How to generate equispaced interpolating values

Question

I have a list of (x,y) values that are not uniformly spaced. Here is the archive used in this question.

I am able to interpolate between the values but what I get ar

Answer 1

Let's first consider a simple case. Suppose your data looked like the blue line, below.

If you wanted to select equidistant points that were r distance apart, then there would be some critical value for r where the cusp at (1,2) is the first equidistant point.

If you wanted points that were greater than this critical distance apart, then the first equidistant point would jump from (1,2) to some place very different -- depicted by the intersection of the green arc with the blue line. The change is not gradual.

This toy case suggests that a tiny change in the parameter r can have a radical, discontinuous affect on the solution.

It also suggests that you must know the location of the ith equidistant point before you can determine the location of the (i+1)-th equidistant point.

So it appears an iterative solution is required:

import numpy as np
import matplotlib.pyplot as plt
import math

x, y = np.genfromtxt('data', unpack=True, skip_header=1)
# find lots of points on the piecewise linear curve defined by x and y
M = 1000
t = np.linspace(0, len(x), M)
x = np.interp(t, np.arange(len(x)), x)
y = np.interp(t, np.arange(len(y)), y)
tol = 1.5
i, idx = 0, [0]
while i < len(x):
    total_dist = 0
    for j in range(i+1, len(x)):
        total_dist += math.sqrt((x[j]-x[j-1])**2 + (y[j]-y[j-1])**2)
        if total_dist > tol:
            idx.append(j)
            break
    i = j+1

xn = x[idx]
yn = y[idx]
fig, ax = plt.subplots()
ax.plot(x, y, '-')
ax.scatter(xn, yn, s=50)
ax.set_aspect('equal')
plt.show()

Note: I set the aspect ratio to 'equal' to make it more apparent that the points are equidistant.