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

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]

def most_frequent(List):

    counter = 0

    num = List[0]

 

    for i in List:

        curr_frequency = List.count(i)

        if(curr_frequency> counter):

            counter = curr_frequency

            num = i


    return num


List = [2, 1, 2, 2, 1, 3]

print(most_frequent(List))

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

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