我需要編寫一個函數(shù)來接收列表并返回最重復(fù)元素的列表。我的問題是我寫的函數(shù)只返回一個重復(fù)的元素，我需要所有最重復(fù)的元素（如果它們重復(fù)相同的時間）。在我放置的示例中，函數(shù)需要返回方塊、黑桃，但結(jié)果我只返回了方塊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))
4 回答
天涯盡頭無女友
TA貢獻1831條經(jīng)驗 獲得超9個贊
以下是如何使用Countercollections 模塊：
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))
輸出：
['diamonds', 'spades']
慕姐4208626
TA貢獻1852條經(jīng)驗 獲得超7個贊
Numpy 是另一種簡短而快速的方法。您找到元素的頻率，然后返回頻率最高的元素：
import numpy as np
def getHighestOcurrence(listAnyKind):
u, freq = np.unique(listAnyKind, return_counts=True)
return u[np.argwhere(freq==freq.max())].ravel().tolist()
輸出：
['diamonds', 'spades']
慕標5832272
TA貢獻1966條經(jīng)驗 獲得超4個贊
嘗試一下：
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))
largeQ
TA貢獻2039條經(jīng)驗 獲得超8個贊
使用字典在其中存儲項目名稱及其頻率
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))
輸出
['diamonds', 'spades']
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