用索引的NumPy数组切片Python列表-有什么快速的方法吗? [英] Slicing a Python list with a NumPy array of indices -- any fast way?

问题描述

我有一个称为a的常规list和一个索引b的NumPy数组. (不，我无法将a转换为NumPy数组.)

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.)

我有什么办法可以有效地达到与"a[b]"相同的效果?明确地说，这意味着由于其性能影响，我不想提取b中的每个单独的int.

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.

(是的，这是我的代码中的瓶颈.这就是为什么我开始使用NumPy数组的原因.)

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

推荐答案

编写cython函数:

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

％timeit的结果:

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

如果两个参数都是ndarray对象，则

numpy.take()最快. cython版本比列表理解速度快5倍.

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

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