用序列中的列表值创建字典的最干净(最Python化)方法是什么?

你可以利用 dict.setdefault ,像这样

dict_stuff = {}
for key, value in stuff:
    dict_stuff.setdefault(key, []).append(value)

它说,如果 key 字典中不存在,则使用第二个参数作为其默认值,否则返回与 钥匙 .

我们还内置了 dict 类,它可以帮助您处理这样的情况,称为 collections.defaultdict .

from collections import defaultdict
dict_stuff = defaultdict(list)
for key, value in stuff:
    dict_stuff[key].append(value)

这里,如果 钥匙 中不存在 defaultdict 对象,传递给 默认词典 构造函数将被调用以创建value对象。

有 defaultdict 在 collections 库。

>>> from collections import defaultdict
>>> dict_stuff = defaultdict(list) # this will make the value for new keys become default to an empty list
>>> stuff = [('key1', 1), ('key2', 2), ('key3', 3), 
...          ('key1', 11), ('key2', 22), ('key3', 33),
...          ('key1', 111), ('key2', 222), ('key3', 333),
...          ]
>>> 
>>> for k, v in stuff:
...     dict_stuff[k].append(v)
... 
>>> dict_stuff
defaultdict(<type 'list'>, {'key3': [3, 33, 333], 'key2': [2, 22, 222], 'key1': [1, 11, 111]})

stuff_dict = {}
for k, v in stuff:
    if stuff_dict.has_key(k):
        stuff_dict[k].append(v)
    else:
        stuff_dict[k] = [v]


print stuff_dict
{'key3': [3, 33, 333], 'key2': [2, 22, 222], 'key1': [1, 11, 111]}