在对象数组中对数组进行排序

Question

if I have a big collection of array, what is the fastest way to sort and return the first 5 items?

Array of Object:

[
  {
    "name" : "placeA",
    "menus" : [ 
        {
            "sold" : 5,
            "name" : "menuA"
        }, 
        {
            "sold" : 20,
            "_id" : "menuB"
        }, 
        {
            "sold" : 1000,
            "_id" : "menuC"
        }
    ]
  },
  {
    "name" : "placeB",
    "menus" : [ 
        {
            "sold" : 300,
            "name" : "menuD"
        }, 
        {
            "sold" : 400,
            "_id" : "menuE"
        }, 
        {
            "sold" : 50,
            "_id" : "menuF"
        }
    ]
  },
  {
    "name" : "placeC",
    "menus" : [ 
        {
            "sold" : 1500,
            "name" : "menuG"
        }, 
        {
            "sold" : 450,
            "_id" : "menuH"
        }, 
        {
            "sold" : 75,
            "_id" : "menuI"
        }
    ]
  }
]

Expected Output:

[
    {
      "sold" : 1500,
      "name" : "menuG"
    },
    {
      "sold" : 1000,
      "_id" : "menuC"
    },
    {
      "sold" : 450,
      "_id" : "menuH"
    },
    {
      "sold" : 400,
      "_id" : "menuE"
    },
    {
      "sold" : 300,
      "name" : "menuD"
    } 
]

The only possible way that I'm able to think of is to create an array that is filled all the menus and then sort the menus and then slice the first 5 items. I don't think this way is good enough for sorting a huge collection of array in real use case.

Answer 1

You can use .map() , .reduce() , .concat() , .sort() , and .slice() .

 var arr = [{ "name": "placeA", "menus": [{ "sold": 5, "name": "menuA" }, { "sold": 20, "_id": "menuB" }, { "sold": 1000, "_id": "menuC" } ] }, { "name": "placeB", "menus": [{ "sold": 300, "name": "menuD" }, { "sold": 400, "_id": "menuE" }, { "sold": 50, "_id": "menuF" } ] }, { "name": "placeC", "menus": [{ "sold": 1500, "name": "menuG" }, { "sold": 450, "_id": "menuH" }, { "sold": 75, "_id": "menuI" } ] } ] var newarr = arr.map(v => v.menus).reduce((a, c) => a.concat(c)); console.log(newarr.sort((a,b) => b.sold - a.sold).slice(0,5));

I'm not sure if this is the fastest way, though.

Answer 2

In addition to what Yousername said, you can also you flat :

array
.map((a) => a.menus) // select menus
.flat() // all sub-array elements concatenated
.sort((a, b) => b.sold - a.sold) // sort by key sold
.slice(0,5) // select the first 5

Answer 3

You can use .reduce() , and clean it up with spread operator to flatten the Arrays as you go, and destructing to get straight to the 'menus' Array

 const data = [{name:"placeA",menus:[{sold:5,name:"menuA"},{sold:20,_id:"menuB"},{sold:1e3,_id:"menuC"}]},{name:"placeB",menus:[{sold:300,name:"menuD"},{sold:400,_id:"menuE"},{sold:50,_id:"menuF"}]},{name:"placeC",menus:[{sold:1500,name:"menuG"},{sold:450,_id:"menuH"},{sold:75,_id:"menuI"}]}]; let newArray = data .reduce((acc, {menus}) => [...acc, ...menus], []) .sort((a, b) => b.sold - a.sold); console.log(newArray);

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