两个数组求差集
let list1 = [1,2,3,4,5,6];
let list2 = [1,2,3];
let dirrence = Array.from(new Set([...list1].filter(x => !new Set(list2).has(x))));
//[4,5,6]let list1 = [1,2,3,4,5,6];
let list2 = [1,2,3,7];
//结果为[4,5,6]
关于几种遍历的比较
forEach:无返回值,对arr操作会改变原来数组 .forEach(item,index,arr),不可return;不能遍历对象
every:判断所有元素是否都符合条件,返回true,false
some:判断至少有一个元素符合条件,返回true,false
sort排序:list.sort((a,b)=>{// return a-b;//正序 return b-a;//倒序});
map:返回新数组,对arr操作不改变原数组.map(item,index,arr) 可以return
filter:过滤,返回满足条件的数组
判断数据是否为数组:Array.isArray(value)
像已知的数组中插入一个变量
let xx = [];
// let oo = _.map(xx, _item => _.defaults({age: 18}, _item));
let oo = _.map(xx, _item => {
let age = 0;
// logic
return {age: age, ..._item};
});