ramda.js如何做groupby，计数，排序

Question

我有一个数据集合，比如：

[{"id":1,"score":4},{"id":2,"score":3},{"id":1,"score":4},{"id":2,"score":3},{"id":3,"score":4},{"id":1,"score":3}]

我希望输出如下：

[{"id":1,"count":3},{"id":2,"count":2},{"id":3,"count":1}]

有没有什么解决方案可以使用Ramda.js来做到这一点？

我试着使用countBy(prop("id"))，但我想不出按计数排序的方法。

Answer 1

使用R.pipe创建一个函数，该函数使用R.countBy获取{ [id]: count }的对象，然后将数据转换为对，并使用R.map和R.applySpec生成对象数组。然后使用R.sortBy对其进行排序。

​

const { pipe, countBy, prop, toPairs, map, applySpec, head, last, sortBy, descend } = R

const fn = pipe(
  countBy(prop('id')),
  toPairs,
  map(applySpec({
    id: pipe(head, Number), // or just id: head if the id doesn't have to be a number
    count: last,
  })),
  sortBy(descend(prop('count'))), // or ascend
)

const arr = [{"id":1,"score":4},{"id":2,"score":3},{"id":1,"score":4},{"id":2,"score":3},{"id":3,"score":4},{"id":1,"score":3}]

const result = fn(arr)

console.log(result)

<script src="https://cdnjs.cloudflare.com/ajax/libs/ramda/0.27.0/ramda.js"></script>

​

Answer 2

显然，如果您使用的是Ramda，那么countBy将是解决方案的一部分。然后，我会选择通过管道将其传递给toPairs和zipObj，以获得最终结果：

​

const collect = pipe (
  countBy (prop ('id')),
  toPairs,
  map (zipObj (['id', 'count']))
) 

const data = [{id: 1, score: 4}, {id: 2, score: 3}, {id: 1, score: 4}, {id: 2, score: 3}, {id: 3, score: 4},{id: 1, score: 3}]

console .log (collect (data))

<script src="https://cdnjs.cloudflare.com/ajax/libs/ramda/0.27.0/ramda.js"></script>
<script> const {pipe, countBy, prop, toPairs, map, zipObj} = R             </script>

​

zipObj接受一个属性名称数组和一个值数组，并将它们压缩成一个对象。

Answer 3

使用vanillaJS，您可以简单地使用reduce和Map

​

const data = [{"id":1,"score":4},{"id":2,"score":3},{"id":1,"score":4},{"id":2,"score":3},{"id":3,"score":4},{"id":1,"score":3}]

const final = [...data.reduce((op,{id,score})=>{
   if(op.has(id)){
    op.set(id, op.get(id)+1)
   } else {
    op.set(id,1)
   }
   return op;
}, new Map()).entries()].map(([id,count])=>({id,count}))

console.log(final)

​