map() inside map() asynchronous in javascript

Question

I am trying to create an array from another array

let a = ["a", "b", "c", ...]

let b = ["22", "33", "44"]

let desireResult = [
    {
    "name": "a", 
    "age": "22"
    },
    {
    "name": "a", 
    "age": "33"
    },
    {
    "name": "a", 
    "age": "44"
    },
    {
    "name": "b", 
    "age": "22"
    },
    {
    "name": "b", 
    "age": "33"
    },
    {
    "name": "b", 
    "age": "44"
    },
    {
    "name": "c", 
    "age": "22"
    },
    {
    "name": "b", 
    "age": "33"
    },
    {
    "name": "b", 
    "age": "44"
    },
    ...
]

this is my code:

let profiles = a.map(name=>{
    b.map(age=>{
        return ({"name": name, "age":age})
    })
})
Promise.all([profiles])

but map() has synchronous functionality. so one way is to use Promise.all([]) . but what about map() inside a map()?

"so one way is to use Promise.all([])" - This won't change anything. Especially not without a .then(). But this won't do anything useful either when there's nothing to wait for... — Andreas
– Andreas, Commented Jun 20, 2020 at 13:34

Rajneesh · Accepted Answer · 2020-06-20 13:50:56Z

4

You can do this by using flatMap and map and Here is the working example:

UPDATE:

let a = ["a", "b", "c",];
let b = ["22", "33", "44"];

var result = a.flatMap(name=>b.map(age=>({name, age})));

console.log(result);

edited Jun 20, 2020 at 13:50

answered Jun 20, 2020 at 13:30

Rajneesh

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Comments

dcangulo · Accepted Answer · 2020-06-20 13:49:44Z

2

I don't know where is the asynchronous part.

It's basically the same code you have but you don't return anything on the first map. I just added a return keyword. I also added flat() to convert array of array of objects to a single array of objects.

But you can do it with this code:

let a = ["a", "b", "c"]

let b = ["22", "33", "44"]

let profiles = a.map(name=>{
    return b.map(age=>{
        return ({"name": name, "age":age})
    })
}).flat()

console.log(profiles)

edited Jun 20, 2020 at 13:49

answered Jun 20, 2020 at 13:35

dcangulo

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1 Comment

ibrahim mahrir Over a year ago

@dcangulo Your output is an array of arrays containing the objects, whereas OP's desired output is an array of objects

ibrahim mahrir · Accepted Answer · 2020-06-20 13:38:10Z

1

You don't need Promise.all as there are no asynchronous stuff going on, what you need is a Cartesian product of the two arrays, which can be achieved by using a reduce/forEach combo like so:

let profiles = a.reduce((acc, name) => {
  b.forEach(age => acc.push({ name, age }));
  return acc;
}, []);

Demo:

let a = ["a", "b", "c"];
let b = [22, 33, 44];

let profiles = a.reduce((acc, name) => {
  b.forEach(age => acc.push({ name, age }));
  return acc;
}, []);

console.log(profiles);

answered Jun 20, 2020 at 13:38

ibrahim mahrir

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Comments

Vahid Alimohamadi · Accepted Answer · 2020-06-20 18:40:11Z

1

let a = ["a", "b", "c"];
let b = ["22", "33", "44"];

Promise.all(a.map((key,index)=>({name: key,age: b[index]})))
.then(r=>console.log(r));

Another example :

let a = ["a", "b", "c"];
let b = ["22", "33", "44"];

const myPromise = wait => new Promise(resolve=>setTimeout(()=>resolve(Date.now()),wait));

Promise.all(a.map(async (key,index)=>({name: key,age: b[index], time: await myPromise(Math.floor(Math.random() * 1000) + 1)})))
.then(r=>console.log(r));

edited Jun 20, 2020 at 18:40

answered Jun 20, 2020 at 18:08

Vahid Alimohamadi

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