#!/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
1.去重 for(var q = 0;q<jsonArr.length;q++){ for(var e=0;e<jsonArr[q].data.length;e++){ var matchCode = jsonArr[q].data[e]; var chua = jsonArr[q].jsonObj; var isExist = false; var isC = false; //去重多少场次 for(var m = 0; m < an.length; m ++){ if(j
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 =
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