python - 使用 map() 函数将字典分配给其他字典中的元组键

Question

假设我们有一个 dict d1 和一个 dict d2。现在，假设 d2 是 d1 中的 tuple-keys 的值，如下所示：

d1 = {}
d2 = {}

for i in range(4):
    for j in range(4):
        d1[(j,i)] = d2

print d1

输出：

{(0, 1): {}, (1, 2): {}, (3, 2): {}, (0, 0): {}, (3, 3): {}, (3, 0): {}, (3, 1): {}, (2, 1): {}, (0, 2): {}, (2, 0): {}, (1, 3): {}, (2, 3): {}, (2, 2): {}, (1, 0): {}, (0, 3): {}, (1, 1): {}}

现在问题来了：

考虑到我无法在 lambda 表达式中分配任何内容，我该如何在不使用 for 循环的情况下做到这一点？我该怎么做这样的事情：

map((lambda x: x =¹ "t"),[(j,i) for j in range(16) for i in range(16)]) # ¹: Inadmissible syntax.

Answer 1

您正在寻找dict 理解：

d1 = {(i, j): {} for i in range(4) for j in range(4)}

此外，您可以使用itertools.product()在单个循环中生成索引：

from itertools import product

d1 = {(i, j): {} for i, j in product(range(4), repeat=2)}

这两个表达式都创建了单独的嵌套字典；key: value每次循环迭代都会评估表达式。

演示：

>>> {(i, j): {} for i in range(4) for j in range(4)}
{(0, 1): {}, (1, 2): {}, (3, 2): {}, (0, 0): {}, (3, 3): {}, (3, 0): {}, (3, 1): {}, (2, 1): {}, (0, 2): {}, (2, 0): {}, (1, 3): {}, (2, 3): {}, (2, 2): {}, (1, 0): {}, (0, 3): {}, (1, 1): {}}
>>> from itertools import product
>>> {(i, j): {} for i, j in product(range(4), repeat=2)}
{(0, 1): {}, (1, 2): {}, (3, 2): {}, (0, 0): {}, (3, 3): {}, (3, 0): {}, (3, 1): {}, (2, 1): {}, (0, 2): {}, (2, 0): {}, (1, 3): {}, (2, 3): {}, (2, 2): {}, (1, 0): {}, (0, 3): {}, (1, 1): {}}