python 将单词列表转换为频率字典的最佳方法
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Best way to turn word list into frequency dict
提问by ???u
What's the best way to convert a list/tuple into a dict where the keys are the distinct values of the list and the values are the the frequencies of those distinct values?
将列表/元组转换为 dict 的最佳方法是什么，其中键是列表的不同值，而值是这些不同值的频率？
In other words:
换句话说：
['a', 'b', 'b', 'a', 'b', 'c']
-->
{'a': 2, 'b': 3, 'c': 1}
(I've had to do something like the above so many times, is there anything in the standard lib that does it for you?)
（我不得不多次做类似上面的事情，标准库中有什么可以为你做的吗？）
EDIT:
编辑：
Jacob Gabrielson points out there is something coming in the standard libfor the 2.7/3.1 branch
回答by SilentGhost
I find that the easiest to understand (while might not be the most efficient) way is to do:
我发现最容易理解（虽然可能不是最有效）的方法是：
{i:words.count(i) for i in set(words)}
回答by S.Lott
Kind of
的种类
from collections import defaultdict
fq= defaultdict( int )
for w in words:
fq[w] += 1
That usually works nicely.
这通常很好用。
回答by Jacob Gabrielson
Just a note that, starting with Python 2.7/3.1, this functionality will be built in to the collectionsmodule, see this bugfor more information. Here's the example from the release notes:
请注意，从 Python 2.7/3.1 开始，此功能将内置到collections模块中，有关更多信息，请参阅此错误。这是发行说明中的示例：
>>> from collections import Counter
>>> c=Counter()
>>> for letter in 'here is a sample of english text':
... c[letter] += 1
...
>>> c
Counter({' ': 6, 'e': 5, 's': 3, 'a': 2, 'i': 2, 'h': 2,
'l': 2, 't': 2, 'g': 1, 'f': 1, 'm': 1, 'o': 1, 'n': 1,
'p': 1, 'r': 1, 'x': 1})
>>> c['e']
5
>>> c['z']
0
回答by YardenR
Actually, the answer of Counter was already mentioned, but we can even do better (easier)!
其实Counter的回答已经提过了，但是我们还可以做得更好（更简单）！
from collections import Counter
my_list = ['a', 'b', 'b', 'a', 'b', 'c']
Counter(my_list) # returns a Counter, dict-like object
>> Counter({'b': 3, 'a': 2, 'c': 1})
回答by Steven Huwig
This is an abomination, but:
这是可憎的，但是：
from itertools import groupby
dict((k, len(list(xs))) for k, xs in groupby(sorted(items)))
I can't think of a reason one would choose this method over S.Lott's, but if someone's going to point it out, it might as well be me. :)
我想不出有什么理由会选择这种方法而不是 S.Lott 的，但如果有人要指出它，那也可能是我。:)
回答by ???u
I have to share an interesting but kind of ridiculous way of doing it that I just came up with:
我必须分享一种我刚刚想出的有趣但有点荒谬的方法：
>>> class myfreq(dict):
... def __init__(self, arr):
... for k in arr:
... self[k] = 1
... def __setitem__(self, k, v):
... dict.__setitem__(self, k, self.get(k, 0) + v)
...
>>> myfreq(['a', 'b', 'b', 'a', 'b', 'c'])
{'a': 2, 'c': 1, 'b': 3}
回答by user8338
I decided to go ahead and test the versions suggested, I found the collections.Counteras suggested by Jacob Gabrielson to be the fastest, followed by the defaultdictversion by SLott.
我决定继续测试建议的版本，我发现collections.CounterJacob Gabrielson 建议的defaultdict版本是最快的，其次是 SLott的版本。
Here are my codes :
这是我的代码：
from collections import defaultdict
from collections import Counter
import random
# using default dict
def counter_default_dict(list):
count=defaultdict(int)
for i in list:
count[i]+=1
return count
# using normal dict
def counter_dict(list):
count={}
for i in list:
count.update({i:count.get(i,0)+1})
return count
# using count and dict
def counter_count(list):
count={i:list.count(i) for i in set(list)}
return count
# using count and dict
def counter_counter(list):
count = Counter(list)
return count
list=sorted([random.randint(0,250) for i in range(300)])
if __name__=='__main__':
from timeit import timeit
print("collections.Defaultdict ",timeit("counter_default_dict(list)", setup="from __main__ import counter_default_dict,list", number=1000))
print("Dict",timeit("counter_dict(list)",setup="from __main__ import counter_dict,list",number=1000))
print("list.count ",timeit("counter_count(list)", setup="from __main__ import counter_count,list", number=1000))
print("collections.Counter.count ",timeit("counter_counter(list)", setup="from __main__ import counter_counter,list", number=1000))
And my results:
我的结果：
collections.Defaultdict
0.06787874956330614
Dict
0.15979115872995675
list.count
1.199258431219126
collections.Counter.count
0.025896202538920665
Do let me know how I can improve the analysis.
请告诉我如何改进分析。
回答by imankalyan
I think using collection library is the easiest way to get it. But If you want to get the frequency dictionary without using it then it's another way,
我认为使用收藏库是获得它的最简单方法。但是如果你想在不使用它的情况下获得频率字典，那么它是另一种方式，
l = [1,4,2,1,2,6,8,2,2]
d ={}
for i in l:
if i in d.keys():
d[i] = 1 + d[i]
else:
d[i] = 1
print (d)
op:
操作：
{1: 2, 4: 1, 2: 4, 6: 1, 8: 1}
