javascript - 从稍微嵌套的数组中删除重复项后保留对象的所有键

Question

我对这个问题有同样的问题，但我的对象有更多的键，例如：

[{
  id: 1
  name: "abcd",
  value: 123,
  type: "foo"
},
{
  id: 1
  name: "abcd",
  value: 321,
  type: "faa"
},
{
  id: 2
  name: "dcba",
  value: 456,
  type: "baa"
}]

我想实现这样的目标：

[{
  id: 1,
  name: "abcd",
  value: [123, 321],
  type: ["foo", "faa"]
},
{
  id: 2
  name: "dcba",
  value: [456],
  type: ["baa"]
}]

额外的键具有相同的值。

标签： javascriptramda.js

Answer 1

这个想法是按 分组id，然后映射每组对象，从第一个对象中选择id和，从组中的所有对象中提取所有和，转置并压缩到另一个对象，然后合并它们。namevaluetype

const { pipe, groupBy, prop, values, map, converge, merge, head, pick, props, transpose, zipObj } = R

const fn = pipe(
  groupBy(prop('id')), // groupBy the id
  values, // convert the object of groups to array of groups
  map(converge(merge, [ // map each group by merging the results of...
    pipe(head, pick(['id', 'name'])), // getting the id and name from the 1st item
    pipe(map(props(['value', 'type'])), transpose, zipObj(['value', 'type'])) // extract the value and type and zipping to an object
  ]))
)

const data = [{
  id: 1,
  name: "abcd",
  value: 123,
  type: "foo"
},
{
  id: 1,
  name: "abcd",
  value: 321,
  type: "faa"
},
{
  id: 2,
  name: "dcba",
  value: 456,
  type: "baa"
}]

const result = fn(data)

console.log(result)

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