Question

我有一个集合，其中包含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})}

我想用值创建字典：素数 - 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})}

你能帮我纠正一下吗？

Answer 1

您进行双重迭代并有效地创建lists和products的笛卡尔积。

要在多个序列上迭代元素，只需使用zip。

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

当然，您可以删除所有中间变量并执行简单的单行程序：

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}

字典理解的例子

3 个答案: