By José Lopes

Though the Python math library has the linear interpolation that, if I'm not mistaken, requires the adjacent points to the point we need the interpolation, I created my own function to do it.
I did it because I needed to do many interactions where the adjacent points changed, and also because I failed to check first if Python had the linear interpolation function!

With this post I share my solution to anyone interested. It may come in handy if you are calculation several points with a bunch of initial data. Have a look!

The linear interpolation is an approximation method to calculate a point between two known points, assuming that a straight line between these points contains the point we want, has you may see with this image:

Knowing X we can calculate Y with:

y = y₀ + (x - x₀) (y₁ - y₀)/ (x₁ - x₀)

Now if we have a set of points (x₀, y₀), (x₁, y₁), ... (x_n, y_n) and we want to calculate the y coordinate for any given x value, then we can use the function that follows.

Let me first point out that for this example:

• I'm assuming that the known points can be given by two tables with the x and y coordinates (one for each). Let's call them xpoints and ypoints.
• The function checks which known points are adjacent to the wanted point in order to apply the interpolation.
• The function returns an error message if the wanted point falls out of the know points universe.

(1)def Linterpolation (xpoints, ypoints, value):
(2)  for n in range(len(xpoints)):
(3)    if (n==0 and value < xpoints[n]) or (n==len(xpoints)-1 and value > xpoints[n]):
(4)      result = "Point outside the known values"
(5)    elif value <= xpoints[n]:
(6)      result = ypoints[n-1] + ((value-xpoints[n-1])*(ypoints[n]-ypoints[n-1]))/(xpoints[n]-xpoints[n-1])
     return result

Let me explain the code:

(1) Declares the function taking as args the x and y coordinates lists and the x value that we want the y coordinate calculated.
(2) To run through the list of known points.
(3) e (4) Tests to see if given coordinate is within the known values and return the error message if not within.
(5) e (6) Checks if it finds the immediate upper adjacent value and runs the interpolation once it finds.

This solved my problem on the moment but we can always find a more elegant solution or adapt to our needs.
In this example I assumed that the known points were described by a list for each coordinate. I could have used as arg a list of tuples like [(x₀, y₀), (x₁, y₁), ... (x_n, y_n)] that is closer to the mathematic language.
Unfortunately my data usually doesn't come with this format and that is why it was more simple just to use two tables (lists).