如果某些元素重复相同的时间，则在列表中出现次数最多

Question

我需要编写一个函数来接收列表并返回最重复元素的列表。我的问题是我写的函数只返回一个重复的元素，我需要所有最重复的元素（如果它们重复相同的时间）。在我放置的示例中，函数需要返回方块、黑桃，但结果我只返回了方块def getHighestOcurrence(listAnyKind):     counter = 0    num = listAnyKind[0]           for i in listAnyKind:         frequency = listAnyKind.count(i)         if (frequency > counter):             counter = frequency             num = i       return num   listAnyKind = ['diamonds', 'spades', 'spades', 'clubs','hearts', 'diamonds']     print(getHighestOcurrence(listAnyKind))

Answer 1

以下是如何使用Countercollections 模块：

from collections import Counter

def getHighestOcurrence(listAnyKind):

    c = Counter(listAnyKind)

    m = max(c.values())

    return [k for k in c if c[k] == m]

listAnyKind = ['diamonds', 'spades', 'spades', 'clubs','hearts', 'diamonds'] 

print(getHighestOcurrence(listAnyKind))

输出：

['diamonds', 'spades']

Answer 2

Numpy 是另一种简短而快速的方法。您找到元素的频率，然后返回频率最高的元素：

import numpy as np

def getHighestOcurrence(listAnyKind): 

    u, freq = np.unique(listAnyKind, return_counts=True)

    return u[np.argwhere(freq==freq.max())].ravel().tolist()

输出：

['diamonds', 'spades']

Answer 3

尝试一下：

def getHighestOcurrence(listAnyKind): 

    counter = 0

    num = listAnyKind[0] 

    frequency = {}  

    for i in set(listAnyKind): 

        frequency[i] = listAnyKind.count(i) 

     

    return [k for k in frequency if frequency[k] == max(frequency.values())]

  

listAnyKind = ['diamonds', 'spades', 'spades', 'clubs','hearts', 'diamonds'] 

print(getHighestOcurrence(listAnyKind))

Answer 4

使用字典在其中存储项目名称及其频率

def getHighestOcurrence(listAnyKind): 

    dic = {}

    for item in listAnyKind:

        if item not in dic:

            dic[item] = 1

        else:

            dic[item]+=1

    max_occu = max(dic.values())

    

    max_item = [k for k, v in dic.items() if v==max_occu]

    return max_item

listAnyKind = ['diamonds', 'spades', 'spades', 'clubs','hearts', 'diamonds'] 

    

print(getHighestOcurrence(listAnyKind))

输出

['diamonds', 'spades']