字典理解示例

Question

我有一个集合，里面有n组素数(3):

>>> sets
{frozenset({3, 13, 23}), frozenset({17, 2, 13}), 
frozenset({19, 2, 3}), frozenset({3, 29, 23}), frozenset({17, 11, 23}),
frozenset({17, 2, 19}), frozenset({11, 17, 3}), frozenset({17, 5, 7})}

我想用以下值创建字典：素数的三元组和键：三个素数的乘积。

这是我的尝试：

lists = [list(i) for i in sets]
products = [reduce(lambda x,y:x*y,i) for i in lists]
dictir = {x:y for x in products for y in sets}

但是 dictir 给了我错误的结果：

{897: frozenset({17, 5, 7}), 114: frozenset({17, 5, 7}), 595: frozenset({17, 5, 7}), 561: frozenset({17, 5, 7}), 646: frozenset({17, 5, 7}), 2001: frozenset({17, 5, 7}), 442: frozenset({17, 5, 7}), 4301: frozenset({17, 5, 7})}

你能帮我改正吗？

Answer 1

您进行了两次迭代并有效地创建了 lists 和 products 的笛卡尔积。

要按元素迭代多个序列，只需使用 zip.

dictir = {x: y for x, y in zip(products, sets)}

当然你可以放弃所有中间变量并做简单的一行：

sets = {frozenset({3, 13, 23}), frozenset({17, 2, 13}),
        frozenset({19, 2, 3}), frozenset({3, 29, 23}), frozenset({17, 11, 23}),
        frozenset({17, 2, 19}), frozenset({11, 17, 3}), frozenset({17, 5, 7})}
dictir = {reduce(lambda x, y: x * y, s): s for s in sets}

Answer 2

您可以使用 zip()，但另一种更有效的方法是：

import operator
tuples = ((reduce(operator.mul, i), i) for i in sets)
dictir = dict(tuples)

# Or as a 1 liner
dictir = {reduce(operator.mul, i): i for i in sets}

Answer 3

我想，你可以简单地使用它：

import operator
dictir = {reduce(operator.mul,numbers):numbers for value in sets}

字典理解示例

Dictionary Comprehension example

python

dictionary

set

dictionary-comprehension