你想要的在统计中被称为模式，Python当然有一个内置函数来为你做这件事:

>>> from statistics import mode
>>> mode([1, 2, 2, 3, 3, 3, 3, 3, 4, 5, 6, 6, 6])
3

请注意，如果没有“最常见元素”，例如前两个元素并列的情况，这将在Python上引发StatisticsError <=3.7，从3.8开始，它将返回遇到的第一个。

一行程序:

def most_common (lst):
    return max(((item, lst.count(item)) for item in set(lst)), key=lambda a: a[1])[0]

#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%。

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

ans  = [1, 1, 0, 0, 1, 1]
all_ans = {ans.count(ans[i]): ans[i] for i in range(len(ans))}
print(all_ans)

all_ans={4: 1, 2: 0}
max_key = max(all_ans.keys())

4

print(all_ans[max_key])

1