數組去重

for循環？NONONO

ES678來襲

先來一個數組

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));

輸出

[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; // 當前元素在數組中第一次出現則返回true
    })
    return res;
  }
  console.log(unique_2(baseArray));

輸出

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

這個把NaN過濾了，由於Array.indexOf(NaN)值爲-1

reduce

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

輸出

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

原理跟普通for循環相同，先聲明一個空數組，若是includes爲false，則push當前元素，只不過逼格高那麼一丟丟丟丟

鍵值對

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))

輸出

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

從某種意義上來講，這個方法最完美

obj[typeof item + item]寫法原因：對象的key必須是字符串