Fill NaNs in array with last value

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Leah
Leah 2013 年 10 月 1 日
編集済み: Shruti Verma 2017 年 8 月 6 日
Hi all, thanks for your help. I have a double array and I want to fill NaN values with the last value.
Example:
[321 nan() nan() nan() nan() 55 nan() nan() nan() 22 nan() 21 nan() nan()]
desired result
[321 321 321 321 321 55 55 55 55 22 22 21 21 21]
Also I want to do this same thing with a cell array of strings with blanks instead of NaN. I'm trying to avoid ugly for loops here. This seems familiar... maybe it was a cody question I couldn't solve. Thanks again.

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Sean de Wolski
Sean de Wolski 2013 年 10 月 1 日
編集済み: Sean de Wolski 2013 年 10 月 1 日
You still have to figure out what to do if the first value is nan.
v = [321 nan() nan() nan() nan() 55 nan() nan() nan() 22 nan() 21 nan() nan()];
idx = (~isnan(v)); %non nans
vr = v(idx); %v non nan
v2 = vr(cumsum(idx)) %use cumsum to build index into vr
  3 件のコメント
Image Analyst
Image Analyst 2013 年 10 月 1 日
Clever!
Leah
Leah 2013 年 10 月 1 日
ya wow... impressive. thank you!

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その他の回答 (1 件)

Jan
Jan 2013 年 10 月 1 日
編集済み: Jan 2013 年 10 月 1 日
What's wrong with FOR loops?
data = [321 nan() nan() nan() nan() 55 nan() nan() nan() 22 nan() 21 nan() nan()];
for k = find(isnan(data))
data(k) = data(k - 1);
end
But here a vectorized version:
notNaN = ~isnan(data);
index = find(notNaN);
fillup = zeros(size(data));
fillup(notNaN) = index - [0, diff(index)];
result = data(cumsum(fillup));
Much uglier than the loop. I cannot test this currently. You have to care for initial NaNs in addition. And Sean's solution is nicer.
  2 件のコメント
Khaing Zin Htwe
Khaing Zin Htwe 2016 年 5 月 10 日
if true
% data = [321 nan() nan() nan() nan() 55 ;nan() nan() nan() 22 nan() 21; nan() nan() 3 66 5 8];
end
I want to fill with integer zeros to NaN values,sir. How can I do it? please help me.
Shruti Verma
Shruti Verma 2017 年 8 月 6 日
編集済み: Shruti Verma 2017 年 8 月 6 日
This should do it :)
data(isnan(data)) = 0;

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