增加 defaultdict 列表的计数

Question

How to increment the count in defaultdict of list like

import collections
dict = collections.defaultdict(list)

dict["i"][0]+=1 throws a type error.

I am expecting the dict to be like

{"i":[1,]}

Any efficient approach for this instead of using any loops statements?

Answer 1

import collections

s = collections.defaultdict(lambda: [0])

s['i'][0] += 1

Answer 2

You need to retrieve the item from the dict and then appending to it

d = collections.defaultdict(list)
d['i'].append(1)

>>> d
defaultdict(<type 'list'>, {'i': [1]})

Also don't use dict as a variable name, it is used to construct dicts.

>>> type(dict)
<type 'type'>

Answer 3

You can use Counter from collections

from collections import Counter
c = Counter()
sample_list = ['a','b','c','a','b','d','e','b']
for l in sample_list:
    c[l] += 1

Now the variable c would give the following,

>>>Counter({'a': 2, 'b': 3, 'c': 1, 'd': 1, 'e': 1})

You can get count of each element as follows:

c['a']
>>>2