如果它们是不可哈希的，您可以对它们进行排序，并对结果进行一次循环，以计数项(相同的项将彼此相邻)。但是使它们可哈希并使用字典可能更快。

def most_common(lst):
    cur_length = 0
    max_length = 0
    cur_i = 0
    max_i = 0
    cur_item = None
    max_item = None
    for i, item in sorted(enumerate(lst), key=lambda x: x[1]):
        if cur_item is None or cur_item != item:
            if cur_length > max_length or (cur_length == max_length and cur_i < max_i):
                max_length = cur_length
                max_i = cur_i
                max_item = cur_item
            cur_length = 1
            cur_i = i
            cur_item = item
        else:
            cur_length += 1
    if cur_length > max_length or (cur_length == max_length and cur_i < max_i):
        return cur_item
    return max_item

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)

最常见的元素应该是在数组中出现超过N/2次的元素，其中N是len(数组)。下面的技术将以O(n)个时间复杂度完成，只消耗O(1)个辅助空间。

from collections import Counter

def majorityElement(arr):        
    majority_elem = Counter(arr)
    size = len(arr)
    for key, val in majority_elem.items():
        if val > size/2:
            return key
    return -1

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%。

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")

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))