关于javascript: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));

输入

[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; // 以后元素在数组中第一次呈现则返回true    })    return res;  }  console.log(unique_2(baseArray));

输入

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

这个把NaN过滤了，因为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));

输入

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

原理跟一般for循环雷同，先申明一个空数组，如果includes为false，则push以后元素，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))

输入

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

从某种意义上来说，这个办法最完满

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