Sorting from the maximum value to minimum value in a vector

A=[2 5 8 7 8 9 3 7 6 5 4 1];
Asorted = sort(A,'descend'); % sorted from max to min
[~,indx] = unique(Asorted);
equalpairs = [(setdiff(1:length(Asorted),indx)-1)' setdiff(1:length(Asorted),indx)']; % each row contains pair of indices with equal values

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mr mo 2017-11-29

Thanks a lot. This is what I want.

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Answer 2

Matt J 2017-11-29

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/370079-sorting-from-the-maximum-value-to-minimum-value-in-a-vector#answer_293822

Using the UNIQUE command.

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mr mo 2017-11-29

In the unique command I lost some of the members and I don't want that.

mr mo 2017-11-29

在 MATLAB Online 中打开

I want to reach this:

sortedA = [9 8 8 7 7 6 5 5 4 3 2 1]

and I want the indices of equal members in the sorted version of A, e.g.

 sortedA(2)=sortedA(3)
 sortedA(4)=sortedA(5)
 sortedA(7)=sortedA(8)

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Answer 3

Image Analyst 2017-11-30

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/370079-sorting-from-the-maximum-value-to-minimum-value-in-a-vector#answer_293985

在 MATLAB Online 中打开

With the accepted solution, if A is

A =[22    25    28    27    28    29    23    27    26    25    22    22    24    21    22]

The accepted solution gives

equalpairs =
   3
   5
   8
  12
  13
  14

However if you use my solution:

A = [2 5 8 7 8 9 3 7 6 5 2 2 4 1 2] + 20
sortedA = sort(A, 'descend')
props = regionprops(sortedA, sortedA, 'MeanIntensity', 'PixelIdxList');
missingNumbers = isnan([props.MeanIntensity]);
props = props(~missingNumbers); % Get rid of nans that occur when there is a missing number.
% Done!  Now print them out so we can see:
for k = 1 : length(props)
  fprintf('The indexes for %d are ', props(k).MeanIntensity);
  fprintf('%d, ', props(k).PixelIdxList);
  fprintf('\n');
  equalIndexes{k} = [props(k).PixelIdxList];
end
celldisp(equalIndexes)

You will get a cell array where each cell has the indexes for a certain number:

A =
    22    25    28    27    28    29    23    27    26    25    22    22    24    21    22
sortedA =
    29    28    28    27    27    26    25    25    24    23    22    22    22    22    21
The indexes for 21 are 15, 
The indexes for 22 are 11, 12, 13, 14, 
The indexes for 23 are 10, 
The indexes for 24 are 9, 
The indexes for 25 are 7, 8, 
The indexes for 26 are 6, 
The indexes for 27 are 4, 5, 
The indexes for 28 are 2, 3, 
The indexes for 29 are 1, 
equalIndexes{1} =
    15
equalIndexes{2} =
    11
    12
    13
    14
equalIndexes{3} =
    10
equalIndexes{4} =
     9
equalIndexes{5} =
     7
     8
equalIndexes{6} =
     6
equalIndexes{7} =
     4
     5
equalIndexes{8} =
     2
     3
equalIndexes{9} =
     1

I think having a complete list is probably more useful and convenient than having a list of pairs which doesn't even include all pairs. For example the accepted solution does not list (11, 14), (12,14), etc. as pairs.