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%。

在这里:

def most_common(l):
    max = 0
    maxitem = None
    for x in set(l):
        count =  l.count(x)
        if count > max:
            max = count
            maxitem = x
    return maxitem

我有一种模糊的感觉，在标准库的某个地方有一个方法可以给你每个元素的计数，但我找不到它。

如果排序和哈希都不可行，这是一个明显的缓慢的解决方案(O(n²))，但相等比较(==)可用:

def most_common(items):
  if not items:
    raise ValueError
  fitems = [] 
  best_idx = 0
  for item in items:   
    item_missing = True
    i = 0
    for fitem in fitems:  
      if fitem[0] == item:
        fitem[1] += 1
        d = fitem[1] - fitems[best_idx][1]
        if d > 0 or (d == 0 and fitems[best_idx][2] > fitem[2]):
          best_idx = i
        item_missing = False
        break
      i += 1
    if item_missing:
      fitems.append([item, 1, i])
  return items[best_idx]

但是，如果你的列表(n)的长度很大，那么让你的项目可哈希或可排序(正如其他答案所建议的那样)几乎总是能更快地找到最常见的元素。哈希时平均为O(n)，排序时最差为O(n*log(n))。

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'

这是O(n)解。

mydict   = {}
cnt, itm = 0, ''
for item in reversed(lst):
     mydict[item] = mydict.get(item, 0) + 1
     if mydict[item] >= cnt :
         cnt, itm = mydict[item], item

print itm

(reversed用于确保它返回最低的索引项)

#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]