Interpolate list to specific length?

Question

I want to increase the length of a list in python, whereas the missing values in between and on the right positions should be interpolated. Example:

[1, 3, 5, 7, 9]

interpolated to the new length of 10 would be:

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

This is just the double size and can be solved with the following mathematical formula for interpolation, whereas every second value is interpolated:

y = y1 + ((y2 - y1) / (x2 - x1)) * (x - x1)

But I am trying to figure out, how to interpolate if the list with a length of 217 needs to be interpolated to a length of 240 for example.

Is there a library, which offers a function for this I don't know about? Or maybe someone can give me an example?

Edit: The data in the list is not linear. So a list like this is probable:

[4.534, 1.2433. 3.353, 2.3452, 6.124, 8.124, 1.232]

if your data are bound to be linear and not just step-wise linear, you can just get their *max* and *min* values and create all the rest — Ma0, May 29 '17 at 09:08
@ChristianKönig It should be `/ (x2 - x1)` like the enumerator — Ma0, May 29 '17 at 09:09
Take a look at [numpy.interp](https://docs.scipy.org/doc/numpy/reference/generated/numpy.interp.html). — Thomas, May 29 '17 at 09:11
I think for interpolation, you both need x and y values. Your first list only makes sense, if you "sample" the range of 1..9 to 5 values. The "interpolated" example has the range of 1..10 "sampled" to 10 values. By only giving the new length, you are forgetting to give the new range. — Christian König, May 29 '17 at 09:14
@martineau Nope, this was just an example. The values are floats. — antarkt1s, May 29 '17 at 09:20
@antarkt1s I still find your question too broad to be honest.. But another thing you can do since the data are not necessarily linear is to sort them, find the biggest gaps iteratively and fill them by assuming your function behaves locally linearly. So for your list with 217 elements that need to become 240, find the 23 biggest gaps and *plant* new values there based on the gap edges. — Ma0, May 29 '17 at 09:36
It really depends on what are your requirements. The simplest way is linear interpolation between 2 adjacent points, not very nice but damn simple. [Cubic splines](https://en.wikipedia.org/wiki/Spline_%28mathematics%29) are often used to give smooth *local* interpolations. On another hand, Fourier transformations (FFT) can be used to get a *global* interpolation function. Said differently, you must first specify what interpolation method you need before asking for a Python implementation. VTC as unclear... — Serge Ballesta, May 29 '17 at 11:55

martineau · Accepted Answer · 2022-03-04T20:27:54.853

A relatively simple way to do this is to map integer index values in the new list onto "fractional indices" in the original one, then use the fractional portion to linearly interpolate between the original list values represented by the integer portion of interpolated value and the following one (unless the fractional portion is zero).

For example, if the fractional index computed into the original list is 2.333..., then the resulting value would be ¹⁄₃ of the way between the values in original[2] and original[3].

Here's sample code implementing the idea:

# For Python 2 compatibility.
from __future__ import division, print_function

def print_list(values):
    print('[' + ', '.join(format(value, '.3f') for value in values) + ']')

def interpolate(inp, fi):
    i, f = int(fi // 1), fi % 1  # Split floating-point index into whole & fractional parts.
    j = i+1 if f > 0 else i  # Avoid index error.
    return (1-f) * inp[i] + f * inp[j]

inp = [1, 3, 5, 7, 9]
new_len = 10

delta = (len(inp)-1) / (new_len-1)
outp = [interpolate(inp, i*delta) for i in range(new_len)]

print_list(inp)
print_list(outp)

Output:

[1.000, 3.000, 5.000, 7.000, 9.000]
[1.000, 1.889, 2.778, 3.667, 4.556, 5.444, 6.333, 7.222, 8.111, 9.000]

Note: I do something similar to this in my answer to another, marginally-related, question (which includes a graphic that may help visualize the process).

This does exactly what I need! Thank you a million times @martineau! — antarkt1s, May 30 '17 at 08:52
@antarkt1s: That's good to hear—because I wasn't entirely sure since the results aren't `[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]` like in your question given the same list of input values. — martineau, May 30 '17 at 13:42
It is not generalized, just try inp of length 27 and new_len =100 — Yacine, Oct 22 '18 at 14:06
@DYEZ: I see what you mean. It's difficult to avoid issues like that when doing floating point arithmetic. A simple workaround would be to just duplicate the last element the `inp` sequence passed to `interpolate()` with a `inp2 = inp + [inp[-1]]` and then just `return inp2[i] + f*(inp2[i+1]-inp2[i])` unconditionally. — martineau, Oct 22 '18 at 15:04

score 0 · Answer 2 · answered May 29 '17 at 09:29

Quoting Wikipedia:

interpolation is a method of constructing new data points within the range of a discrete set of known data points.

So traditionally interpolation is used to calculate new values of y for given x between known values of f(x1) and f(x2). Your formula is the formula for linear interpolation, which assumes at least linear behaviour between x1 and x2.

In your example, give a list with values, which you want to expand by interpolated values, so I assume they are your y-values:

[1, 3, 5, 7, 9]

What you are not giving are the corresponding x-values. According to your expected result, I assume they should be the same:

f(1)=1, f(3)=3...

Now you extend your range to 10, which is a problem for classic linear interpolation, since you have no known upper f(x) to do your calculation.

Interpreting your expected result, you are assuming general linearity, which means the base function of your given y-values is globally linear and can be assumed to be f(x)=x. If that is the case for your data, there is no need to calculate your formula for every point: calculate your linear function once (using any two different points) and use this to calculate any desired f(x).

If your target function is not globally linear, you cannot interpolate any values larger than your biggest f(x) or smaller than the smallest f(x) - and for any x inbetween you would not only need the f(x) values, but additionally the x-values that produced them.

Interpolate list to specific length?

2 Answers2