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

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!

Answer 1

[Matt J]

If you mark missing data by putting NaNs in "b" instead of in "a", then you will be able to interpolate freely.

[Adam Pigon]

What do you mean ? I could mark missing values in vector b but what to do then with values of vector a ? They must be of equal length. I could impose some artificial values for them that just maintain the ordering of values in vector b. But howo to do it ? For example by giving any consequitive values for NaN's in a that just lie between surrounding not NaN'.

[Matt J]

You should restore the NaNs in "a" to whatever values they had before you overwrote them with NaNs.

[Adam Pigon]

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!