Indexing with conditions for certain columns

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Yaser Khojah
Yaser Khojah el 22 de Mzo. de 2019
Comentada: Walter Roberson el 25 de Mzo. de 2019
I have a huge matrix where I want to find the indexes that meet each column median only and the rest of the column median should not be included. A manual way to do it is written as below but I need an easier way since my matrix is huge. Thank you so much in advanced.
Matrix_All = rand(1000,3) * 100;
Median_All = median(Matrix_All);
idx_1 = find( Matrix_All(:,1) == Median_All(1) & Matrix_All(:,2) ~= Median_All(2) & Matrix_All(:,3) ~= Median_All(3));
idx_2 = find( Matrix_All(:,1) ~= Median_All(1) & Matrix_All(:,2) == Median_All(2) & Matrix_All(:,3) ~= Median_All(3));
idx_3 = find( Matrix_All(:,1) ~= Median_All(1) & Matrix_All(:,2) ~= Median_All(2) & Matrix_All(:,3) == Median_All(3));
Mat_1 = Matrix_All(idx_1,:);
Mat_2 = Matrix_All(idx_2,:);
Mat_3 = Matrix_All(idx_3,:);
  7 comentarios
Guillaume
Guillaume el 22 de Mzo. de 2019
There are no limitation based on the number of columns to doing what you want... whatever that is...
You haven't answered Walter's questions, so we're in the dark about what exactly you're trying to do:
  • what if the median is not found anywhere? e.g. median([1 2 3 4]) is 2.5.
  • what if the median is found multiple time? e,.g median([1 1 2 2 3 3]) is 2
Perhaps, you should explain what the purpose of all this is. Maybe it's not the median that you actually need.
Note that find was completely unnecessary in your code so far.
idx = Matrix_All(:,1) == Median_All(1) & Matrix_All(:,2) ~= Median_All(2) & Matrix_All(:,3) ~= Median_All(3)
Mat_1 = Matrix_All(idx_1,:);
would have produced the same result (or error).
Yaser Khojah
Yaser Khojah el 25 de Mzo. de 2019
Editada: Yaser Khojah el 25 de Mzo. de 2019
Dear Guiliaume, Sorry I could not get back to you since I did not have access to my MATLAB. Below is what i want to do but not manually since I have to find do that for different criteria.
idx_1 = MaT_All(:,1) == Median_All(1) & MaT_All(:,2) ~= Median_All(2)...
& MaT_All(:,3) ~= Median_All(3) & MaT_All(:,4) ~= Median_All(4)...
& MaT_All(:,5) ~= Median_All(5) & MaT_All(:,6) ~= Median_All(6)...
& MaT_All(:,7) ~= Median_All(7) & MaT_All(:,8) ~= Median_All(8);
Mat_1 = MaT_All(idx_1,:);
scatter(MaT_All(idx_1,18),MaT_All(idx_1,17)); hold on;
idx_2 = MaT_All(:,1) ~= Median_All(1) & MaT_All(:,2) == Median_All(2)...
& MaT_All(:,3) ~= Median_All(3) & MaT_All(:,4) ~= Median_All(4)...
& MaT_All(:,5) ~= Median_All(5) & MaT_All(:,6) ~= Median_All(6)...
& MaT_All(:,7) ~= Median_All(7) & MaT_All(:,8) ~= Median_All(8);
Mat_1 = MaT_All(idx_2,:);
scatter(MaT_All(idx_2,18),MaT_All(idx_2,17)); hold on;
Is it because if the median is found multiple time you only pick one answer where in my case I select all of them?

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Walter Roberson
Walter Roberson el 22 de Mzo. de 2019
matches_median = bsxfun(@eq, MaT_All, Median_All);
matches_one = find(sum(matches_median,2) == 1);
matches_which = 1 + sum( cumprod(~matches_median(matches_one,:), 2), 2 );
Mat = cell(8,1);
for G = 1 : 8
Mat{G} = MaT_All(matches_one(matches_which==G),:);
end
  2 comentarios
Yaser Khojah
Yaser Khojah el 25 de Mzo. de 2019
Dear Walter thank you so much for coding this problem. It is not really easy to do it and thank you so much. I'm sorry for coming back late since I could not access MATLAB. Thanks again :).
I’m just wondering when your answer as below, I do not get the same answer my code. Any idea?
figures (yours)
scatter(Mat{1,1}(:,18),Mat{1,1}(:,17))
figures (mine)
idx_1 = MaT_All(:,1) == Median_All(1) & MaT_All(:,2) ~= Median_All(2)...
& MaT_All(:,3) ~= Median_All(3) & MaT_All(:,4) ~= Median_All(4)...
& MaT_All(:,5) ~= Median_All(5) & MaT_All(:,6) ~= Median_All(6)...
& MaT_All(:,7) ~= Median_All(7) & MaT_All(:,8) ~= Median_All(8);
Mat_1 = MaT_All(idx_1,:);
scatter(MaT_All(idx_1,18),MaT_All(idx_1,17)); hold on;
Walter Roberson
Walter Roberson el 25 de Mzo. de 2019
The data you have provided us only has 10 columns, so looking at column 18 vs column 17 does not make any sense to us.

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