How to efficiently generate an array using 2 arrays and a formula as input with NumPy

Question

I have two arrays, x and t with n elements (t's elements are in strict ascending order, so no dividing by 0) and a formula on which the creation of my new array, v is based:

v[i] = (x[i+1] - x[i]) / (t[i+1] - t[i])

How can I write this in NumPy? I tried using numpy.fromfunction but didn't manage to make it work.

I did manage to do it using a for loop - but I feel like there's a better way of doing this:

n = 100000
x = np.random.rand(n)
t = np.random.randint(1, 10, n)
t = t.cumsum()

def gen_v(x, t):
    v = np.zeros(n - 1)
    for i in range(0, n - 1):
        v[i] = (x[i+1] - x[i])/(t[i+1]-t[i])
    return v

v = gen_v(x, t)
%timeit gen_v(x, t)

Outputs

156 ms ± 15 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

obchardon · Accepted Answer · 2022-10-25 12:24:47Z

2

You can use np.diff():

def gen_v(x,t):
    return np.diff(x)/np.diff(t)

The benchmark give us:

# Your function:
8.45 ms +- 557 us per loop (mean +- std. dev. of 7 runs, 100 loops each)
# My function:
63.4 us +- 1.62 us per loop (mean +- std. dev. of 7 runs, 10000 loops each)

answered Oct 25, 2022 at 12:24

obchardon

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Comments

Invarianz · Accepted Answer · 2022-10-25 12:37:26Z

1

You could use array slicing

def gen_v(x, t):
    return (x[1:] - x[:-1])/(t[1:] - t[:-1])

Benchmarking yields

# Your Function
62.4 ms ± 1.52 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
# Slicing
277 µs ± 3.34 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)

On my meager hardware. ;)

edited Oct 25, 2022 at 12:37

answered Oct 25, 2022 at 12:29

Invarianz

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