你想要的在统计中被称为模式，Python当然有一个内置函数来为你做这件事:

>>> from statistics import mode
>>> mode([1, 2, 2, 3, 3, 3, 3, 3, 4, 5, 6, 6, 6])
3

请注意，如果没有“最常见元素”，例如前两个元素并列的情况，这将在Python上引发StatisticsError <=3.7，从3.8开始，它将返回遇到的第一个。

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。

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

如果没有最低索引的要求，您可以使用集合。计数器:

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'

ans  = [1, 1, 0, 0, 1, 1]
all_ans = {ans.count(ans[i]): ans[i] for i in range(len(ans))}
print(all_ans)

all_ans={4: 1, 2: 0}
max_key = max(all_ans.keys())

4

print(all_ans[max_key])

1

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”