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I am trying to interpolate a set of rainfall data in order to find the rainfall at an unknown point.

I have been using the inverse distance weighting interpolation method (details given here: http://en.wikipedia.org/wiki/Inverse_distance_weighting), and the main problem I am having is that where the rainfall at some of the points is equal to zero, these points are effectively ignored because the weight for that point is being multiplied by zero. This means that the estimated rainfall given by the interpolation is way too high.

It may be that I am being stupid, but I can't work out how to apply the method when the data at some of the input points is zero. I have seen other people interpolating the rainfall anomaly instead, but this won't work as the data I am interpolating is the monthly climatology data. Basically some points have never had any rainfall in a particular month.

Can anyone suggest a way I can incorporate this zero rainfall data?

emmalgale
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  • Use a data value of zero is not the same as "ignoring" that data. The weight given to a nearby point with zero rainfall may be large, which tends to produce a weighted average near zero (the effect that one intuitively expects to see). Points where no measurements are available are a different problem; these one is force to "ignore" as there is no data to incorporate. – hardmath May 12 '14 at 11:34

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The zero-rainfall points are not being ignored at all! They are contributing to the denominator in the equation for $u$(x).

In fact the equation is linear in the $u_i$, which means that if you add a constant $c$ to all readings, the result is increased by $c$. So if you are doing it right, whether a reading is zero or not is irrelevant $-$ it's just the relative values of the readings that is important.

TonyK
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  • It may be that I am not doing it right then. The value I got out is a lot higher than I would expect it to be based on the surrounding values. I will have another go and doing it. – emmalgale May 12 '14 at 11:33
  • Turns out I was doing the method slightly differently (standardising the weights before multiplying them by the rainfall), which meant the zero values were having slightly less influence. The value I have got now is more reasonable. – emmalgale May 12 '14 at 12:15