简单的一行代码:

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

>>> 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'

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

# use Decorate, Sort, Undecorate to solve the problem

def most_common(iterable):
    # Make a list with tuples: (item, index)
    # The index will be used later to break ties for most common item.
    lst = [(x, i) for i, x in enumerate(iterable)]
    lst.sort()

    # lst_final will also be a list of tuples: (count, index, item)
    # Sorting on this list will find us the most common item, and the index
    # will break ties so the one listed first wins.  Count is negative so
    # largest count will have lowest value and sort first.
    lst_final = []

    # Get an iterator for our new list...
    itr = iter(lst)

    # ...and pop the first tuple off.  Setup current state vars for loop.
    count = 1
    tup = next(itr)
    x_cur, i_cur = tup

    # Loop over sorted list of tuples, counting occurrences of item.
    for tup in itr:
        # Same item again?
        if x_cur == tup[0]:
            # Yes, same item; increment count
            count += 1
        else:
            # No, new item, so write previous current item to lst_final...
            t = (-count, i_cur, x_cur)
            lst_final.append(t)
            # ...and reset current state vars for loop.
            x_cur, i_cur = tup
            count = 1

    # Write final item after loop ends
    t = (-count, i_cur, x_cur)
    lst_final.append(t)

    lst_final.sort()
    answer = lst_final[0][2]

    return answer

print most_common(['x', 'e', 'a', 'e', 'a', 'e', 'e']) # prints 'e'
print most_common(['goose', 'duck', 'duck', 'goose']) # prints 'goose'

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'

简单的一行解决方案

moc= max([(lst.count(chr),chr) for chr in set(lst)])

它会返回频率最高的元素。