Replace NaNs with previous values
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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?
2 Comments
Answers (5)
Moshe Flam
on 3 Dec 2017
Edited: Moshe Flam
on 4 Dec 2017
ROWBYROW = 2;
B = fillmissing(A,'previous',ROWBYROW);
2 Comments
Namrata Goswami
on 11 Dec 2020
Edited: Namrata Goswami
on 11 Dec 2020
This worked for me partially, since I need to replace missing values withing group. How to use fillmising within a group, like with splitapply ?
Matt Fig
on 9 Oct 2012
Edited: 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
7 Comments
Timothy Jackson
on 1 Apr 2016
Is there a way to do this both before and after values? For instance changing
A= NaN NaN 2 4 8 NaN NaN to A= 2 2 2 4 8 8 8 ?
Wayne King
on 9 Oct 2012
Edited: 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;
1 Comment
Matt Fig
on 9 Oct 2012
Johannes comments:
"Solution there:
A =
4 5 6 7 8
32 5 6 21 8
12 5 12 7 8
34 5 6 7 8
Not good, would need the following: 4 5 6 7 8 32 5 6 21 8 12 5 12 21 8 34 5 12 21 8
Still thanks for you help!"
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
2 Comments
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.
Carlos Vladimir Rodriguez Caballero
on 18 Jul 2017
I found your procedure much more elegant and efficient. It was very helpful man.
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