不改变原数组，返回新数组的方法

concat()

合并两个数组

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
let arr1 = [1,2,3];
let arr2 = [4,5,6];
let twoArr = [1,[2]]
let arr3 = arr1.concat(arr2);
let arr4 = arr2.concat(arr1);
let arr5 = arr1.concat(twoArr);
console.log(arr1,arr2,arr3,arr4,arr5);
// [ 1, 2, 3 ] [ 4, 5, 6 ] [ 1, 2, 3, 4, 5, 6 ] [ 4, 5, 6, 1, 2, 3 ] [ 1, 2, 3, 1, [ 2 ] ]
``` 


### join()

> 合并为字符串
``` JavaScript
let joinArr1 = [1,2,'A','B']
let joinArr2 = [3,4,'c','d']
let join1 = joinArr1.join('');
let join2 = joinArr2.join('');
console.log(join1,join2);
// 12AB 34cd

slice()

1
2
3
4
5
6
7
8
9
let sliceArr1 = [1,2,3,4,5,6];
let sliceArr2 = ['A','B','C','D'];
let sliceStr = 'abcdef'
let slice1 = sliceArr1.slice(1,2);
let slice2 = sliceArr2.slice(1,3);
let slice3 = sliceStr.slice(1,3);
//字符串的则是单个数如果是负数从后面截取
let slice4 = sliceStr.slice(-2);
console.log(slice1,slice2,slice3,slice4);

every()

执行一个回调函数，如果数组中有一个不满足，则every返回false，都满足则返回true；

1
2
3
4
5
6
7
let evArr = [1,2,3];
let every1 = evArr.every((e)=> {
    console.log(e,'e');
    // 在这里记得是需要return的，e表示数组循环出来的每个值
    return e < 4
})
console.log(every1);

map()

1
2
3
4
5
6
7
let mapArr = [1,2,3,4];
let map1 = mapArr.map((a,index,c) => {
    // 参数一为值，参数二为索引，参数三为原数组
    console.log(a,index,c)
    return a + 1;
})
console.log(map1)

some()

除非全部不满足，否则则还是为true

1
2
3
4
5
6
let someArr = [1,2,3];
let some1 = someArr.some((e)=> {
    return e <2;
});
console.log(some1);
// true

filter()

返回了满足的新数组

1
2
3
let filterArr = [1,2,3];
let filter1 = filterArr.filter((e)=> e <2);
console.log(filter1);//[ 1 ]

改变原数组的方法

forEach()

pop()

push()

reverse()

shift()

unshift()

sort()

splice()

数组去重

reduce去重

1
2
3
4
5
6
7
8
9
10
const distinct = arr => arr.sort().reduce( (init, current) => {
    
    if (init.length === 0 || init[init.length - 1] !== current) {
        init.push( current );
    }
    return init;
}, []);

let arr = [1,2,1,2,3,5,4,5,3,4,4,4,4];
distinct(arr); // [1, 2, 3, 4, 5]

filter去重

1
2
3
4
5
6
const distinct = arr => arr.filter( (element, index, self) => {
    return self.indexOf( element ) === index;
});

let arr = [1,2,1,2,3,5,4,5,3,4,4,4,4];
distinct(arr); // [1, 2, 3, 5, 4]

多维数组降维

1
2
3
4
5
const flattenDeep = arr => Array.isArray(arr)
  ? arr.reduce( (a, b) => [...a, ...flattenDeep(b)] , [])
  : [arr]

flattenDeep([1, [[2], [3, [4]], 5]]); // [1, 2, 3, 4, 5]