This question already has an answer here:
['a','a','b','c','c','c'] to
[2, 2, 1, 3, 3, 3] and
{'a': 2, 'c': 3, 'b': 1} 可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试):
问题:
回答1:
>>> x=['a','a','b','c','c','c'] >>> map(x.count,x) [2, 2, 1, 3, 3, 3] >>> dict(zip(x,map(x.count,x))) {'a': 2, 'c': 3, 'b': 1} >>> 回答2:
This coding should give the result:
from collections import defaultdict myDict = defaultdict(int) for x in mylist: myDict[x] += 1 Of course if you want the list inbetween result, just get the values from the dict (mydict.values()).
回答3:
Use a set to only count each item once, use the list method count to count them, store them in a dict with the item as key and the occurrence is value.
l=["a","a","b","c","c","c"] d={} for i in set(l): d[i] = l.count(i) print d Output:
{'a': 2, 'c': 3, 'b': 1} 回答4:
On Python ≥2.7 or ≥3.1, we have a built-in data structure collections.Counter to tally a list
>>> l = ['a','a','b','c','c','c'] >>> Counter(l) Counter({'c': 3, 'a': 2, 'b': 1}) It is easy to build [2, 2, 1, 3, 3, 3] afterwards.
>>> c = _ >>> [c[i] for i in l] # or map(c.__getitem__, l) [2, 2, 1, 3, 3, 3] 回答5:
a = ['a','a','b','c','c','c'] b = [a.count(x) for x in a] c = dict(zip(a, b)) I've included Wim answer. Great idea
回答6:
Second one could be just
dict(zip(['a','a','b','c','c','c'], [2, 2, 1, 3, 3, 3])) 回答7:
For the first one:
l = ['a','a','b','c','c','c']
map(l.count,l)
回答8:
d=defaultdict(int) for i in list_to_be_counted: d[i]+=1 l = [d[i] for i in list_to_be_counted]