def most_common(lst):
    if max([lst.count(i)for i in lst]) == 1:
        return False
    else:
        return max(set(lst), key=lst.count)

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)

#This will return the list sorted by frequency:

def orderByFrequency(list):

    listUniqueValues = np.unique(list)
    listQty = []
    listOrderedByFrequency = []
    
    for i in range(len(listUniqueValues)):
        listQty.append(list.count(listUniqueValues[i]))
    for i in range(len(listQty)):
        index_bigger = np.argmax(listQty)
        for j in range(listQty[index_bigger]):
            listOrderedByFrequency.append(listUniqueValues[index_bigger])
        listQty[index_bigger] = -1
    return listOrderedByFrequency

#And this will return a list with the most frequent values in a list:

def getMostFrequentValues(list):
    
    if (len(list) <= 1):
        return list
    
    list_most_frequent = []
    list_ordered_by_frequency = orderByFrequency(list)
    
    list_most_frequent.append(list_ordered_by_frequency[0])
    frequency = list_ordered_by_frequency.count(list_ordered_by_frequency[0])
    
    index = 0
    while(index < len(list_ordered_by_frequency)):
        index = index + frequency
        
        if(index < len(list_ordered_by_frequency)):
            testValue = list_ordered_by_frequency[index]
            testValueFrequency = list_ordered_by_frequency.count(testValue)
            
            if (testValueFrequency == frequency):
                list_most_frequent.append(testValue)
            else:
                break    
    
    return list_most_frequent

#tests:
print(getMostFrequentValues([]))
print(getMostFrequentValues([1]))
print(getMostFrequentValues([1,1]))
print(getMostFrequentValues([2,1]))
print(getMostFrequentValues([2,2,1]))
print(getMostFrequentValues([1,2,1,2]))
print(getMostFrequentValues([1,2,1,2,2]))
print(getMostFrequentValues([3,2,3,5,6,3,2,2]))
print(getMostFrequentValues([1,2,2,60,50,3,3,50,3,4,50,4,4,60,60]))

Results:
[]
[1]
[1]
[1, 2]
[2]
[1, 2]
[2]
[2, 3]
[3, 4, 50, 60]

我这样做使用scipy统计模块和lambda:

import scipy.stats
lst = [1,2,3,4,5,6,7,5]
most_freq_val = lambda x: scipy.stats.mode(x)[0][0]
print(most_freq_val(lst))

结果:

 most_freq_val = 5

# 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'