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)

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

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

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

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