标签：相同 好的 设置 itertools new end log app import
#需求：#对列表去重:lis = [2,3,5,3,2,4,8,5,6,7,5](目前为三种方法，持续更新。。。。方法思路来源于https://www.cnblogs.com/nyist-xsk/p/7473236.html，感谢了解去重之后解决了一个问题)
lis = [2,3,5,3,2,4,8,5,6,7,5] #方法一：使用set() 集合，这种方法利用set() 集合的去重。出来的结果是进行升序排好的 lis1 = list(set(lis)) print(lis1) #方法二：使用for循环和not in判断 该方法获取到的列表跟之前的顺序是相同的 lis2 = [] for i in lis: if i not in lis2: lis2.append(i) print(lis2) #方法三：使用类itertools中groupby() 的方法 该方法需要先进行排序（根据排序的规则最后输出的结果就是） import itertools lis = [2,3,5,3,2,4,8,5,6,7,5] lis2 = [] lis.sort() #不设置reverse=True 默认是升序，设置是降序，这一步是必须的 lis1 = itertools.groupby(lis) for k,v in lis1: lis2.append(k) print(lis2)#[2, 3, 4, 5, 6, 7, 8]升序
#其他方法之后更新
感谢https://www.cnblogs.com/nyist-xsk/p/7473236.html作者提供的思路解决了下面的问题。
#需求：对lis = [2,3,5,3,2,4,8,5,6,7,5]计算元素出现的次数。
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) #计数 print(dic)
#需求如下：给出一个购物车列表，对列表进行计数输出结果
"""
goods = [{"name": "电脑", "price": 1999},
{"name": "罗技鼠标", "price": 10},
{"name": "游艇", "price": 20},
{"name": "美女", "price": 998},
{"name": "罗技鼠标", "price": 20},
{"name":"电脑","price":1999},
{"name": "电脑", "price": 2999}]
#输出结果：
new_goods = [{"name":"电脑","price":1999,"count":2},
{"name": "罗技鼠标", "price": 10, "count": 1},
{"name": "罗技鼠标", "price": 20, "count": 1},
{"name": "美女", "price": 998,"count":1},
{"name": "游艇", "price": 20,"count":1},
{"name": "电脑", "price": 2999,"count":1}]
"""
goods = [{"name": "电脑", "price": 1999},
{"name": "罗技鼠标", "price": 10},
{"name": "游艇", "price": 20},
{"name": "美女", "price": 998},
{"name": "罗技鼠标", "price": 20},
{"name":"电脑","price":1999},
{"name": "电脑", "price": 2999}]
#先进行去重
new_goods = []
for good in goods:
if good not in new_goods:
new_goods.append(good)
#进行计数
for new_good in new_goods:
new_good["count"] = goods.count(new_good)
print(new_goods)
标签：相同 好的 设置 itertools new end log app import
原文地址：https://www.cnblogs.com/yangshier/p/9463213.html
