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

Show 2 older comments

Jakob Hannibal
on 16 Nov 2014

Jakob Hannibal
on 16 Nov 2014

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

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.

Opportunities for recent engineering grads.

## 2 Comments

## Matt Fig (view profile)

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?

## Johannes (view profile)

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