Use `np.diff` but assume the input starts with an extra zero

Question

Given a series of event times v, I can create their interval durations using np.diff(v). Is there a way to have np.diff assume the ser

Answer 1

Given for example:

t = np.array([1.1, 2.0, 4.5, 4.9, 5.2])

We want to compute the consecutive differences in t, including the diff from 0. to the first element in t.

The question gave this way of accomplishing this:

>>> np.diff(np.hstack((0, t)))

And it could be this too:

>>> np.hstack((t[0], np.diff(t)))

But the obscurely-named function ediff1d can do it in one function call:

>>> np.ediff1d(t, to_begin=t[0])
array([ 1.1,  0.9,  2.5,  0.4,  0.3])

Prepending t[0] to the result is the same as computing the difference t[0] - 0., of course. (Assuming t is nonempty).

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Timings (not the motivation of the question, but I was curious)

import numpy as np
t = np.random.randn(10000)
%timeit np.diff(np.concatenate(([0], t)))
10000 loops, best of 3: 23.1 µs per loop
%timeit np.diff(np.hstack((0, t)))
10000 loops, best of 3: 31.2 µs per loop
%timeit np.ediff1d(t, to_begin=t[0])
10000 loops, best of 3: 92 µs per loop