从list-object为dict创建密钥

Question

我有清单:

k = ["key1", "subkey2", "subsubkey3"]

我确信这d是一个d["key1"]["subkey2"]["subsubkey3"]有效的字典.

如何将列表转换k为dict的键d以便返回d[k[0]][k[1]]...？

Answer 1

您可以尝试使用reduce()带operator.getitem:

>>> from operator import getitem
>>> 
>>> d = {'key1': {'subkey2': {'subsubkey3': 'value'}}}
>>> k = ["key1", "subkey2", "subsubkey3"]
>>> 
>>> reduce(getitem, k, d)
'value'

在Python 3.x中,您应该使用functools.reduce().

reduce()简单地采用2参数函数并连续地将其应用于列表的元素,累积结果.还有一个可选的初始化参数,我们在这里使用它.正如文档所述,reduce()大致相当于:

def reduce(function, iterable, initializer=None):
    it = iter(iterable)
    if initializer is None:
        try:
            initializer = next(it)
        except StopIteration:
            raise TypeError('reduce() of empty sequence with no initial value')
    accum_value = initializer
    for x in it:
        accum_value = function(accum_value, x)
    return accum_value

在我们的例子中,我们正在传递一个initializer所以它不会None.因此我们拥有的是:

def reduce(function, iterable, initializer=None):
    it = iter(iterable)
    accum_value = initializer
    for x in it:
        accum_value = function(accum_value, x)
    return accum_value

我们function在这种情况下getitem(a, b)(见上面的链接)只返回a[b].而且,我们iterable是k和我们initializer的d.所以reduce()上面的调用相当于:

accum_value = d
for x in k:
    accum_value = accum_value[x]