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")

>>> li  = ['goose', 'duck', 'duck']

>>> def foo(li):
         st = set(li)
         mx = -1
         for each in st:
             temp = li.count(each):
             if mx < temp:
                 mx = temp 
                 h = each 
         return h

>>> foo(li)
'duck'

我在最近的一个项目中需要这样做。我承认，我无法理解Alex的回答，所以这就是我最后得到的答案。

def mostPopular(l):
    mpEl=None
    mpIndex=0
    mpCount=0
    curEl=None
    curCount=0
    for i, el in sorted(enumerate(l), key=lambda x: (x[1], x[0]), reverse=True):
        curCount=curCount+1 if el==curEl else 1
        curEl=el
        if curCount>mpCount \
        or (curCount==mpCount and i<mpIndex):
            mpEl=curEl
            mpIndex=i
            mpCount=curCount
    return mpEl, mpCount, mpIndex

我根据Alex的解决方案计时，对于短列表，它要快10-15%，但一旦超过100个或更多元素(测试多达20万个)，它就会慢20%。

这是一个很简单的解，时间复杂度是线性的

L =['鹅'，'鸭'，'鸭']

def most_common (L):

current_winner = 0
max_repeated = None
for i in L:
    amount_times = L.count(i)
    if amount_times > current_winner:
        current_winner = amount_times
        max_repeated = i

return max_repeated

打印(most_common (L)

“鸭子”

number是列表中重复次数最多的元素吗

def mostCommonElement(list):
  count = {} // dict holder
  max = 0 // keep track of the count by key
  result = None // holder when count is greater than max
  for i in list:
    if i not in count:
      count[i] = 1
    else:
      count[i] += 1
    if count[i] > max:
      max = count[i]
      result = i
  return result

主要组合(“a”、“b”、“a”、“c”)->“a”

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)