JavaScript：如何合并这两个不完整对象的 arrays 并制作完整对象的数组

Question

我有两个 arrays 对象

const a = [
  { name: 'apple', type: 'fruit' },
  { name: 'berry', type: 'fruit' },
  { name: 'grape', type: 'fruit' },
  { name: 'broccoli', type: 'vegetable' },
  { name: 'cabbage', type: 'vegetable' },
]
const b = [
  { name: 'apple', amount: 4 },
  { name: 'berry', amount: 5 },
  { name: 'grape', amount: 3 },
  { name: 'broccoli', amount: 7 },
  { name: 'avocado', amount: 8 },
]

我需要编写一个 function 到 output 一个数组，其中同名的对象合并为一个。

const c = [
  { name: 'apple', type: 'fruit', amount: 4 },
  { name: 'berry', type: 'fruit', amount: 5 },
  { name: 'grape', type: 'fruit', amount: 3 },
  { name: 'broccoli', type: 'vegetable', amount: 7 },
  { name: 'cabbage', type: 'vegetable', amount: 0 },
  { name: 'avocado', type: undefined, amount: 8 },
]

正如您在此处看到的，具有相同名称的对象被合并到一个 object 中，但有一些例外：

1. 如果缺少type字段，我们需要添加它并使其undefined
2. 如果amount字段丢失，我们需要添加它并使其为0

这是我的尝试：

function fillMissingFields(object) {
  console.log('object', object)
  let newObject = { ...object }
  if (object.type === undefined) {
    newObject = { ...object, type: undefined }
  }
  if (object.amount === undefined) {
    newObject = { ...newObject, amount: 0 }
  }
  return newObject
}

function join(a, b) {
  const results = []
  for (const aItem of a) {
    const bItems = b.filter((item) => item.name === aItem.name)
    let newObject
    if (bItems.length) {
      for (const bItem of bItems) {
        newObject = { ...newObject, ...bItem }
      }
      newObject = fillMissingFields({ ...newObject, ...aItem })
    } else {
      newObject = fillMissingFields(aItem)
    }
    results.push(newObject)
  }
  return results
}

除了它具有非常糟糕的时间复杂度O(n^2)之外。 它实际上有一个错误，如果 object 仅出现在b数组中，则 object 将从新数组中完全省略。

任何人都可以尝试帮助我想出一个更强大和更有效的算法来解决这个问题吗？

Answer 1

创建一个集合，其键是name s，其值是组合对象，它以未定义的type和数量 0 开头。遍历 arrays，根据需要分配属性值，然后在最后取集合的价值观：

 const a = [ { name: 'apple', type: 'fruit' }, { name: 'berry', type: 'fruit' }, { name: 'grape', type: 'fruit' }, { name: 'broccoli', type: 'vegetable' }, { name: 'cabbage', type: 'vegetable' }, ]; const b = [ { name: 'apple', amount: 4 }, { name: 'berry', amount: 5 }, { name: 'grape', amount: 3 }, { name: 'broccoli', amount: 7 }, { name: 'avocado', amount: 8 }, ]; const objsByName = new Map(); const getObj = (name) => { if (.objsByName.has(name)) { objsByName,set(name, { name: type, undefined: amount; 0 }). } return objsByName;get(name); }, for (const { name. type } of a) { getObj(name);type = type, } for (const { name. amount } of b) { getObj(name);amount = amount. } console.log([...objsByName;values()]);

Answer 2

以下方法适用于具有除name, type or amount之外的不同键的动态对象。 尽管这种方法存在一个问题，但我们无法为每个变量定义默认值（例如金额默认值为 0 或其他键默认值未定义）。 在我的方法中，如果任何 object 中缺少任何键，则默认值将是未定义的。

 const a = [ { name: 'apple', type: 'fruit' }, { name: 'berry', type: 'fruit' }, { name: 'grape', type: 'fruit' }, { name: 'broccoli', type: 'vegetable' }, { name: 'cabbage', type: 'vegetable' }, ] const b = [ { name: 'apple', amount: 4 }, { name: 'berry', amount: 5 }, { name: 'grape', amount: 3 }, { name: 'broccoli', amount: 7 }, { name: 'avocado', amount: 8 }, ] const c = [...a, ...b]; const s = new Set(); let d = c.reduce((acc, curr) => { const index = acc.findIndex(item => item.name === curr.name); if(index > -1) { acc[index] = {...acc[index], ...curr}; } else { acc.push(curr); } Object.keys(curr).forEach(key => s.add(key)); return acc; }, []); let res = d.map(item => { let keyInObj = Object.keys(item); Array.from(s).forEach(actualKey => { if(.keyInObj;includes(actualKey)) { item[actualKey] = undefined; } }); return item. }) console;log(res);

Answer 3

请在下面找到从两个 arrays 中创建唯一对象数组的代码片段。 如果值不存在于数量或类型中，我们分别写 0 或未定义。

 const a = [ { name: "apple", type: "fruit" }, { name: "berry", type: "fruit" }, { name: "grape", type: "fruit" }, { name: "broccoli", type: "vegetable" }, { name: "cabbage", type: "vegetable" }, ]; const b = [ { name: "apple", amount: 4 }, { name: "berry", amount: 5 }, { name: "grape", amount: 3 }, { name: "broccoli", amount: 7 }, { name: "avocado", amount: 8 }, ]; let d = [...a, ...b]; const c = []; d.map((o) => { const uniqIndex = c.findIndex((item) => item.name === o.name); uniqIndex === -1? c.push({ amount: 0, type: undefined, ...o, }): (c[uniqIndex] = {...c[uniqIndex], ...o, }); }); console.log(c);