Python：获取列表中最常见的项目

Question

我有一个元组列表，我想获取最频繁出现的元组，但是如果有“联合获胜者”，它应该随机选择它们。

tups = [ (1,2), (3,4), (5,6), (1,2), (3,4) ] 

所以我想要的东西，会在回报（1,2）或随机对以上列举

Answer 1

您可以先使用Counter来查找最重复的元组。然后找到所需的元组并最终随机获得第一个值。

from collections import Counter 
import random 

tups = [ (1,2), (3,4), (5,6), (1,2), (3,4) ] 
lst = Counter(tups).most_common() 
highest_count = max([i[1] for i in lst]) 
values = [i[0] for i in lst if i[1] == highest_count] 
random.shuffle(values) 
print values[0] 

Answer 2

（3,4）你可以先排序列表以获得通过频率排序的元组。之后，线性扫描可以从列表中获得最频繁的元组。总体时间O(nlogn)

>>> tups = [ (1,2), (3,4), (5,6), (1,2), (3,4) ] 
>>> 
>>> sorted(tups) 
[(1, 2), (1, 2), (3, 4), (3, 4), (5, 6)] 

Answer 3

使用collections.Counter：

>>> collections.Counter([ (1,2), (3,4), (5,6), (1,2), (3,4) ]).most_common()[0] 
((1, 2), 2) 

这是O(n log(n))。

Answer 4

这一个应该做你的任务在o(n)时间：

>>> from random import shuffle 
>>> from collections import Counter 
>>> 
>>> tups = [(1,2), (3,4), (5,6), (1,2), (3,4)] 
>>> c = Counter(tups)       # count frequencies 
>>> m = max(v for _, v in c.iteritems())   # get max frq 
>>> r = [k for k, v in c.iteritems() if v == m] # all items with highest frq 
>>> shuffle(r)         # if you really need random - shuffle 
>>> print r[0] 
(3, 4) 

Answer 5

最常见的collections.Counter，然后随机选择计数：

import collections 
import random 

lis = [ (1,2), (3,4), (5,6), (1,2), (3,4) ] # Test data 
cmn = collections.Counter(lis).most_common() # Numbering based on occurrence 
most = [e for e in cmn if (e[1] == cmn[0][1])] # List of those most common 
print(random.choice(most)[0]) # Print one of the most common at random