如何计算表中的对象数量

Question

How can I get count of objects in the table. I want all objects looks like the same. Eg in element "b" I don't have any objects, but in output I want to get all used objects with count 0.

INPUT DATA
{
  a: {
    obj3: [{...}, {...}]
  },
  b: { },
  c: {
    obj1: [{...}, {...}, {...}]
    obj2: [{...}, {...}]
  }
}

OUTPUT DATA

{
  a: {
    obj1: 0,
    obj2: 0,
    obj3: 2
  },
  b: {
    obj1: 0,
    obj2: 0,
    obj3: 0
  },
  c: {
    obj1: 3
    obj2: 2,
    obj3: 0
  }
}

Answer 1

You can do something like this only if you want specific number of objX as output and the key name is in the same format.

 let input = { a: { obj3: [{x:1}, {x:2}] }, b: { }, c: { obj1: [{x:1}, {x:2}, {x:3}], obj2: [{x:1}, {x:2}] } } let output = {}; for (const [key, value] of Object.entries(input)) { let result = {} for (let i = 1; i < 4; i++) { result['obj'+i] = value['obj'+i]? value['obj'+i].length: 0; } output[key] = result; } console.log(output);

Answer 2

To keep the code generic and a bit easier to manage and understand, I would loop over your data twice: first to discover all the unique keys you need to handle, and second to fill the structure with the right data.

Getting the right keys

To get the right keys, I look at the values of the input object (we can ignore the a , b and c keys).

We're interested in all keys of the objects in this layer. We can create a flat list of them using flatMap(Object.keys) .

Because this list might contain duplicate keys, I store them in a Set . This ensures all keys appear only once. With the example data you provided, we now have a Set of "obj1", "obj2", "obj3" .

Transforming an object

We can now transform any object to an object that has all keys. I capture this transformation in a Result constructor function.

This function creates a list of all keys ( [...allKeys] ), map s over them, and checks our input object for the existence of the key (obj[k]? ). If the key exists, we use its length. If not, we default to 0 .

Transforming the whole input

To transform your whole object, I defined a mapObj helper. This takes an object, applies a function to each of its values, and returns those in a new object with the same keys.

 const input = { a: { obj3: [{}, {}] }, b: { }, c: { obj1: [{}, {}, {}], obj2: [{}, {}] } }; // Set(3) {"obj1", "obj2", "obj3"} const allKeys = new Set( Object.values(input).flatMap(Object.keys) ); const Result = obj => Object.fromEntries( [...allKeys].map( (k) => [ k, obj[k]?.length || 0 ] ) ); console.log( mapObj(Result, input) ) // Utils function mapObj(f, o) { return Object.fromEntries( Object.entries(o).map( ([k, v]) => [k, f(v)] ) ) }

Answer 3

This is my solution:

 let input = { a: { obj3: [{x:1}, {x:2}] }, b: {}, c: { obj1: [{x:3}, {x:5}, {x:6}], obj2: [{x:7}, {x:8}] } } let output={}; Object.keys(input).forEach(key=>{ let item={obj1:0,obj2:0,obj3:0}; ["obj1","obj2","obj3"].forEach(itemType=>{ if (input[key][itemType]){ item[itemType]= input[key][itemType].length; } }) output[key]=item; }); console.log(output);