Matlab：以独特的方式对矩阵进行排序

Question

I have a problem with sorting some finance data based on firmnumbers. So given is a matrix that looks like:

[1 3 4 7;
1 2 7 8;
2 3 7 8;]

On Matlab i would like the matrix to be sorted as follows:

[1 0 3 4 7 0;
1 2 0 0 7 8;
0 2 3 0 7 8;]

So basically every column needs to consist of 1 type of number.

I have tried many things but i cant get the matrix sorted properly.

Answer 1

A = [1 3 4 7;
     1 2 7 8;
     2 3 7 8;]

%// Get a unique list of numbers in the order that you want them to appear as the new columns
U = unique(A(:))'

%'//For each column (of your output, same as columns of U), find which rows have that number. Do this by making A 3D so that bsxfun compares each element with each element
temp1 = bsxfun(@eq,permute(A,[1,3,2]),U)    

%// Consolidate this into a boolean matrix with the right dimensions and 1 where you'll have a number in your final answer
temp2 = any(temp1,3)

%// Finally multiply each line with U
bsxfun(@times, temp2, U)

So you can do that all in one line but I broke it up to make it easier to understand. I suggest you run each line and look at the output to see how it works. It might seem complicated but it's worthwhile getting to understand bsxfun as it's a really useful function. The first use which also uses permute is a bit more tricky so I suggest you first make sure you understand that last line and then work backwards.

Answer 2

What you are asking can also be seen as an histogram

 A = [1 3 4 7;
     1 2 7 8;
     2 3 7 8;]
uniquevalues  = unique(A(:))   
N = histc(A,uniquevalues'  ,2)                             %//'               
B = bsxfun(@times,N,uniquevalues')                         %//'
%// bsxfun can replace the following instructions: 
%//(the instructions are equivalent only when each value appears only once per row )
%// B = repmat(uniquevalues', size(A,1),1)                    
%// B(N==0) = 0

Answer 3

Answer without assumptions - Simplified

I did not feel comfortable with my old answer that makes the assumption of everything being an integer and removed the possibility of duplicates, so I came up with a different solution based on @lib's suggestion of using a histogram and counting method.

The only case I can see this not working for is if a 0 is entered. you will end up with a column of all zeros, which one might interpret as all rows initially containing a zero, but that would be incorrect. you could uses nan instead of zeros in that case, but not sure what this data is being put into, and if it that processing would freak out.

EDITED

Includes sorting of secondary matrix, B, along with A.

A = [-1 3 4 7 9; 0 2 2 7 8.2; 2 3 5 9 8];
B = [5 4 3 2 1; 1 2 3 4 5; 10 9 8 7 6];

keys = unique(A);
[counts,bin] = histc(A,transpose(unique(A)),2);
A_sorted = cell(size(A,1),1);
for ii = 1:size(A,1)
    for jj = 1:numel(keys)
        temp = zeros(1,max(counts(:,jj)));
        temp(1:counts(ii,jj)) = keys(jj);
        A_sorted{ii} = [A_sorted{ii},temp];
    end
end

A_sorted = cell2mat(A_sorted);
B_sorted = nan(size(A_sorted));
for ii = 1:size(bin,1)
   for jj = 1:size(bin,2)
       idx = bin(ii,jj);
       while ~isnan(B_sorted(ii,idx))
          idx = idx+1; 
       end
       B_sorted(ii,idx) = B(ii,jj);
   end
end
B_sorted(isnan(B_sorted)) = 0

Answer 4

You can create at the beginning a matrix with 9 columns , and treat the values in your original matrix as column indexes.

A = [1 3 4 7;
1 2 7 8;
2 3 7 8;]

B = zeros(3,max(A(:)))
for i = 1:size(A,1)
    B(i,A(i,:)) = A(i,:)
end
B(:,~any(B,1)) = []