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 Counterdef 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 npdef 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_itemlistAnyKind = ['diamonds', 'spades', 'spades', 'clubs','hearts', 'diamonds']     print(getHighestOcurrence(listAnyKind))输出['diamonds', 'spades']