相当于itertools.product的数字

问题描述

我了解itertools.product用于迭代关键字的多个维度的列表.例如，如果我有这个:

I know about itertools.product for iterating on a list of several dimensions of keywords. For instance if I have this:

categories = [
    [ 'A', 'B', 'C', 'D'],
    [ 'E', 'F', 'G', 'H'],
    [ 'I', 'J', 'K', 'L']
]

，在它上面使用itertools.product()，我有类似的东西:

and I use itertools.product() over it, I have something like:

>>> [ x for x in itertools.product(*categories) ]
('A', 'E', 'I'),
('A', 'E', 'J'),
('A', 'E', 'K'),
('A', 'E', 'L'),
('A', 'F', 'I'),
('A', 'F', 'J'),
# and so on...

有没有一种等效的，直接的方法可以对numpy的数组执行相同的操作?

Is there an equivalent, straightforward way of doing the same thing with numpy's arrays?

推荐答案

已经问了几次这个问题:

This question has been asked a couple of times already:

使用numpy构建两个数组的所有组合的数组

itertools产品加速

第一个链接具有一个有效的numpy解决方案，尽管没有提供基准测试，但据称它比itertools快几倍.这段代码是由一个名为pv的用户编写的.请点击链接，并在他认为有用的情况下支持他的回答:

The first link has a working numpy solution, that is claimed to be several times faster than itertools, though no benchmarks are provided. This code was written by a user named pv. Please, follow the link and support his answer if you find it useful:

import numpy as np

def cartesian(arrays, out=None):
    """
    Generate a cartesian product of input arrays.

    Parameters
    ----------
    arrays : list of array-like
        1-D arrays to form the cartesian product of.
    out : ndarray
        Array to place the cartesian product in.

    Returns
    -------
    out : ndarray
        2-D array of shape (M, len(arrays)) containing cartesian products
        formed of input arrays.

    Examples
    --------
    >>> cartesian(([1, 2, 3], [4, 5], [6, 7]))
    array([[1, 4, 6],
           [1, 4, 7],
           [1, 5, 6],
           [1, 5, 7],
           [2, 4, 6],
           [2, 4, 7],
           [2, 5, 6],
           [2, 5, 7],
           [3, 4, 6],
           [3, 4, 7],
           [3, 5, 6],
           [3, 5, 7]])

    """

    arrays = [np.asarray(x) for x in arrays]
    dtype = arrays[0].dtype

    n = np.prod([x.size for x in arrays])
    if out is None:
        out = np.zeros([n, len(arrays)], dtype=dtype)

    m = n / arrays[0].size
    out[:,0] = np.repeat(arrays[0], m)
    if arrays[1:]:
        cartesian(arrays[1:], out=out[0:m,1:])
        for j in xrange(1, arrays[0].size):
            out[j*m:(j+1)*m,1:] = out[0:m,1:]
    return out

尽管如此，在同一篇文章中，Alex Martelli(他是SO的一位伟大的Python专家)写道，itertools是完成此任务的最快方法.因此，这是一个快速基准，证明了亚历克斯的话.

Nevertheless, in the same post Alex Martelli - he is a great Python guru at SO - wrote, that itertools was the fastest way to get this task done. So here is a quick benchmark, that proves Alex's words.

import numpy as np
import time
import itertools


def cartesian(arrays, out=None):
    ...


def test_numpy(arrays):
    for res in cartesian(arrays):
        pass


def test_itertools(arrays):
    for res in itertools.product(*arrays):
        pass


def main():
    arrays = [np.fromiter(range(100), dtype=int), np.fromiter(range(100, 200), dtype=int)]
    start = time.clock()
    for _ in range(100):
        test_numpy(arrays)
    print(time.clock() - start)
    start = time.clock()
    for _ in range(100):
        test_itertools(arrays)
    print(time.clock() - start)

if __name__ == '__main__':
    main()

输出:

0.421036
0.06742

因此，您绝对应该使用itertools.

So, you should definitely use itertools.

这篇关于相当于itertools.product的数字的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！