根据属性值列表快速切换数组中对象的布尔属性

Question

我有一个对象数组，例如：

[{type: 'marker', name: 'somename', id: '123abc321', lat: 123.00, lng: 32.00, educationRegion: 'someregion', seDistrict: 'sometype', sector: 'somesector', selected: false, shownOnMap: true, listed: false}, {type: 'marker', name: 'somename', id: '123abc322', lat: 123.00, lng: 32.00, educationRegion: 'someregion', seDistrict: 'sometype', sector: 'somesector', selected: false, shownOnMap: true, listed: false}, {type: 'marker', name: 'somename', id: '123abc323', lat: 123.00, lng: 32.00, educationRegion: 'someregion', seDistrict: 'sometype', sector: 'somesector', selected: false, shownOnMap: true, listed: false}]

数组中的最大对象数约为 7600

我正在努力寻找一种快速的方法，将其 id 包含在 id 值数组中的每个对象的“选定”属性从 true 翻转为 false，反之亦然：

['123abc322', '123abc323']

id 值的供应商数组可能包括所有对象 id，通常是 ~2000 个 id。

所以如果 object.id 字段在提供的数组中，我想将 object.selected 交换为 !object.selected

任何帮助将不胜感激。

Answer 1

我建议将您的 ID 数组转换为 Set。然后，只需遍历您的数组，如果 Set 包含元素的 ID，则切换 selected 属性。

请注意，此方法会修改数组中的对象。如果您不需要改变输入，您可以使用map。

改变输入：

const arr = [{type: 'marker', name: 'somename', id: '123abc321', lat: 123.00, lng: 32.00, educationRegion: 'someregion', seDistrict: 'sometype', sector: 'somesector', selected: false, shownOnMap: true, listed: false}, {type: 'marker', name: 'somename', id: '123abc322', lat: 123.00, lng: 32.00, educationRegion: 'someregion', seDistrict: 'sometype', sector: 'somesector', selected: false, shownOnMap: true, listed: false}, {type: 'marker', name: 'somename', id: '123abc323', lat: 123.00, lng: 32.00, educationRegion: 'someregion', seDistrict: 'sometype', sector: 'somesector', selected: false, shownOnMap: true, listed: false}];

const ids = new Set(['123abc322', '123abc323']);

arr.forEach(el => {
  if (ids.has(el.id)) {
    el.selected = !el.selected;
  }
});

console.log(arr);

不改变输入：

const arr = [{type: 'marker', name: 'somename', id: '123abc321', lat: 123.00, lng: 32.00, educationRegion: 'someregion', seDistrict: 'sometype', sector: 'somesector', selected: false, shownOnMap: true, listed: false}, {type: 'marker', name: 'somename', id: '123abc322', lat: 123.00, lng: 32.00, educationRegion: 'someregion', seDistrict: 'sometype', sector: 'somesector', selected: false, shownOnMap: true, listed: false}, {type: 'marker', name: 'somename', id: '123abc323', lat: 123.00, lng: 32.00, educationRegion: 'someregion', seDistrict: 'sometype', sector: 'somesector', selected: false, shownOnMap: true, listed: false}];

const ids = new Set(['123abc322', '123abc323']);

const newArr = arr.map(el => {
  if (!ids.has(el.id)) return el;

  return {
    ...el,
    selected: !el.selected
  }
});

console.log(newArr);

Answer 2

您可以使用一行代码完成所有操作，下面是一个代码示例：

let array = [{
    type: "marker",
    name: "somename",
    id: "123abc321",
    lat: 123.0,
    lng: 32.0,
    educationRegion: "someregion",
    seDistrict: "sometype",
    sector: "somesector",
    selected: false,
    shownOnMap: true,
    listed: false,
  },
  {
    type: "marker",
    name: "somename",
    id: "123abc322",
    lat: 123.0,
    lng: 32.0,
    educationRegion: "someregion",
    seDistrict: "sometype",
    sector: "somesector",
    selected: false,
    shownOnMap: true,
    listed: false,
  },
  {
    type: "marker",
    name: "somename",
    id: "123abc323",
    lat: 123.0,
    lng: 32.0,
    educationRegion: "someregion",
    seDistrict: "sometype",
    sector: "somesector",
    selected: false,
    shownOnMap: true,
    listed: false,
  },
];

let ids = ['123abc322', '123abc323'];

// .filter() create a new array with all elements that pass the test implemented by the provided function.
// .map() run a function for everyone of the remaining objects and flip the "selected" value.

array.filter(x => ids.includes(x.id)).map(x => x.selected = !x.selected);
console.log(array);

希望这就是你要找的东西。