Indexing with conditions for certain columns

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,:);

[Walter Roberson]

You have an even number of rows. The median would be defined as the mean of the values adjacent to the center in a sorting. There might not be any matrix values that exactly equal that.

[Walter Roberson]

What would you want to happen in the case where a row happens to match more than one of the column medians ?

[Yaser Khojah]

That was only an example so I wanted to apply that on this code

idx_1      = find( 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) ); 

and i get somehting as "Index exceeds array bounds."

So, it is like selecting the median for the first column only and that will follow for the rest of the idx_2....Idx_8

[Walter Roberson]

You show Matrix_All being initialized to 3 columns. Your newer version has a MaT_All indexed up to 8 columns. We as outsiders do not know that MaT_All has at least 8 columns.

[Yaser Khojah]

Yeah i did that earlier only to give an example but then in my comment I added something from my code. Do you think there is a way to do that with 8 columns or it is a limitation in MATLAB? Here is my data to give you an idea. Thank you

[Guillaume]

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]

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?

Answer 1

[Walter Roberson]

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

[Yaser Khojah]

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]

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.