数组去重

for循环？NONONO

ES678来袭

先来一个数组

var baseArray = [1, 1, '1', '1', null, null,
    undefined, undefined,
    new String('1'), new String('1'),
    /a/, /a/,
    NaN, NaN
  ];

set

let unique_1 = arr => [...new Set(arr)];
  // let unique_1 = arr => Array.from(new Set(arr))
  console.log(unique_1(baseArray));

输出

[1, "1", null, undefined, String, String, /a/, /a/, NaN]

filter

let unique_2 = arr => {
    let res = arr.filter((item, index, array) => {
      return array.indexOf(item) === index; // 当前元素在数组中第一次出现则返回true
    })
    return res;
  }
  console.log(unique_2(baseArray));

输出

[1, "1", null, undefined, String, String, /a/, /a/]

这个把NaN过滤了，由于Array.indexOf(NaN)值为-1

reduce

let unique_3 = arr => arr.reduce((pre, cur) => pre.includes(cur) ? pre : [...pre, cur], []);
  console.log(unique_3(baseArray));

输出

[1, "1", null, undefined, String, String, /a/, /a/, NaN]

原理跟普通for循环相同，先声明一个空数组，若是includes为false，则push当前元素，只不过逼格高那么一丢丢丢丢

键值对

let unique_4 = arr => {
    let obj = {};
    return  arr.filter(item => Reflect.has(obj, typeof item + item) ? false : (obj[typeof item + item] = true));
  }
  console.log(unique_4(baseArray))

输出

[1, "1", null, undefined, String, /a/, NaN]

从某种意义上来讲，这个方法最完美

obj[typeof item + item]写法原因：对象的key必须是字符串