以Luiz的回答为基础，但满足“在抽取索引最低的项目时应返回”的条件:

from statistics import mode, StatisticsError

def most_common(l):
    try:
        return mode(l)
    except StatisticsError as e:
        # will only return the first element if no unique mode found
        if 'no unique mode' in e.args[0]:
            return l[0]
        # this is for "StatisticsError: no mode for empty data"
        # after calling mode([])
        raise

例子:

>>> most_common(['a', 'b', 'b'])
'b'
>>> most_common([1, 2])
1
>>> most_common([])
StatisticsError: no mode for empty data

简单的一行代码:

def most_common(lst):
    return max(set(lst), key=lst.count)

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

对列表的一个副本排序并找到运行时间最长的。您可以在用每个元素的索引对列表排序之前对其进行修饰，然后在并列的情况下选择从最低索引开始的运行。

如果排序和哈希都不可行，这是一个明显的缓慢的解决方案(O(n²))，但相等比较(==)可用:

def most_common(items):
  if not items:
    raise ValueError
  fitems = [] 
  best_idx = 0
  for item in items:   
    item_missing = True
    i = 0
    for fitem in fitems:  
      if fitem[0] == item:
        fitem[1] += 1
        d = fitem[1] - fitems[best_idx][1]
        if d > 0 or (d == 0 and fitems[best_idx][2] > fitem[2]):
          best_idx = i
        item_missing = False
        break
      i += 1
    if item_missing:
      fitems.append([item, 1, i])
  return items[best_idx]

但是，如果你的列表(n)的长度很大，那么让你的项目可哈希或可排序(正如其他答案所建议的那样)几乎总是能更快地找到最常见的元素。哈希时平均为O(n)，排序时最差为O(n*log(n))。

>>> 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'