#!/usr/bin/env python # encoding: utf-8 #字典去重小代码 import sys import os import platform try: pass except: print '''you have something wrong this is a simple jiaoben ''' sys.exit() why = 'why.txt' for i in xrange(len(sys.argv)): if(i>=1): other = sys.ar…
一.方法1 代码如下 复制代码 ids = [1,2,3,3,4,2,3,4,5,6,1] news_ids = [] for id in ids: if id not in news_ids: news_ids.append(id) print news_ids 思路看起来比较清晰简单 ,也可以保持之前的排列顺序. 二.方法2 通过set方法进行处理 代码如下 复制代码 ids = [1,4,3,3,4,2,3,4,5,6,1] ids = list(set(ids…
一.方法1 代码如下 复制代码 ids = [1,2,3,3,4,2,3,4,5,6,1] news_ids = [] for id in ids: if id not in news_ids: news_ids.append(id) print news_ids 思路看起来比较清晰简单 ,也可以保持之前的排列顺序. 二.方法2 通过set方法进行处理 代码如下 复制代码 ids = [1,4,3,3,4,2,3,4,5,6,1] ids = list(set(ids…
今天实习的web大表哥说帮我看环境不过前提是要我帮他写个python合并列表的demo, 大概思路就是利用zip库进行keys和values的遍历,然后在输出就行 key1={'name1':'小明','name2':'小红'} key2={'小明':'[men,20]','小红':'[women,30]'} for k,v in zip(key1.values(),key1.keys()): for i, j in zip(key2.values(), key2.keys()): if k =…
1. 列表去重 li = [] for item in my_list: if item not in li: li.append(item) 2.集合去重 list(set(my_list)) 3.字典去重 dict1 = dict.fromkeys(my_list, '') li = dict1.keys…
1 a = [,,,,,,,,,,] a1 = [] for i in a: if i not a1: a1.append(i) else: continue 2 a = [,,,,,,,,,] a1 = set(a) 3 a = [,,,,,,,,,,] for i in a: : del a[a.index(i)] 4 import itertools a = [,,,,,,,] a1 = [] for i in itertools.groupby(a): a1.append(i)…
一.一个列表中可能含有重复元素,使用set()可以实现列表的去重处理,但是无法知道哪些元素是重复的,下面的函数用于找出哪些元素重复了,以及重复的次数. 代码: from collections import Counter #引入Counter a = [1, 2, 3, 3, 4, 4] b = dict(Counter(a)) print(b) print ([key for key,value in b.items() if value > 1]) #只展示重复元素 print ({key…
1.如何实现对python列表去重,并保持原来顺序 li = [1,2,5,3,1,6,3,8,0,3,2,4] l = [] for i in li: if i not in l: l.append(i) print(l) 方法一 li = [1,2,5,3,1,6,3,8,0,3,2,4] l = list(set(li)) l.sort(key=lambda i:li.index(i)) print(l) # 先去重,再按照li的索引进行排序 方法二 li = [1,2,5,3,1,6,3…
