Question

我想从数组中删除对象的每个副本：

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

最后我得像：

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

如何在我的代码中执行此操作？

Answer 1

您可以使用2个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

无论您希望在数组中保留每个元素的副本数量是多少，都应根据定义Map<Location, number>创建一个Map并列出每个Location对象的出现次数。然后，获取每个元素并将其仅附加到数组一次。


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];

注意：为清晰起见，这是TypeScript，但概念相同。

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);

Answer 4

您可以使用ES6集-

const unique = [...new Set(object.map(({value}) => value))].map(e => object.find(({value}) => value == e));

console.log("unique",unique);

//[{label: "SUA", value: "sua"},{label: "Florida", value: "florida"}]

对于您的问题，您可以简单地使用reduce-

const uniq =  object.reduce((a,b)=>{return a.value === b.value ? {} :[b]},[])

console.log(uniq);

//[{label: "Florida", value: "florida"}]

Answer 5

lodash的简单解决方案

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

result = _.xor(_.uniqWith(object,_.isEqual),object)

从阵列中删除重复项

5 个答案: