使用函数式编程在 Typescript 中过滤数组 [filter、map、some、reduce 等]

Question

I'm trying to work with functional programming

I have two arrays

arr1=[{prodId:2}{prodId:4}]

arr2=[{id:1, name:"Test1"},
      {id:2, name:"Test2"},
      {id:3, name:"Test3"},
      {id:4, name:"Test4"},
      {id:5, name:"Test5"}]

using a combination of

filter, map, some, reduce functions

I want to extract items from arr2

where arr2.id === arr1.prodId 

my output arr would be:

  [{id:2, name:"Test2"},
  {id:4, name:"Test4"}]

I'm trying to avoid forEach and use functional programming.

Answer 1

You should use filter method in combination with includes and map .

 let arr1=[{prodId:2},{prodId:4}], arr2=[{id:1, name:"Test1"}, {id:2, name:"Test2"}, {id:3, name:"Test3"}, {id:4, name:"Test4"}, {id:5, name:"Test5"}]; let ids = arr1.map(({prodId}) => prodId); let result = arr2.filter(({id}) => ids.includes(id)); console.log(result);

Another approach is to use some method.

 let arr1=[{prodId:2},{prodId:4}], arr2=[{id:1, name:"Test1"}, {id:2, name:"Test2"}, {id:3, name:"Test3"}, {id:4, name:"Test4"}, {id:5, name:"Test5"}]; let result = arr2.filter(({id}) => arr1.some(({prodId}) => prodId == id)); console.log(result);

Answer 2

Solutions using .includes , .find , or .some are array operations and cost linear time. When used inside .filter , another linear time operation, the result is a quadratic time computation. If the input lists are large, this impact is not insignificant.

Instead, first collect the ids to match in a Set then take advantage of constant time lookup in the filter -

 const arr1 = [ { prodId: 2 } , { prodId: 4 } ] const arr2 = [ { id:1, name:"Test1" } , { id:2, name:"Test2" } , { id:3, name:"Test3" } , { id:4, name:"Test4" } , { id:5, name:"Test5" } ] const find = (whitelist, list) => { const ids = new Set (whitelist.map(x => x.prodId)) // create a set return list.filter(x => ids.has(x.id)) // Set#has uses constant time } console.log(find(arr1,arr2)) // [ { id: 2, name: "Test2" }, { id: 4, name: "Test4" } ]

Answer 3

Is this what you are looking for.

 arr1=[{prodId:2},{prodId:4}] arr2=[{id:1, name:"Test1"}, {id:2, name:"Test2"}, {id:3, name:"Test3"}, {id:4, name:"Test4"}, {id:5, name:"Test5"}] let k=arr2.reduce((o,a)=>{ if(arr1.map(a=>a.prodId).indexOf(a.id)!=-1) { o.push(a) } return o; },[]) console.log(k)