使用NumPy索引数组切片Python列表 - 任何快速方式？

Question

I have a regular list called a , and a NumPy array of indices b .
(No, it is not possible for me to convert a to a NumPy array.)

Is there any way for me to the same effect as " a[b] " efficiently? To be clear, this implies that I don't want to extract every individual int in b due to its performance implications.

(Yes, this is a bottleneck in my code. That's why I'm using NumPy arrays to begin with.)

Answer 1

a = list(range(1000000))
b = np.random.randint(0, len(a), 10000)

%timeit np.array(a)[b]
10 loops, best of 3: 84.8 ms per loop

%timeit [a[x] for x in b]
100 loops, best of 3: 2.93 ms per loop

%timeit operator.itemgetter(*b)(a)
1000 loops, best of 3: 1.86 ms per loop

%timeit np.take(a, b)
10 loops, best of 3: 91.3 ms per loop

I had high hopes for numpy.take() but it is far from optimal. I tried some Numba solutions as well, and they yielded similar times--around 92 ms.

So a simple list comprehension is not far from the best here, but operator.itemgetter() wins, at least for input sizes at these orders of magnitude.

Answer 2

Write a cython function:

import cython
from cpython cimport PyList_New, PyList_SET_ITEM, Py_INCREF

@cython.wraparound(False)
@cython.boundscheck(False)
def take(list alist, Py_ssize_t[:] arr):
    cdef:
        Py_ssize_t i, idx, n = arr.shape[0]
        list res = PyList_New(n)
        object obj

    for i in range(n):
        idx = arr[i]
        obj = alist[idx]
        PyList_SET_ITEM(res, i, alist[idx])
        Py_INCREF(obj)

    return res

The result of %timeit:

import numpy as np

al= list(range(10000))
aa = np.array(al)

ba = np.random.randint(0, len(a), 10000)
bl = ba.tolist()

%timeit [al[i] for i in bl]
%timeit np.take(aa, ba)
%timeit take(al, ba)

1000 loops, best of 3: 1.68 ms per loop
10000 loops, best of 3: 51.4 µs per loop
1000 loops, best of 3: 254 µs per loop

numpy.take() is the fastest if both of the arguments are ndarray object. The cython version is 5x faster than list comprehension.