从这里借鉴，这可以在Python 2.7中使用:

from collections import Counter

def Most_Common(lst):
    data = Counter(lst)
    return data.most_common(1)[0][0]

比Alex的解决方案快4-6倍，比newacct提出的一行程序快50倍。

在CPython 3.6+(任何Python 3.7+)上，上面将选择第一个看到的元素。如果你在旧的Python上运行，为了检索列表中第一个出现的元素，你需要进行两次传递来保持顺序:

# Only needed pre-3.6!
def most_common(lst):
    data = Counter(lst)
    return max(lst, key=data.get)

我这样做使用scipy统计模块和lambda:

import scipy.stats
lst = [1,2,3,4,5,6,7,5]
most_freq_val = lambda x: scipy.stats.mode(x)[0][0]
print(most_freq_val(lst))

结果:

 most_freq_val = 5

def popular(L):
C={}
for a in L:
    C[a]=L.count(a)
for b in C.keys():
    if C[b]==max(C.values()):
        return b
L=[2,3,5,3,6,3,6,3,6,3,7,467,4,7,4]
print popular(L)

numbers = [1, 3, 7, 4, 3, 0, 3, 6, 3]
max_repeat_num = max(numbers, key=numbers.count)     *# which number most* frequently
max_repeat = numbers.count(max_repeat_num)           *#how many times*
print(f" the number {max_repeat_num} is repeated{max_repeat} times")

从这里借鉴，这可以在Python 2.7中使用:

from collections import Counter

def Most_Common(lst):
    data = Counter(lst)
    return data.most_common(1)[0][0]

比Alex的解决方案快4-6倍，比newacct提出的一行程序快50倍。

在CPython 3.6+(任何Python 3.7+)上，上面将选择第一个看到的元素。如果你在旧的Python上运行，为了检索列表中第一个出现的元素，你需要进行两次传递来保持顺序:

# Only needed pre-3.6!
def most_common(lst):
    data = Counter(lst)
    return max(lst, key=data.get)

def most_frequent(List):

    counter = 0

    num = List[0]

 

    for i in List:

        curr_frequency = List.count(i)

        if(curr_frequency> counter):

            counter = curr_frequency

            num = i


    return num


List = [2, 1, 2, 2, 1, 3]

print(most_frequent(List))