怎么把两个数组合并成一个php

假设我们有两个list,分别是: (def a [“one” “two” “three”]) (def b [1 2 3]) 我们要把它们合为一个键值对应的map,做法很简单: 1. 先将a和b合为一个一一对应的list: (map vector a b) => (["one" 1] ["two" 2] ["three" 3]) 2. 然后再将list转化成为map: (into {} (map vector a b)) => {&q

using System;using System.Collections.Generic;using System.Linq;using System.Collections;using System.Text;using System.Diagnostics; namespace Hecha.Test{ class Program { static void Main(string[] args) { List<string>[] aa = new List<string>[5

JavaArray合并数组 目录(?)[+] 一apache-commons 二Systemarraycopy 三ArrayscopyOf 四ArraynewInstance http://freewind.me/blog/20110922/350.html 在Java中,如何把两个String[]合并为一个? 看起来是一个很简单的问题.但是如何才能把代码写得高效简洁,却还是值得思考的.这里介绍四种方法,请参考选用. 一.apache-commons 这是最简单的办法.在apache-commo

需求 有这么两个数组 1 let metrodates = [ 2 "2008-01", 3 "2008-02", 4 "2008-03",..ect 5 ]; 6 let figures = [ 7 0, 8 0.555, 9 0.293,..ect 10 ] 想要这样的结果 1 let result = [ 2 {data: 0, date: "2008-01"}, 3 {data: 0.555, date: "

解决方案: NSMutableDictionary *targetMutableDictionary = [mutableDictionary1 copy]; [targetMutableDictionary addEntriesFromDictionary:mutableDictionary2];

//调用方法: var targetObject = [{name:"11",age:11}]; var sourceObject = [{name:"22",age:22}]; var result = $.extend(true,targetObject,sourceObject); //手写拼接: var str1=[{name:"11",age:11}]; var str2=[{name:"22",age:22}];

1, add 2, 用list的extend方法,L1.extend(L2),该方法将参数L2的全部元素添加到L1的尾部,例如: 3, 用切片(slice)操作,L1[len(L1):len(L1)] = L2和上面的方法等价, 但切片方法用起来更灵活,可以插入到头部,或其他任意部位,也可以加到中间. 原文地址:https://www.cnblogs.com/qingyuanjushi/p/8409949.html

select c.ClientID,c.Name,case MAX(c.Grade) when 0 then '厂家直营' when 1 then '总代理' when 2 then '省级代理'when 3 then '省级代理' end Grade,MAX(d.CooperationIntention) as CooperationIntention,COUNT(*) as number,ISNULL(c.IsEmphasis,0) IsEmphasis from SG_Client c i

两个已经从小到大的数组合并成为一个从小到大排序的数组 1 #include<stdio.h> 2 int main() 3 { 4 int m,n,i,j,k,tem=0; 5 printf("这两个数组分别有多少个数:\n"); 6 scanf("%d%d",&m,&n); 7 int a[m],b[n],c[m+n]; 8 printf("从小到大输入%d个数:\n",m); 9 for(i=0;i<m;i+