New array of objects, but with changed keys

Question

I have a problem with picking some data with array of objects and pushing it into new object (in different array), but in different way.

const fruits = [
{name: "apple"},
{name: "kiwi"},
{name: "banana"},
{name: "orange"},
{name: "pineapple"},
{name: "coconut"},
{name: "peach"}, 
{name: "lemon"}
]

I want to pick a few items from this and paste them into completely new array, so it could look like this:

const manyFruits = [
{name: "apple-banana-kiwi-coconut"},
{name: "orange-pineapple-peach-lemon"}]

It would be great to do a function which takes an argument of how many fruits we want to pick into new array and they shouldn't repeat.

Below is my code. Firstly I create new array with names, then I push them into new array which depends on "length" and finally I try to create new array with objects, but it fails.

const fruitArray = length => {


const fruits = [
    {name: "apple"},
    {name: "kiwi"},
    {name: "banana"},
    {name: "orange"},
    {name: "pineapple"},
    {name: "coconut"},
    {name: "peach"},
    {name: "lemon"}
]

const allFruits = []

for (let i = 0; i < fruits.length; i++) {
    allFruits.push(fruits[i].name)
}

const newFruits =[]

for (let i = 0; i < length; i++) {
    newFruits.push(allFruits[i])
}

const manyFruitsInOneArr = []

for (let i = 0; i < 2; i++) {
    let newArr = {
        name: newFruits.join("-"),
    }

    manyFruitsInOneArr[i] = (newArr)
}

console.log(manyFruitsInOneArr)

}
fruitArray(2)

It generates new objects in new array, but items are the same all the time.

shouldnt repeat .. in one of the results or in all of them? So is apple-kiwi-banana and apple-kiwi-orange desired or not (for length 3)? — Jonas Wilms
– Jonas Wilms, Commented Feb 11, 2018 at 13:53
no, they shouldn't repeat at all. If we have apple-kiwi-banana, no apple, kiwi nor banana should appear again. — Martz89
– Martz89, Commented Feb 11, 2018 at 14:08

Nenad Vracar · Accepted Answer · 2018-02-11 14:07:29Z

2

You can create function for this using for loop and inside map() method to get array of names and join() to make a string from values.

const fruits = [{"name":"apple"},{"name":"kiwi"},{"name":"banana"},{"name":"orange"},{"name":"pineapple"},{"name":"coconut"},{"name":"peach"},{"name":"lemon"}]

const pick = (arr, n) => {
  const r = [];
  for (var i = 0; i < arr.length; i += n) {
    const name = arr
      .slice(i, i + n)
      .map(({name}) => name)
      .join('-');

    r.push({name})
  }
  return r;
}

const result = pick(fruits, 2)
console.log(result)

Update: to get names and sum of energy for each slice you can use map() and reduce() methods.

 fruits = [ {name: "apple", energy: 100}, {name: "kiwi", energy: 126}, {name: "banana", energy: 150}, {name: "orange", energy: 118}, {name: "pineapple", energy: 98}, {name: "coconut", energy: 83}, {name: "peach", energy: 65}, {name: "lemon", energy: 36} ] 

const pick = (arr, n) => {
  const r = [];
  for (var i = 0; i < arr.length; i += n) {
    const slice = arr.slice(i, i+n);
    const name = slice.map(({name}) => name).join('-')
    const energy = slice.reduce((r, {energy}) => r + energy, 0);
    r.push({name, energy})
  }
  return r;
}

const result = pick(fruits, 2)
console.log(result)

edited Feb 11, 2018 at 14:07

answered Feb 11, 2018 at 12:59

Nenad Vracar

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6 Comments

Martz89 Over a year ago

Thanks! That really helped! I have one more question. If "fruits" looked like this const fruits = [ {name: "apple", energy: 100}, {name: "kiwi", energy: 126}, {name: "banana", energy: 150}, {name: "orange", energy: 118}, {name: "pineapple", energy: 98}, {name: "coconut", energy: 83}, {name: "peach", energy: 65}, {name: "lemon", energy: 36} ] how can I add "energy" position to the result, which sums up all energies from "used" fruits?

Martz89 Over a year ago

Sorry, I wasn't precise enough. The result should look like this: [ {name: "apple-kiwi", energy: 226}, {name: "banana-orange", energy: 268} ], where energy is a total from energy of apple and kiwi from "fruits"

Martz89 Over a year ago

Thanks! You are the best

Martz89 Over a year ago

Sorry to bother you again, but I have another problem. fruits database contains only 8 items, so if n = 3 the last item in result array is going to be "half-empty". For example n = 3, so the result is:

{name: "apple-kiwi-banana", energy: 376},  {name: "orange-pineapple-coconut", energy: 299},  {name: "peach-lemon", energy: 101}

As you can see peach-lemon has only 2 items. Items, which are not correct (doesn't contain n names) shouldn't be returned. I want the result to look like this: {name: "apple-kiwi-banana", energy: 376}, {name: "orange-pineapple-coconut", energy: 299}

Martz89 Over a year ago

And how to make an exception? For example I don't want apple to match with kiwi, so the position in the result should look like this: {name: "apple-banana-orange", energy: something}

|

Jonas Wilms · Accepted Answer · 2018-02-11 13:57:30Z

1

That could be done using recursion:

 function subset(arr, n){
   if(!(n - 1)) return [arr[0].name];
   const result = [];
   for(let i = 0; i < arr.length - n; i++){
       for(const sub of subset(arr.slice(i + 1), n - 1)){
          result.push(arr[i].name + "-" + sub);
       }
   }
   return result;
 }

So you can do:

 subset(fruits, 5);

edited Feb 11, 2018 at 13:57

answered Feb 11, 2018 at 13:51

Jonas Wilms

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