Question

我有重复模式的数据 -

years cf1  years  cf2
2010   45  2010   37
2011   39  2011   29
2012   51  2012   31
2013   25  2013   33 
2014   35  2014   28

我需要在＆＃34; cfX＆＃34;之后对每个其他列进行排序后，数据或数组看起来像这样。变量

years cf1  years cf2
2013  25   2014  28
2014  35   2011  29
2011  39   2012  31
2010  45   2013  33
2012  51   2010  37

非常感谢你的帮助！

Answer 1

对于每组两列，您可以使用sortrows。

for idx=1:2:size(M,2)
    M(:,idx:idx+1)=sortrows(M(:,idx:idx+1),2)
end

Answer 2

矢量化方法

这假定A是输入矩阵。

[m,n] = size(A);  %// size of input matrix
[~,id] = sort(A(:,2:2:end),1); %// sorted IDs

%// Use id to get linear indices of all elements based on asked sorting criteria
%// and index into A for the final output
Aout = A(bsxfun(@plus,reshape(repmat(permute(id,[1 3 2]),1,2),m,n),[0:n-1]*m));

最后一行可以替换为以下代码，这似乎为小数据集提供了边际运行时改进 -

Aout = A(bsxfun(@plus,reshape(repmat(id,2,1),m,n),[0:n-1]*m));

基准

本节对@Daniel's answer和late addition by @Luis中列出的基于循环的方法提出的矢量化方法进行了基准测试。

基准代码

%// Random huge input array
A = rand(10000);

disp('----------------------------------------  With vectorized approach')
tic
[m,n] = size(A);  %// size of input matrix
[~,id] = sort(A(:,2:2:end),1); %// sorted IDs

%// Use id to get linear indices of all elements based on asked sorting criteria 
Aout = A(bsxfun(@plus,reshape(repmat(permute(id,[1 3 2]),1,2),m,n),[0:n-1]*m));
toc
clear Aout m n id

disp('----------------------------------------  With loop based approach')
tic
Aout2 = zeros(size(A));
for idx=1:2:size(A,2)
    Aout2(:,idx:idx+1)=sortrows(A(:,idx:idx+1),2);
end
toc
clear Aout2 idx

disp('----------------------------------------  With Luis Vectorized approach')
tic
[m, n] = size(A);
[~, rows] = sort(A(:,2:2:n)); %// indices to sort columns 2, 4,...
ind = bsxfun(@plus, rows, (0:n/2-1)*2*m); %// convert to linear index
y = NaN(m,n);
y(:,2:2:n) = A(ind+m); %// fill columns 2, 4,... sorted
y(:,1:2:n) = A(ind);  %// fill columns 1, 3,... with the same order
toc

<强>运行时

----------------------------------------  With vectorized approach
Elapsed time is 2.244272 seconds.
----------------------------------------  With loop based approach
Elapsed time is 3.255867 seconds.
----------------------------------------  With Luis Vectorized approach
Elapsed time is 2.800249 seconds.

Answer 3

这是另一种矢量化方法。让x表示您的矩阵。

[m, n] = size(x);
[~, rows] = sort(x(:,2:2:n)); %// indices to sort columns 2, 4,...
ind = bsxfun(@plus, rows, (0:n/2-1)*2*m); %// convert to linear index
y = NaN(m,n); %// you can remove this line if `y` is assured not to exist,
              %// because in that case the next line serves as preallocation
y(:,2:2:n) = x(ind+m); %// fill columns 2, 4,... sorted
y(:,1:2:n) = x(ind);  %// fill columns 1, 3,... with the same order

Answer 4

另一种基于循环的方法，仅使用sort：

disp('----------------------------------------  With other loop based approach')
tic
Aout3 = zeros(size(A));
for i=0:size(A,2)/2-1
  [ord iord]=sort(A(:,2*i+2),'ascend');
  Aout3(:,2*i+1)=A(iord,2*i+1);
  Aout3(:,2*i+2)=ord;
end
toc

将其添加到benchmark，A=rand(5000)我得到：

----------------------------------------  With vectorized approach
Elapsed time is 1.415872 seconds.
----------------------------------------  With loop based approach
Elapsed time is 1.997568 seconds.
----------------------------------------  With Luis Vectorized approach
Elapsed time is 1.560120 seconds.
----------------------------------------  With other loop based approach
Elapsed time is 1.566022 seconds.

Matlab对其他列进行排序

4 个答案:

矢量化方法

基准