python列表的去重

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#需求：#對列表去重:lis = [2,3,5,3,2,4,8,5,6,7,5](目前為三種方法，持續更新。。。。方法思路來源於https://www.cnblogs.com/nyist-xsk/p/7473236.html，感謝瞭解去重之後解決了一個問題)

lis = [2,3,5,3,2,4,8,5,6,7,5]#方法一：使用set() 集合，這種方法利用set() 集合的去重。出來的結果是進行升序排好的lis1 = list(set(lis))print(lis1) #方法二：使用for迴圈和not in判斷 該方法擷取到的列表跟之前的順序是相同的lis2 = []for i in lis:    if i not in lis2:       lis2.append(i)print(lis2)#方法三：使用類itertools中groupby() 的方法 該方法需要先進行排序（根據排序的規則最後輸出的結果就是）import itertoolslis = [2,3,5,3,2,4,8,5,6,7,5]lis2 = []lis.sort() #不設定reverse=True 預設是升序，設定是降序，這一步是必須的lis1 = itertools.groupby(lis)for k,v in lis1:    lis2.append(k)print(lis2)#[2, 3, 4, 5, 6, 7, 8]升序

#其他方法之後更新

感謝https://www.cnblogs.com/nyist-xsk/p/7473236.html作者提供的思路解決了下面的問題。

#需求：對lis = [2,3,5,3,2,4,8,5,6,7,5]計算元素出現的次數。

lis = [2,3,5,3,2,4,8,5,6,7,5]#去重lis1 = list(set(lis))dic = {}for i in lis1:    dic[i] = lis.count(i) #計數print(dic)

#需求如下：給出一個購物車列表，對列表進行計數輸出結果

"""
goods = [{"name": "電腦", "price": 1999},
         {"name": "羅技滑鼠", "price": 10},
         {"name": "遊艇", "price": 20},
         {"name": "美女", "price": 998},
         {"name": "羅技滑鼠", "price": 20},
         {"name":"電腦","price":1999},
         {"name": "電腦", "price": 2999}]
#輸出結果：
new_goods = [{"name":"電腦","price":1999,"count":2},
         {"name": "羅技滑鼠", "price": 10, "count": 1},
         {"name": "羅技滑鼠", "price": 20, "count": 1},
         {"name": "美女", "price": 998,"count":1},
         {"name": "遊艇", "price": 20,"count":1},
         {"name": "電腦", "price": 2999,"count":1}]
"""

goods = [{"name": "電腦", "price": 1999},         {"name": "羅技滑鼠", "price": 10},         {"name": "遊艇", "price": 20},         {"name": "美女", "price": 998},         {"name": "羅技滑鼠", "price": 20},         {"name":"電腦","price":1999},         {"name": "電腦", "price": 2999}]#先進行去重new_goods = []for good in goods:    if good not in new_goods:        new_goods.append(good)#進行計數for new_good in new_goods:    new_good["count"] = goods.count(new_good)print(new_goods)

 

python列表的去重