如何从python中的列表中获取最常见的元素

Question

我有一个如此处所示的列表。 a=[1936,2401,2916,4761,9216,9216,9604,9801]我想得到更多重复的值。 在这里它是'9216'我怎么能得到这个值？ 谢谢

Answer 1

您可以使用collections.Counter ：

from collections import Counter

a = [1936, 2401, 2916, 4761, 9216, 9216, 9604, 9801] 

c = Counter(a)

print(c.most_common(1)) # the one most common element... 2 would mean the 2 most common
[(9216, 2)] # a set containing the element, and it's count in 'a'

来自文档：

Answer 2

https://docs.python.org/2.7/library/collections.html#counter

来自收藏夹进口柜台柜台（a）.most_common（1）

Answer 3

这是另一个不使用计数器的人

a=[1936,2401,2916,4761,9216,9216,9604,9801]
frequency = {}
for element in a:
    frequency[element] = frequency.get(element, 0) + 1
# create a list of keys and sort the list
# all words are lower case already
keyList = frequency.keys()
keyList.sort()
print "Frequency of each word in the word list (sorted):"
for keyElement in keyList:
    print "%-10s %d" % (keyElement, frequency[keyElement])

Answer 4

有两种标准库方法可以做到这一点：

statistics.mode ：

from statistics import mode
most_common = mode([3, 2, 2, 2, 1])  # 2
most_common = mode([3, 2])  # StatisticsError: no unique mode

collections.Counter.most_common ：

from collections import Counter
most_common, count = Counter([3, 2, 2, 2, 1]).most_common(1)[0]  # 2, 3
most_common, count = Counter([3, 2]).most_common(1)[0]  # 3, 1

两者在性能方面都是相同的，但是当没有唯一的最常见元素时，第一个引发异常，第二个也返回频率。