javascript - 从js中的两个数组中删除重复的哈希？

Question

有两个数组:

arr1 = [
        {name: "gunner", id: 639, label: "group"}, 
        {name: "Gunners", id: 1313, label: "channel", isPrivate: false}
       ]

arr2 = [
        {name: "gunner", id: 639, label: "group"}
       ]

需要删除 arr1 中存在于 arr2 中的所有值？

最终 arr1 将是，因为 arr2 已经有 {name: "gunner", id: 639, label: "group"}:

arr1 = [
        {name: "Gunners", id: 1313, label: "channel", isPrivate: false}
       ]

我试过了，但没有按预期工作。

arr1 = arr1.filter(function(val) {
  return arr2.indexOf(val) == -1;
});

Answer 1

如果属性顺序是可预测的，首先将 arr2 中的字符串化项目组成一个 Set，然后根据 arr1 中的字符串化项目是否存在于该集合中进行过滤:

const arr1 = [
  {name: "gunner", id: 639, label: "group"}, 
  {name: "Gunners", id: 1313, label: "channel", isPrivate: false}
];
const arr2 = [
  {name: "gunner", id: 639, label: "group"}
];
const arr2Set = new Set(arr2.map(JSON.stringify));

const output = arr1.filter(obj => !arr2Set.has(JSON.stringify(obj)));
console.log(output);

如果您不想或不能进行字符串化，则改为遍历对象的每个 Object.entries，并检查是否不是每个条目都与另一个数组匹配:

const arr1 = [
  {name: "gunner", id: 639, label: "group"}, 
  {name: "Gunners", id: 1313, label: "channel", isPrivate: false}
];
const arr2 = [
  {name: "gunner", id: 639, label: "group"}
];

const output = arr1.filter((obj1) => {
  const entries1 = Object.entries(obj1);
  return arr2.every(obj2 => (
    entries1.length !== Object.keys(obj2).length ||
    entries1.some(([key, val]) => obj2[key] !== val)
  ));
});

console.log(output);