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 个评论
回答(5 个)
Moshe Flam
2017-12-3
编辑:Moshe Flam
2017-12-4
ROWBYROW = 2;
B = fillmissing(A,'previous',ROWBYROW);
2 个评论
Namrata Goswami
2020-12-11
编辑:Namrata Goswami
2020-12-11
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
2012-10-9
编辑:Matt Fig
2012-10-9
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 个评论
Timothy Jackson
2016-4-1
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
2012-10-9
编辑:Wayne King
2012-10-9
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 个评论
Matt Fig
2012-10-9
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
2012-10-9
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 个评论
Matt Fig
2012-10-9
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
2012-10-9
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
2017-7-18
I found your procedure much more elegant and efficient. It was very helpful man.
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