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Interpolation with NaN: how to interpolate a vector so it DOES contract the NaN

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Adam Pigon
Adam Pigon 2017-11-25
评论： Adam Pigon 2017-11-26
Dear All, I have the following problem:
There are the following vectors:
a = [-1, 0.15, 0.7, 0.9, 1.1, 1.98, NaN, NaN, 2.7, 2.9, 3.1, NaN, NaN, NaN],
b = randn(length(a))
x = 0:5
I would like to interpolate x on a:
V = interp1(a,b,x)
Unfortunately, vector a contains NaN, which makes it impossible. I could use
V = interp1(a(~isnan(a)),b(~isnan(a)),x)
but then I get crude interpolation for values, in this case, between 1.98 and 2.7 while they should in fact do not exist, i.e. they should get NaN. How to do it ? How can I 'transfer' NaN values from vector a to vector x so that all x values that are next to NaN in a turn into NaNs and I can use interp1(a(~isnan(a)),b(~isnan(a)),x) getting proper results ?
Any help would be appreciated!

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回答（1 个）

Matt J
Matt J 2017-11-26
If you mark missing data by putting NaNs in "b" instead of in "a", then you will be able to interpolate freely.
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Matt J
Matt J 2017-11-26
编辑：Matt J 2017-11-26
You should restore the NaNs in "a" to whatever values they had before you overwrote them with NaNs.
Adam Pigon
Adam Pigon 2017-11-26
if it were that easy I would have already done it :) These values do not exist. Fortunately, there is a solution to this problem: you can perform a linear interpolation of these values.
a = [1 2 3 NaN NaN 5 NaN 6 7 NaN 8];
t = 1:length(a);
d = interp1(t(~isnan(a)),a(~isnan(a)),t);
Then it is enough to give vector b NaN's using indices from vector a and perform the interpolation using interp1. Problem solved!

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