用序列中的列表值创建字典的最简洁（最Pythonic）方法是什么？

Question

I have a collection that looks like this:

stuff = [('key1', 1), ('key2', 2), ('key3', 3), 
         ('key1', 11), ('key2', 22), ('key3', 33),
         ('key1', 111), ('key2', 222), ('key3', 333),
         ]
# Note: values aren't actually that nice. That would make this easy.

I want to turn it into a dictionary that looks like this:

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

What's the nicest way to convert this data? The first method that comes to mind is:

dict_stuff = {}
for k,v in stuff:
    dict[k] = dict.get(k, [])
    dict[k].append(v)

Is that the cleanest way to do this?

Answer 1

You can make use of dict.setdefault , like this

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

It says that, if the key doesn't exist in the dictionary, then use the second parameter as the default value for it, otherwise return the actual value corresponding to the key .

We also have a built-in dict class, which helps you deal with cases like this, called collections.defaultdict .

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

Here, if the key doesn't exist in the defaultdict object, the factory function passed to the defaultdict constructor will be called to create the value object.

Answer 2

There is defaultdict in the collections lib.

>>> 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]})

Answer 3

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]}