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Asked by Johannes on 9 Oct 2012

Hello,

I have the following problem. I like to replace NaNs with the previous values.

A =

4 5 6 7 8 32 NaN NaN 21 NaN 12 NaN 12 NaN NaN 34 NaN NaN NaN NaN

B =

4 5 6 7 8 32 5 6 21 8 12 5 12 21 8 34 5 12 21 8

I sloved it like this:

for i = 2:5

[r,c] = find(isnan(A(:,i)));

while sum(isnan(A(:,i)))>0

A(r,i) = A(r-1,i);

end

end

I'm sure there is a way avoiding the for and the while statement. I search for an "elegant" solution.

Someone's able to help me?

Answer by Matt Fig on 9 Oct 2012

Edited by Matt Fig on 9 Oct 2012

Johannes, notice that your solution will fail if the first value in a column is nan. Rather than looking for a vectorized solution that may end up being rather convoluted (and being **slower**!), I would simply write a good FOR loop function that can handle all cases. For example, the following solution does not use the FIND function, and only uses simple loops and thus should be very fast:

function A = fill_nans(A) % Replaces the nans in each column with % previous non-nan values.

for ii = 1:size(A,2) I = A(1,ii); for jj = 2:size(A,1) if isnan(A(jj,ii)) A(jj,ii) = I; else I = A(jj,ii); end end end

owr on 9 Oct 2012

This is really nice, readable and makes sense. I especially like the fact that you were able to implement it as an in-place function. In a couple quick tests, a "find" based solution doesnt seem to be any worse performance wise, but I still think I like this better because it is really clean. I may use it for myself, thanks for sharing!

Mohammad Sayeed on 14 Jan 2014

Hi I tried to apply your codes but it showed following error: Error using fill_nans (line 4) Not enough input arguments. Can you please tell me how can I correct it?

Answer by Wayne King on 9 Oct 2012

Edited by Wayne King on 9 Oct 2012

How about:

A = [ 4 5 6 7 8 32 NaN NaN 21 NaN 12 NaN 12 NaN NaN 34 NaN NaN NaN NaN]; indices = isnan(A); A(indices) = 0; B = repmat([4 5 6 7 8],size(A,1),1); A = A+B.*indices;

Answer by owr on 9 Oct 2012

I do this all the time, my code uses for loops, but I dont see anything wrong with for loops. Im sure there are more elegent solutions but this does the trick for me and is more than fast enough:

function datai = backfillnans(data)

% Dimensions [numRow,numCol] = size(data);

% First, datai is copy of data datai = data;

% For each column for c = 1:numCol % Find first non-NaN row indxFirst = find(~isnan(data(:,c)),1,'first'); % Find all NaN rows indxNaN = find(isnan(data(:,c))); % Find NaN rows beyond first non-NaN indx = indxNaN(indxNaN > indxFirst); % For each of these, copy previous value for r = (indx(:))' datai(r,c) = datai(r-1,c); end end

Matt Fig on 9 Oct 2012

This seems to fail when a whole column of data is nan.

A = [25 NaN 54 99 20 3 NaN 92 74 89 7 NaN NaN NaN 82 75 NaN 43 65 77 NaN NaN 15 NaN 38]

owr on 9 Oct 2012

Ah, good catch Matt, thanks for that. Ive been using this for almost 2 years multiple times a day and thats never come up - I guess I never have a full column of nans. It can be fixed I guess by putting an:

if( ~isempty(indxFirst) )

after the line that calculates "indxFirst". Part of me would actually like the whole process to fail so I can figure out why I passed a full column of nans in the first place - that would be symptomatic of a much bigger issue...

Anyways, thanks for taking the time to run and test the code.

## 2 Comments

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/50298#comment_104042

What if a whole column is nan? Which value will fill it?

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/50298#comment_104048

If the first value is NaN, everything should be NaN untill a different value appears in the column.

Thanks, Johannes