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Filtering Columns of Array by Number of Row Contents

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Justin Delano 2020-3-27

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评论： Ameer Hamza 2020-4-2

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I have arrays that look like this:

A =
 NaN   NaN
   3     4
   5   NaN

I want to remove columns that contain rows with less than n non-NaN entries. In this case, with n=2, the first column of A would be removed, since the first row contains only 1 non-NaN value. Is there a compact way to do this? I hope this explanation makes sense!

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Ameer Hamza 2020-3-27

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编辑：Ameer Hamza 2020-3-27

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A = [1   NaN   NaN
     2     3     4
     2     5   NaN];
n = 2;
A(sum(~isnan(A), 2) < n, :) = [];

Result

A =
     2     3     4
     2     5   NaN

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Justin Delano 2020-3-27

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Thank you so much! I knew I was close!

Ameer Hamza 2020-3-27

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Glad to be of help.

Justin Delano 2020-3-27

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I'm sorry, I just realized this wasn't the result I'm looking for. I'm looking for the result:

A = [NaN NaN
       3   4
       5 NaN]

Is there a solution for this problem?

Ameer Hamza 2020-3-30

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Are you looking for something like this

A = [1   NaN   NaN
     2     3     4
     2     5   NaN];
n = 2;
A(:, sum(~isnan(A), 2) < n) = [];

Justin Delano 2020-3-30

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Hello, I can give another example.

Input = 
    1     2   NaN
    10   NaN     5
     1     3     5
   NaN   NaN     6
Output = 
    1     2   
    10   NaN  
     1     3  
   NaN   NaN  

Column 3 was removed due to it having a value in row 4, which contains more than n=2 NaNs.

Ameer Hamza 2020-3-30

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try this:

A = [1     2   NaN
    10   NaN     5
     1     3     5
   NaN   NaN     6];
idx = sum(isnan(A),2) >= 2;
idx2 = find(~isnan(A(idx, :)));
A(:,idx2) = [];

Justin Delano 2020-3-30

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Thank you, that seems closer. I didn't expect this to be such a tough problem. If need be I'll do this with loops instead of vectorized functions. That attempt works on that test case, but throws an error on this one:

A =
     1   NaN   NaN
    10   NaN     5
     1     3     5
   NaN   NaN     6

Ameer Hamza 2020-3-30

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I realized I overlooked a things in my last code. Does following code work

A = [1   NaN   NaN
    10   NaN     5
     1     3     5
   NaN   NaN     6];
idx = sum(isnan(A),2) >= 2;
[~, idx2] = find(~isnan(A(idx, :)));
A(:,idx2) = [];

Justin Delano 2020-3-30

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编辑：Justin Delano 2020-3-30

ToTrim.mat
ToTrim2.mat

Thank you, that is better. Unfortunately, for my data, there are a lot more NaNs. I have attached two example matrices. When I use your code, the first matrix is turned completely blank, which I don't think should be the case.

Ameer Hamza 2020-3-30

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I am still not entirely sure what is the correct rule to delete columns. Can you write a simple for loop version for a small matrix. I will try to see if it can be vactorized.

Justin Delano 2020-3-30

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编辑：Justin Delano 2020-3-30

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Here is a rough loop version:

%scan each column of full matrix
for column = matrix_columns
    %scan each row of each column
    for entry = column_rows
        %assume entry is from row = entry_row in full matrix
        if ~isnan(entry)
            if sum(~isnan(entry_row)) < 10
                %delete this column from full matrix
                continue
            end
        end
    end
end

Ameer Hamza 2020-4-1

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Justin, there some to be some issue in this loop version too. For example If I apply it to your previous example, it still does not give the result you want

x =  [1     2   NaN
    10   NaN     5
     1     3     5
   NaN   NaN     6];
col_delete = [];
%scan each column of full matrix
for col = 1:size(x,2)
    %scan each row of each column
    for row = 1:size(x,1)
        entry = x(row, col);
        %assume entry is from row = entry_row in full matrix
        if ~isnan(entry)
            entry_row = x(row, :);
            if sum(~isnan(entry_row)) < 10
                %delete this column from full matrix
                col_delete = [col_delete col];
            end
        end
    end
end
x(:,unique(col_delete)) = [];

Final value of x should be

   2   
 NaN  
   3  
   NaN   NaN

But your loop version make it an empty matrix.

Justin Delano 2020-4-1

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I apologize, you are correct. I set the sum in the final if statement to be less than 10, but in reality it should be less than n, which we have set to 2.

Ameer Hamza 2020-4-1

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Ok. Now it is clear. Please try

x = fate1nums; % variable from file you shared
n = 10;
[r, c] = find(~isnan(x));
x(:, c(sum(~isnan(x(r,:)), 2) < n)) = [];

Justin Delano 2020-4-2

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That looks good! Thank you so much!

Ameer Hamza 2020-4-2

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Glad to be of help.

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Filtering Columns of Array by Number of Row Contents

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