#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

从这里借鉴，这可以在Python 2.7中使用:

from collections import Counter

def Most_Common(lst):
    data = Counter(lst)
    return data.most_common(1)[0][0]

比Alex的解决方案快4-6倍，比newacct提出的一行程序快50倍。

在CPython 3.6+(任何Python 3.7+)上，上面将选择第一个看到的元素。如果你在旧的Python上运行，为了检索列表中第一个出现的元素，你需要进行两次传递来保持顺序:

# Only needed pre-3.6!
def most_common(lst):
    data = Counter(lst)
    return max(lst, key=data.get)

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

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'

你可能不再需要这个了，但这是我对一个类似问题所做的。(因为评论，它看起来比实际要长。)

itemList = ['hi', 'hi', 'hello', 'bye']

counter = {}
maxItemCount = 0
for item in itemList:
    try:
        # Referencing this will cause a KeyError exception
        # if it doesn't already exist
        counter[item]
        # ... meaning if we get this far it didn't happen so
        # we'll increment
        counter[item] += 1
    except KeyError:
        # If we got a KeyError we need to create the
        # dictionary key
        counter[item] = 1

    # Keep overwriting maxItemCount with the latest number,
    # if it's higher than the existing itemCount
    if counter[item] > maxItemCount:
        maxItemCount = counter[item]
        mostPopularItem = item

print mostPopularItem