interpolate a series over another series in R

Question

Suppose if I have a random time series that I want to interpolate over another time series. How would I do this in R?

# generate time interval series from exponential distribution
s = sort(rexp(10))
# scale between 0 and 1
scale01 = function(x){(x-min(x))/(max(x)-min(x))}
s = scale01(s)
> s
 [1] 0.00000000 0.02804113 0.05715588 0.10630185 0.15778932 0.20391987 0.26066608 0.27265697 0.39100373
[10] 1.00000000

# generate random normal series
x = rnorm(20) 
> x
 [1] -0.82530658  0.92289557  0.39827984 -0.62416117 -1.69055539 -0.28164232 -1.32717654 -1.36992509
 [9] -1.54352202 -1.09826247 -0.68260576  1.07307043  2.35298180 -0.41472811  0.38919315 -0.27325343
[17] -1.52592682  0.05400849 -0.43498544  0.73841106

# interpolate 'x' over 's' ?
> approx(x,xout=s)
$x
 [1] 0.00000000 0.02804113 0.05715588 0.10630185 0.15778932 0.20391987 0.26066608 0.27265697 0.39100373
[10] 1.00000000

$y
 [1]         NA         NA         NA         NA         NA         NA         NA         NA         NA
[10] -0.8253066
>

I want to interpolate the series 'x' over the series 's'. Lets assume time interval series for the 'x' series has 20 elements distributed uniformly over the interval [0,1]. Now I want to interpolate those 10 elements from 'x' that occur at time intervals described by 's'.

EDIT: I think this does the job.

> approx(seq(0,1,length.out=20), x, xout=s)
$x
 [1] 0.00000000 0.02804113 0.05715588 0.10630185 0.15778932 0.20391987 0.26066608 0.27265697 0.39100373
[10] 1.00000000

$y
 [1] -0.8253066  0.1061033  0.8777987  0.3781018 -0.6221134 -1.5566990 -0.3483466 -0.4703429 -1.4444105
[10]  0.7384111

Thanks for your help guys. I think I now understand how to use interpolation functions in R now. I should really use a time series data structure here.

Why does it not work? The code works, and it does what you told it to do. If you want us to help you, you'll need to be much more specific about what you expect. For example, rather than illustrating your problem with a random vector of 10K points, make up some data with a handful of points and show us the results you expect with that data. Also, read the very good help supplied by ?approx. — Andrie
but your x isn't uniform on 0,1, it is a random normal deviate. You refer to 10 and 20 elements? Which is it? Also, what do you mean by "interpolate over". You are certainly not using approx correctly here. — Gavin Simpson
Yes x is random normal but the time intervals at which points in 'x' occur are uniform over (0,1) — user236215

Gavin Simpson Gavin Simpson · Accepted Answer · 2011-03-29T10:35:11

This isn't meant as a direct answer to the OP's Q but rather to illustrate how approx() works so the OP can formulate a better Q

Your Q makes next to no sense. approx() works by taking a reference set of x, and y coordinates and then interpolating to find y at n locations over the range of x, or at the specified xout locations supplied by the user.

So in your call, you don't provide y and x doesn't contain a y component so I don't see how this can work.

If you want to interpolate s, so you can find time intervals for any value over range of s then:

> approx(s, seq_along(s), n = 20)
$x
 [1] 0.00000000 0.05263158 0.10526316 0.15789474 0.21052632 0.26315789
 [7] 0.31578947 0.36842105 0.42105263 0.47368421 0.52631579 0.57894737
[13] 0.63157895 0.68421053 0.73684211 0.78947368 0.84210526 0.89473684
[19] 0.94736842 1.00000000

$y
 [1]   1.00000  26.25815  42.66323  54.79831  64.96162  76.99433  79.67388
 [8]  83.78458  86.14656  89.86223  91.98513  93.36233  93.77353  94.19731
[15]  94.63652  95.26239  97.67724  98.74056  99.40548 100.00000

Here $y contains the interpolated values for s at n = 20 equally spaced locations on the range of s (0,1).

Edit: If x represents the series at unstated time intervals uniform on 0,1 and you want the interpolated values of x at the time intervals s, then you need something like this:

> set.seed(1)
> x <- rnorm(20)
> s <- sort(rexp(10))
> scale01 <- function(x) {
+     (x - min(x)) / (max(x) - min(x))
+ }
> s <- scale01(s)
> 
> ## interpolate x at points s
> approx(seq(0, 1, length = length(x)), x, xout = s)
$x
 [1] 0.00000000 0.04439851 0.11870795 0.14379236 0.20767388 0.21218632
 [7] 0.25498856 0.29079300 0.40426335 1.00000000

$y
 [1] -0.62645381  0.05692127 -0.21465011  0.94393053  0.39810806  0.29323742
 [7] -0.64197207 -0.13373472  0.62763207  0.59390132

Is that closer to what you want?

interpolate a series over another series in R

1 Answers