Self-difference of a list i(difference of sequential list values)

Question

I have a list, a time series, s. I would like to efficiently take the "self difference" of the list: output[n] = s(n)-s(n+1). What's an efficient (in terms of time) way of doing this?

My eventual goal is to have a list representing this below, with this question implementing the first step:

∑[∡s1​(t)−∡s1​(t−1)]⋅[∡s2​(t+τ)−∡s2​(t−1+τ)]

Answer 1

Consider a third-party tool, more_itertools.difference.

import more_itertools as mit


list(mit.difference([0, 1, 3, 6, 10]))
# [0, 1, 2, 3, 4]

Answer 2

If you want an efficient way, use numpy: https://docs.scipy.org/doc/numpy-1.12.0/reference/generated/numpy.diff.html

import numpy as np
s = np.array([1, 2, 4, 7, 0])
output = np.diff(s)
print(output)    #  [ 1  2  3 -7]

Answer 3

Use a simple list comprehension:

output = [s[n] - s[n+1] for n in range(len(n)-1)]

Let the Python run-time system do any optimizations it wants; this method saves your time as programmer and maintainer.