Remove rows or cols whose elements are all NaN

How can I remove rows or cols whose elements are all NaN ? Withouot any dirty iterations?
For example,
A = [1 1 1 1 1 1 1 1 1 1;
NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN;
1 1 1 1 1 1 1 1 1 1;
NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN;];
should turned into
A = [1 1 1 1 1 1 1 1 1 1;
1 1 1 1 1 1 1 1 1 1];
This is just an example. Actually I have a very big matrix. So I want a solution of this question to work well with my big matrix.

2 Comments

Hi! You can find some clues here: A discussion of the opposite problem on SE. I personally achieved what you were attempting using:
A(~any(~isnan(A), 2),:)=[];
Michal Gajewski commented
works

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 Accepted Answer

out = A(all(~isnan(A),2),:); % for nan - rows
out = A(:,all(~isnan(A))); % for nan - columns

6 Comments

A can be a row-vector, such that all will operate along the 2nd dimension automatically. Therefore this is safer:
out = A(:, all(~isnan(A), 1));
Why do you write 1 for columns and 2 for rows?
@Hugo: See doc all for details. The numbers specify the dimension to operate on.
Can I obtain a logical matrix I of indices to be held / removed something like
I = something here
and then use assign an altered matrices:
out1 = out1(I);
out2 = out2(I);
% ... etc.
Thanks
This doesn't really do what the question asks for. This selects all the columns or rows with none (zero) NaN values. So, this is answering the question: "Remove rows or cols whose elements have any (at least one) NaN"
Adding on what @gringer45 said. The code to remove columns and/or rows with all NaN's is below.
A(:,~all(isnan(A))) % for all nan - columns
A(~all(isnan(A),2),:) % for all nan - rows

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More Answers (6)

To remove only ALL-NaN columns, do this instead:
A = A(:,~all(isnan(A)));
Saman
Saman on 23 Oct 2013
Edited: Saman on 23 Oct 2013
Use this :
out = A(:,any(~isnan(A))); % for columns
out = A(any(~isnan(A),2),:); %for rows
A(isnan(A))=[]

6 Comments

This does not solve the wanted: "delete rows or cols whose elements are all NaN"
I am not sure what he want, from his example his rows are all 1 or all nan! In this case the answer solve his problem.
Your answer will create an array. he wants to keep the matrix.
To delete rows with all NaN's, maybe try this:
A = A(find(sum(~isnan(A)'))',:)
A(all(isnan(A),2),:) = []; %rows that are all nan
A(:, all(isnan(A),1)) = []; %cols that are all nan
You proposed too simple a solution and you were wrong.

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input varibale : data
output variable : row_index( index of rows with all the value as NaN)
row_index( index of rows with all the value as NaN)
column_index( index of columns with all the value as NaN)
if(iscell(data))
x =find(cell2mat((cellfun(@(data) any(isnan(data),2),data,'UniformOutput',false))));
[ia,ib] = ind2sub(size(data),x);
rows_unique = unique(ia);
rows_unique(:,2)=histc(ia,rows_unique);
row_index = rows_unique(find(rows_unique(:,2)==size(data,2)),1);
columns_unique = unique(ib);
columns_unique(:,2)=histc(ib,columns_unique);
column_index = rows_unique(find(columns_unique(:,2)==size(data,1)),1);
else
row_index =find(~any(~isnan(data), 2)); % row with all NaN values
column_index =find(~any(~isnan(data), 1)); %column with all NaN values
end

2 Comments

Seems like a bit of a bother to just remove the rows or columns ?
I notice that you are using cellfun on the data, implying that the data is a cell array; in the original question it was a plain array.
yeah its for cell array. for plain array you can remove cellfun and just simply keep any(isnan(data),2) instead of x =find(cell2mat((cellfun(@(data) any(isnan(data),2),data,'UniformOutput',false))));
this can you work for cell array of different data type.
Also I have updated solution.

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Haven't tried this, but it should works:
A(isnan(A))=[];

1 Comment

This does not solve the wanted: "delete rows or cols whose elements are all NaN"

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Nike
Nike on 25 Mar 2013
Simplest is
A(isnan(A))= [];

1 Comment

This has been posted twice already. But it still does not solve the original question:
delete rows or cols whose elements are all NaN
For e.g. A = [1, NaN, 1; NaN, 1, NaN] nothing should be deleted.

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Asked:

on 25 Mar 2013

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on 28 Jan 2025

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