vector substitution in numpy/python [duplicate]

Question

Given two vectors v=['a','b','c'] and i=np.random.randint(len(v),size=10), I can get the "substitution" vector

vi = [v[i[x]] for x in range(len(i))]

E.g., vi is

['a', 'a', 'c', 'c', 'b', 'a', 'c', 'a', 'c', 'a']

if

i = array([0, 0, 2, 2, 1, 0, 2, 0, 2, 0])

Is there a vectorized operation for this?

Answer 1

You can simply use numpy indexing (note that you need to convert v to a numpy.array for this to work):

v = np.array(['a','b','c'])
i = np.random.randint(len(v),size=10)

>>> v[i]
array(['c', 'b', 'a', 'b', 'c', 'b', 'a', 'a', 'b', 'b'], dtype='<U1')

Timings

In [26]: i = np.random.randint(len(v),size=1000000)

In [27]: %timeit [v[i[x]] for x in range(len(i))]
554 ms ± 6.41 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [28]: %timeit v[i]
4.85 ms ± 12.5 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [29]: %timeit [v[s] for s in i]
505 ms ± 1.95 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)