在 javascript (lodash) 中合并部分重复的数组答案

【问题标题】：merge partially duplicated arrays in javascript (lodash)在 javascript (lodash) 中合并部分重复的数组
【发布时间】：2019-05-01 06:35:11
【问题描述】：

我有很多人在不同年份买了一辆车。简化后的数组是这样的：

const owners = [{
  name: "john",
  hasCar: true,
  yearBought: 2002
}, {
  name: "john",
  hasCar: true,
  yearBought: 2005
}, {
  name: "mary",
  hasCar: true,
  yearBought: 2015
}, {
  name: "john",
  hasCar: true,
  yearBought: 2018
}]

如果一个人拥有不止一辆汽车（如本例中的约翰），那么对于他购买汽车的不同年份会有不同的对象。我想合并属于每个人的对象，最终结果应该是这样的：

const owners = [{
  name: "john",
  hasCar: true,
  yearBought: [2002, 2005, 2018]
}, {
  name: "mary",
  hasCar: true,
  yearBought: 2018
}]

【问题讨论】：

我认为您不需要 Lodash。只需检查具有给定名称的元素是否存在，如果存在，将年份推送到yearsBought，如果不存在，只需将新元素推送到列表中

标签： javascript arrays lodash

【解决方案1】：

您可以使用reduce 数组并首先根据name 对它们进行分组。累加器是一个对象，每个唯一的name 作为键。如果name 已经存在，请使用concat 将它们推送到数组中。否则，在累加器中创建一个新键并将其设置为当前对象。然后，使用Object.values() 将数组的值作为数组获取

const owners = [{name:"john",hasCar:!0,yearBought:2002},{name:"john",hasCar:!0,yearBought:2005},{name:"mary",hasCar:!0,yearBought:2015},{name:"john",hasCar:!0,yearBought:2018}];

const merged = owners.reduce((r, o) => {
  if(r[o.name])
    r[o.name].yearBought = [].concat(r[o.name].yearBought, o.yearBought)
  else
    r[o.name] = { ...o };
  return r;
},{})

console.log(Object.values(merged))

【讨论】：

【解决方案2】：

只需使用reduce:

const owners = [{name:"john",hasCar:true,yearBought:2002},{name:"john",hasCar:true,yearBought:2005},{name:"mary",hasCar:true,yearBought:2015},{name:"john",hasCar:true,yearBought:2018}];

const newOwners = Object.values(owners.reduce((acc, curr) => {
  acc[curr.name] = acc[curr.name] ? { ...acc[curr.name], yearBought: [].concat(acc[curr.name].yearBought, curr.yearBought) } : curr;
  return acc;
}, {}));

console.log(newOwners);

【讨论】：

【解决方案3】：

我希望这有助于使用 PURE lodash 函数.. 它看起来干净且可读。

var array = [{
  name: "john",
  hasCar: true,
  yearBought: 2002
}, {
  name: "john",
  hasCar: true,
  yearBought: 2005
}, {
  name: "mary",
  hasCar: true,
  yearBought: 2015
}, {
  name: "john",
  hasCar: true,
  yearBought: 2018
}]

function mergeNames(arr) {
  return Object.values(_.chain(arr).groupBy('name').mapValues((g) => (_.merge(...g, {
    yearBought: _.map(g, 'yearBought')
  }))).value());
}
console.log(mergeNames(array));

&lt;script src="https://cdn.jsdelivr.net/npm/lodash@4.17.11/lodash.min.js"&gt;&lt;/script&gt;

谢谢:)

【讨论】：

这也会为mary 创建一个yearBought 数组
我相信它应该在每种情况下都保持一致，即我们对yearBought 有一个或多个值。这样你就可以轻松管理了

【解决方案4】：

您可以使用唯一键（在您的情况下为name）对项目进行分组，然后映射组，并合并每个组中的项目：

const { flow, partialRight: pr, groupBy, map, mergeWith, concat, isUndefined } = _

const mergeDuplicates = (isCollected, key) => flow(
  pr(groupBy, key), // group by the unique key
  pr(map, group => mergeWith({}, ...group, 
    (o, s, k) => isCollected(k) && !isUndefined(o) ? concat(o, s) : s
  )) // merge each group to a new object
)

const owners = [{name:"john",hasCar:true,yearBought:2002},{name:"john",hasCar:true,yearBought:2005},{name:"mary",hasCar:!0,yearBought:2015},{name:"john",hasCar:true,yearBought:2018}]

const isCollected = key => key === 'yearBought'

const result = mergeDuplicates(isCollected, 'name')(owners)

console.log(result)

&lt;script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.js"&gt;&lt;/script&gt;

还有 lodash/fp 版本：

const { flow, groupBy, map, mergeAllWith, cond, nthArg, concat } = _

const mergeDuplicates = (isCollected, key) => flow(
  groupBy(key),
  map(mergeAllWith(cond([[
    flow(nthArg(2), isCollected),
    concat,
    nthArg(1)
  ]])))
)

const owners = [{name:"john",hasCar:!0,yearBought:2002},{name:"john",hasCar:!0,yearBought:2005},{name:"mary",hasCar:!0,yearBought:2015},{name:"john",hasCar:!0,yearBought:2018}]

const isCollected = key => key === 'yearBought'

const result = mergeDuplicates(isCollected, 'name')(owners)

console.log(result)

&lt;script src='https://cdn.jsdelivr.net/g/lodash@4(lodash.min.js+lodash.fp.min.js)'&gt;&lt;/script&gt;

【讨论】：