How to implement linear interpolation?

Question

Say I am given data as follows:

x = [1, 2.5, 3.4, 5.8, 6]
y = [2, 4, 5.8, 4.3, 4]

I want to design a function that will interpolate linearly between 1 and 2.5, 2.5 to 3.4, and so on using Python.

I have tried looking through this Python tutorial, but I am still unable to get my head around it.

Answer 1

import scipy.interpolate
y_interp = scipy.interpolate.interp1d(x, y)
print y_interp(5.0)

scipy.interpolate.interp1d does linear interpolation by and can be customized to handle error conditions.

Answer 2

def interpolate(x1: float, x2: float, y1: float, y2: float, x: float):
    """Perform linear interpolation for x between (x1,y1) and (x2,y2) """

    return ((y2 - y1) * x + x2 * y1 - x1 * y2) / (x2 - x1)

Answer 3

Building on Lauritz` answer, here's a version with the following changes

• Updated to python3 (the map was causing problems for me and is unnecessary)
• Fixed behavior at edge values
• Raise exception when x is out of bounds
• Use __call__ instead of __getitem__

from bisect import bisect_right

class Interpolate:
    def __init__(self, x_list, y_list):
        if any(y - x <= 0 for x, y in zip(x_list, x_list[1:])):
            raise ValueError("x_list must be in strictly ascending order!")
        self.x_list = x_list
        self.y_list = y_list
        intervals = zip(x_list, x_list[1:], y_list, y_list[1:])
        self.slopes = [(y2 - y1) / (x2 - x1) for x1, x2, y1, y2 in intervals]

    def __call__(self, x):
        if not (self.x_list[0] <= x <= self.x_list[-1]):
            raise ValueError("x out of bounds!")
        if x == self.x_list[-1]:
            return self.y_list[-1]
        i = bisect_right(self.x_list, x) - 1
        return self.y_list[i] + self.slopes[i] * (x - self.x_list[i])

Example usage:

>>> interp = Interpolate([1, 2.5, 3.4, 5.8, 6], [2, 4, 5.8, 4.3, 4])
>>> interp(4)
5.425

Answer 4

As I understand your question, you want to write some function y = interpolate(x_values, y_values, x), which will give you the y value at some x? The basic idea then follows these steps:

1. Find the indices of the values in x_values which define an interval containing x. For instance, for x=3 with your example lists, the containing interval would be [x1,x2]=[2.5,3.4], and the indices would be i1=1, i2=2
2. Calculate the slope on this interval by (y_values[i2]-y_values[i1])/(x_values[i2]-x_values[i1]) (ie dy/dx).
3. The value at x is now the value at x1 plus the slope multiplied by the distance from x1.

You will additionally need to decide what happens if x is outside the interval of x_values, either it's an error, or you could interpolate "backwards", assuming the slope is the same as the first/last interval.

Did this help, or did you need more specific advice?

Answer 5

Instead of extrapolating off the ends, you could return the extents of the y_list. Most of the time your application is well behaved, and the Interpolate[x] will be in the x_list. The (presumably) linear affects of extrapolating off the ends may mislead you to believe that your data is well behaved.

• Returning a non-linear result (bounded by the contents of x_list and y_list) your program's behavior may alert you to an issue for values greatly outside x_list. (Linear behavior goes bananas when given non-linear inputs!)

• Returning the extents of the y_list for Interpolate[x] outside of x_list also means you know the range of your output value. If you extrapolate based on x much, much less than x_list[0] or x much, much greater than x_list[-1], your return result could be outside of the range of values you expected.

  def __getitem__(self, x):
      if x <= self.x_list[0]:
          return self.y_list[0]
      elif x >= self.x_list[-1]:
          return self.y_list[-1]
      else:
          i = bisect_left(self.x_list, x) - 1
          return self.y_list[i] + self.slopes[i] * (x - self.x_list[i])
  

Answer 6

Your solution did not work in Python 2.7. There was an error while checking for the order of the x elements. I had to change to code to this to get it to work:

from bisect import bisect_left
class Interpolate(object):
    def __init__(self, x_list, y_list):
        if any([y - x <= 0 for x, y in zip(x_list, x_list[1:])]):
            raise ValueError("x_list must be in strictly ascending order!")
        x_list = self.x_list = map(float, x_list)
        y_list = self.y_list = map(float, y_list)
        intervals = zip(x_list, x_list[1:], y_list, y_list[1:])
        self.slopes = [(y2 - y1)/(x2 - x1) for x1, x2, y1, y2 in intervals]
    def __getitem__(self, x):
        i = bisect_left(self.x_list, x) - 1
        return self.y_list[i] + self.slopes[i] * (x - self.x_list[i])

Answer 7

I thought up a rather elegant solution (IMHO), so I can't resist posting it:

from bisect import bisect_left

class Interpolate(object):
    def __init__(self, x_list, y_list):
        if any(y - x <= 0 for x, y in zip(x_list, x_list[1:])):
            raise ValueError("x_list must be in strictly ascending order!")
        x_list = self.x_list = map(float, x_list)
        y_list = self.y_list = map(float, y_list)
        intervals = zip(x_list, x_list[1:], y_list, y_list[1:])
        self.slopes = [(y2 - y1)/(x2 - x1) for x1, x2, y1, y2 in intervals]

    def __getitem__(self, x):
        i = bisect_left(self.x_list, x) - 1
        return self.y_list[i] + self.slopes[i] * (x - self.x_list[i])

I map to float so that integer division (python <= 2.7) won't kick in and ruin things if x1, x2, y1 and y2 are all integers for some iterval.

In __getitem__ I'm taking advantage of the fact that self.x_list is sorted in ascending order by using bisect_left to (very) quickly find the index of the largest element smaller than x in self.x_list.

Use the class like this:

i = Interpolate([1, 2.5, 3.4, 5.8, 6], [2, 4, 5.8, 4.3, 4])
# Get the interpolated value at x = 4:
y = i[4]

I've not dealt with the border conditions at all here, for simplicity. As it is, i[x] for x < 1 will work as if the line from (2.5, 4) to (1, 2) had been extended to minus infinity, while i[x] for x == 1 or x > 6 will raise an IndexError. Better would be to raise an IndexError in all cases, but this is left as an exercise for the reader. :)