python - 如何找到等于 x 的不同长度数组的加权组合？

Question

我有两个不同长度的数组(a 和 b)。我还有一组权重，我需要找到等于 x 的 a + b 的加权组合。权重之和必须始终等于 1。

我尝试过以下方法:

import numpy as np

a = np.arange(1.2, 1.7, 0.1)
b = np.arange(0.0, 0.9, 0.1)
x = 0.6
weights = np.arange(0.05, 1.05, 0.05)

a.resize(b.shape)

lw = []
for w in weights:
    for y in weights:
        np.where((w * a + y * b) == x, lw.append(w), np.nan)

这给了我一个 20*20 权重的列表，即所有权重。

我需要一个权重列表，其中:w * a + y * b = x

知道我该怎么做吗？

Answer 1

IIUC 这是一种方法 -

sums = weights[:,None, None, None] * a[:,None] + weights[:,None,None]*b
idx = np.argwhere(np.isclose(sums,x))
out_idx = idx[np.isclose(weights[idx[:,0]] + weights[idx[:,1]], 1)]

这里，前两列是权重索引的组合，最后两列分别是a和b的索引。

验证给定样本 -

In [165]: sums = weights[:,None, None, None] * a[:,None] + weights[:,None,None]*b

In [166]: idx = np.argwhere(np.isclose(sums,x))

In [167]: out_idx = idx[np.isclose(weights[idx[:,0]] + weights[idx[:,1]], 1)]

In [168]: out_idx
Out[168]: 
array([[ 1, 17,  3,  5],
       [ 3, 15,  2,  4],
       [ 4, 14,  0,  4],
       [ 4, 14,  3,  3],
       [ 4, 14,  5,  8],
       [ 4, 14,  6,  8],
       [ 4, 14,  7,  8],
       [ 4, 14,  8,  8],
       [ 5, 13,  1,  3],
       [ 7, 11,  0,  2],
       [ 7, 11,  3,  0],
       [ 9,  9,  0,  0]])

In [169]: weights[1]*a[3] + weights[17]*b[5]
Out[169]: 0.60000000000000009

In [170]: weights[3]*a[2] + weights[15]*b[4]
Out[170]: 0.60000000000000009

In [171]: weights[1] + weights[17]
Out[171]: 1.0000000000000002

In [172]: weights[3] + weights[15]
Out[172]: 1.0