这将为你解决它，问题在于字典的赋值。

import numpy as np

elasticity = np.array([[1,2],[3,4]])
genders = ['female', 'male']
income = ['i0','i1']

table = {}
for i in range(len(genders)):
    for j in range(len(income)):
        key = genders[i]+"_"+income[j]
        table[key] = elasticity[i,j]
print(table)

则输出为：

{'female_i0': 1, 'female_i1': 2, 'male_i0': 3, 'male_i1': 4}

从python 3.9开始，你可以使用update操作符：

table = {}

elasticity = np.array([[1,2],[3,4]])
elasticity.shape

genders = ['female', 'male']
income = ['i0','i1']

for i in range(len(genders)):
    for j in range(len(income)):
        key = f"{genders[i]}_{income[j]}"
        table |= { key: elasticity[i,j]}

在这个例子中，我还使用了一个f字符串来生成密钥，这是我不久前做的一个基准测试中最快的选项。

您还可以使用字典理解在一行中创建字典：

import numpy as np

elasticity = np.array([[1,2],[3,4]])

genders = ['female', 'male']
income = ['i0', 'i1']

table = {f"{gen}_{inc}": elasticity[i, j] for i, gen in enumerate(genders) for j, inc in enumerate(income)}

print(table)

输出：

{'female_i0': 1, 'female_i1': 2, 'male_i0': 3, 'male_i1': 4}

使用单字节理解：

table = {k:v for i, g in enumerate(genders) 
         for k, v in zip(map('_'.join, zip([g] * len(income), income)), elasticity[i])}
print(table)

{'female_i0': 1, 'female_i1': 2, 'male_i0': 3, 'male_i1': 4}