从数组中删除重复项

Question

我想从数组中删除object的每个副本：

​

  const object = [
    { label: "SUA", value: "sua" },
    { label: "SUA", value: "sua" },
    { label: "Florida", value: "florida" }
  ];
  console.log(object.map(i=> [...new Set(i)]))

​

最后，我必须这样做：

const object = [
    { label: "Florida", value: "florida" }
  ];

如何在我的代码中做到这一点？

Answer 1

您可以使用2 for循环来实现您的功能。

let result = [];
for(let i=0;i<object.length ;i++) {
    let isDuplicate = false;
    for(let j=0; j<object.length;j++) {
        if(object[i].label === object[j].label && object[i].value === object[j].value && i !== j) {
            isDuplicate = true;   
        }
    }
    if(!isDuplicate) result.push(object[i]);
}

Answer 2

不管您希望在数组中保留多少个元素的副本，都可以通过定义创建一个映射Map<Location, number>，并将每个Location对象出现的次数列成表格。然后，只取每个元素并将其附加到数组中一次。

type Location = {label: string, value: string};

const object: Location[] = [
    { label: "SUA", value: "sua" },
    { label: "SUA", value: "sua" },
    { label: "Florida", value: "florida" }
];
const objects: Map<Location, number> = new Map();

for (const location of object)
    if (objects.has(location))
        objects.set(location, objects.get(location) + 1);
    else
        objects.set(location, 1);

const filtered: Location[] = [];

for (const location of objects)
    if (location[1] === 1) // You change 1 to any value depending on how many copies of each you'd like.
        filtered.push(location[0];

备注为了清晰起见，这是TypeScript，但概念是相同的。

Answer 3

快捷方式：

​

const object = [
  { label: "SUA", value: "sua" },
  { label: "SUA", value: "sua" },
  { label: "Florida", value: "florida" }
];

const a = object.filter((v, i, a) => a.findIndex(t => (t.label === 
v.label && t.value === v.value)) === i);

console.log(a);

​