In python, how do I take the highest occurrence of something in a list, and sort it that way?

[3, 3, 3, 4, 4, 2]

Would be:

[ (3, 3), (4, 2), (2, 1) ]

开发者_JAVA百科

The output should be sorted by highest count first to lowest count. In this case, 3 to 2 to 1.

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You can use a Counter in Python 2.7+ (this recipe works on 2.5+):

from collections import Counter
print Counter([3, 3, 3, 4, 4, 2]).most_common()
# [(3, 3), (4, 2), (2, 1)]

---

data = [3, 3, 3, 4, 4, 2]
result = []
for entry in set(data):
    result.append((entry, data.count(entry)))
result.sort(key = lambda x: -x[1])
print result

>>[(3, 3), (4, 2), (2, 1)]

---

Try using a collections.Counter:

from collections import Counter
data = [3,4,2,3,4,3]
Counter(data).most_common()

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Why would you choose an O(n**2) algorithm to do this. The alternative to Counter (if you have <2.7) is not too difficult

>>> from operator import itemgetter
>>> from collections import defaultdict
>>> L=[3, 3, 3, 4, 4, 2]
>>> D=defaultdict(int)
>>> for i in L:
...     D[i]+=1
... 
>>> sorted(D.items(), key=itemgetter(1), reverse=True)
[(3, 3), (4, 2), (2, 1)]

---

def myfun(x,y):
    return x[1]-y[1]

list1 = [3, 3, 3, 4, 4, 2]
s1 = set(list1)
newlist = []
for e in s1:
    newlist.append((e,list1.count(e)))
print sorted(newlist,cmp=myfun)

I think, this is what you asked for. Sorry for hurry with the first answer. But just note that cmp argument for sorted is not available in python3