Why does interp1d in scipy give a NaN when the first 2 values of the x-array are identical? (fill_value = 0)

Your problem is that you are trying to interpolate points that are outside the interval, this causes that scipy.interpolate.interp1d launches a RuntimeWarning when it tries to calculate the slope between two points (it happens in interpolate.py around line 416):

slope = (y_hi - y_lo) / (x_hi - x_lo)[:, None]

See what happens when you move your points within the interval:

>>> import numpy as np
>>> from scipy.interpolate import interp1d
>>> x = np.array([ 5,  5,  10,  10,  10,  20])
>>> y = np.array([ 0,  0,   0,  0,   0,  30])
>>> X = np.array([5.1,5.1,5.1,6,10,11,20, 19.999])
>>> f = interp1d(x,y,'linear', 0, True, False, 0)
>>> Y = f(X)
 [  0.      0.      0.      0.      0.      3.     30.     29.997]

If you plot it you could see that all makes sense:

This is how interp1d works:

1. You pass x and yto interp1d and it creates a f callable method

2. Then you pass the new x_new values in which you want to evaluate f and it performs the following steps:

   • Find where in the original data, the values to interpolate would be inserted.

     >>> x_new_indices = np.searchsorted(x, X)
     

   • Clip x_new_indices so that they are within the range of x indices and at least 1. Removes mis-interpolation of x_new[n] = x[0]

     >>> x_new_indices = x_new_indices.clip(1, len(x)-1).astype(int)
     

   • Calculate the slope of regions that each x_new value falls in.

     >>> lo = x_new_indices - 1
     >>> hi = x_new_indices
     >>> x_lo = x[lo]
     >>> x_hi = x[hi]
     >>> y_lo = y[lo]
     >>> y_hi = y[hi]
     

   • Calculate the actual value for each entry in x_new.

     >>> slope = (y_hi - y_lo) / (x_hi - x_lo)[:, None]
     >>> y_new = slope*(x_new - x_lo)[:, None] + y_lo