基于列表组合的numpy乘法

Question

我有 n 个 numpy 数组的列表和相应名称的列表。一个例子：

import numpy as np
np.random.seed(0)
i = 3
x = [np.array(np.random.randint(-1,high=1, size=100)).T for k in range(i)]
names = ['n1', 'n2', 'n3']

其中i可以任意取值，但对应名称的长度要相等

我想要的是找到可能的组合，这样：

import itertools
combinations = list(itertools.combinations(names,2))
new_names = ['*'.join(i) for i in combinations]
new_names
Out[]: ['n1*n2', 'n1*n3', 'n2*n3']

[[names.index(i) for i in [*combinations[k]]] for k in range(len(names))]
Out[]: [[0, 1], [0, 2], [1, 2]]

根据组合乘以 x 的元素的最简单方法是什么？所以这个例子的输出是：

[x[0]*x[1], x[0]*x[2], x[1]*x[2]]

输出：

[array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0,
        0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0,
        1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1]),
 array([0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0,
        0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0,
        0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0]),
 array([0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0,
        0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1,
        0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0])]

Answer 1

像您对名称所做的那样简单地使用 combinations 怎么样？

from itertools import combinations
out = [a*b for a,b in combinations(x, 2)]

输出：

[array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0,
        0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0,
        1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0,
        0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1]),
 array([0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0,
        0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0,
        0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0]),
 array([0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0,
        0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0,
        0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1,
        0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0])]

基于列表组合的numpy乘法

numpy multiplication based on combinations of a list

python

numpy