Python 从两个屈服函数更新字典

假设我有两个函数，它们都

生成
字典：

def one_two_three():
    myDict1 = {}
    myList1 = range(1, 4)
    for i in myList1:
        myDict1['number'] = i
        yield myDict1

def four_five_six():
    myDict2 = {}
    myList2 = range(4, 7)
    for i in myList2:
        myDict2['other_number'] = i
        yield myDict2

是否有方法调用这两个函数并更新字典以返回如下数据结构：

{'number': 0, 'other_number': 3}
{'number': 1, 'other_number': 4}
{'number': 2, 'other_number': 5}

是的，但作为一种更好的方法，您可以生成键的元组并将其值
zip
generators，并创建具有列表理解功能的预期词典：

def one_two_three():
    myList1 = range(0, 4)
    for i in myList1:
        yield 'number',i

def four_five_six():
    myList2 = range(3, 6)
    for i in myList2:
        yield 'other_number',i


print [{i:j,k:z} for (i,j),(k,z) in zip(one_two_three(),four_five_six())]

>>> [{'number':i,'other_number':j} for i,j in zip(range(0,4),range(3,6))]
[{'number': 0, 'other_number': 3},
 {'number': 1, 'other_number': 4},
 {'number': 2, 'other_number': 5}]

输出：

[{'number': 0, 'other_number': 3},
 {'number': 1, 'other_number': 4},
 {'number': 2, 'other_number': 5}]

作为一种不使用生成器函数的更具python风格的方法，您可以压缩范围对象并使用列表创建词典：

def one_two_three():
    myList1 = range(0, 4)
    for i in myList1:
        yield 'number',i

def four_five_six():
    myList2 = range(3, 6)
    for i in myList2:
        yield 'other_number',i


print [{i:j,k:z} for (i,j),(k,z) in zip(one_two_three(),four_five_six())]

>>> [{'number':i,'other_number':j} for i,j in zip(range(0,4),range(3,6))]
[{'number': 0, 'other_number': 3},
 {'number': 1, 'other_number': 4},
 {'number': 2, 'other_number': 5}]

最后一个示例不是单个数据结构，而是三个单独的字典。你想把那三本字典列在一个列表里，还是什么？似乎您最好只生成值（
i
），而不是字典本身，然后您可以在调用函数的代码中创建字典。