Array.prototype.getUnique = function() {
var o = {}, a = [], i, e;
for (i = 0; e = this[i]; i++) {o[e] = 1};
for (e in o) {a.push (e)};
return a;
}
为去除重复项，可能有两种情况。首先，所有数据都不是对象；其次，所有数据都是对象。
如果所有数据都是任何一种原始数据类型，如int、float、string等，则可以按照以下方式进行操作：
const uniqueArray = [...new Set(oldArray)]
但是假设你的数组包含像下面这样的JS对象
{
id: 1,
name: 'rony',
email: 'rony@example.com'
}
let uniqueIds = [];
const uniqueUsers = oldArray.filter(item => {
if(uniqueIds.includes(item.id)){
return false;
}else{
uniqueIds.push(item.id);
return true;
}
})
uniqueIds数组上保留跟踪键即可。使用ES6（一行代码）
基本类型值的数组
let originalArr= ['a', 1, 'a', 2, '1'];
let uniqueArr = [...new Set(originalArr)];
对象数组
let uniqueObjArr = [...new Map(originalObjArr.map((item) => [item["propertyName"], item])).values()];
const ObjArray = [
{
name: "Eva Devore",
character: "Evandra",
episodes: 15,
},
{
name: "Alessia Medina",
character: "Nixie",
episodes: 15,
},
{
name: "Kendall Drury",
character: "DM",
episodes: 15,
},
{
name: "Thomas Taufan",
character: "Antrius",
episodes: 14,
},
{
name: "Alessia Medina",
character: "Nixie",
episodes: 15,
},
];
let uniqueObjArray = [...new Map(ObjArray.map((item) => [item["id"], item])).values()];
var a = [1,5,1,6,4,5,2,5,4,3,1,2,6,6,3,3,2,4];
// note: jQuery's filter params are opposite of javascript's native implementation :(
var unique = $.makeArray($(a).filter(function(i,itm){
// note: 'index', not 'indexOf'
return i == $(a).index(itm);
}));
// unique: [1, 5, 6, 4, 2, 3]
最初的回答位于：jQuery函数获取数组中所有唯一元素？
eq函数会阻止我们以有效的方式利用Set来确定重复项，因为Set会退化到===。正如你所知道的那样，===不适用于引用类型。所以我们有点陷入困境了，对吧？Set查找的内容即可。例如，我们可以使用哈希函数，或者JSON.stringify数据结构（如果它不包含任何函数）。Object的引用。const dedupeOn = k => xs => {
const s = new Set();
return xs.filter(o =>
s.has(o[k])
? null
: (s.add(o[k]), o[k]));
};
const dedupeBy = f => xs => {
const s = new Set();
return xs.filter(x => {
const r = f(x);
return s.has(r)
? null
: (s.add(r), x);
});
};
const xs = [{foo: "a"}, {foo: "b"}, {foo: "A"}, {foo: "b"}, {foo: "c"}];
console.log(
dedupeOn("foo") (xs)); // [{foo: "a"}, {foo: "b"}, {foo: "A"}, {foo: "c"}]
console.log(
dedupeBy(o => o.foo.toLowerCase()) (xs)); // [{foo: "a"}, {foo: "b"}, {foo: "c"}]这并不是对原问题的直接字面回答，因为我更喜欢在第一次生成数组时就没有重复的值。所以这里是我的UniqueArray：
class UniqueArray extends Array {
constructor(...args) {
super(...new Set(args));
}
push(...args) {
for (const a of args) if (!this.includes(a)) super.push(a);
return this.length;
}
unshift(...args) {
for (const a of args.reverse()) if (!this.includes(a)) super.unshift(a);
return this.length;
}
concat(...args) {
var r = new UniqueArray(...this);
for (const a of args) r.push(...a);
return r;
}
}
> a = new UniqueArray(1,2,3,1,2,4,5,1)
UniqueArray(5) [ 1, 2, 3, 4, 5 ]
> a.push(1,4,6)
6
> a
UniqueArray(6) [ 1, 2, 3, 4, 5, 6 ]
> a.unshift(1)
6
> a
UniqueArray(6) [ 1, 2, 3, 4, 5, 6 ]
> a.unshift(0)
7
> a
UniqueArray(7) [
0, 1, 2, 3,
4, 5, 6
]
> a.concat(2,3,7)
UniqueArray(8) [
0, 1, 2, 3,
4, 5, 6, 7
]
var myArray = ["a",2, "a", 2, "b", "1"];
const uniques = [];
myArray.forEach((t) => !uniques.includes(t) && uniques.push(t));
console.log(uniques);const removeDuplicates = (arr) =>
arr.reduce(
([set, acc], item) => set.has(item) ? [set, acc] : [set.add(item), (acc.push(item), acc)],
[new Set(), []]
)[1][...new Set(array)]一样。如果项是包含id属性的对象，则应该这样做：
const removeDuplicates = (arr) =>
arr.reduce(
([set, acc], item) => set.has(item.id) ? [set, acc] : [set.add(item.id), (acc.push(item), acc)],
[new Set(), []]
)[1]let ar = [1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 2, 1];
let unique = ar.filter((value, index) => {
return ar.indexOf(value) == index;
});
console.log(unique);使用lodash和身份lambda函数来完成，只需在使用对象之前定义即可
const _ = require('lodash');
...
_.uniqBy([{a:1,b:2},{a:1,b:2},{a:1,b:3}], v=>v.a.toString()+v.b.toString())
_.uniq([1,2,3,3,'a','a','x'])
[{a:1,b:2},{a:1,b:3}]
[1,2,3,'a','x']
(这是最简单的方法)
o表示对象，a表示数组，i表示索引，而e则表示嗯，某个东西:P - Mottie