这篇关于升序排序，但最后保留零的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！