Return nested array by object property

Question

I have got an array of object: And would like to map it to another array witch structure should look like this:

const desireOutput = [
  { seq: 0, level: 1, code: '20', value: 'lorem', connectedByCode: [] },
  { seq: 2, level: 1, code: '30A', value: 'lorem', connectedByCode: [      
      { seq: 3, level: 2, code: '30A', value: 'lorem' },
      { seq: 4, level: 2, code: '30A', value: 'lorem' },
      { seq: 5, level: 2, code: '30A', value: 'lorem' }] },
  { seq: 8, level: 1, code: '40', value: 'lorem', connectedByCode: []  },
  { seq: 13, level: 1, code: '50', value: 'lorem', connectedByCode: []  },
  { seq: 15, level: 1, code: '60', value: 'lorem', connectedByCode: []  },
];

I was able to manage it by this code:

 const array1 = [{ seq: 0, level: 1, code: '20', value: 'lorem' }, { seq: 2, level: 1, code: '30A', value: 'lorem' }, { seq: 3, level: 2, code: '30A', value: 'lorem' }, { seq: 4, level: 2, code: '30A', value: 'lorem' }, { seq: 5, level: 2, code: '30A', value: 'lorem' }, { seq: 8, level: 1, code: '40', value: 'lorem' }, { seq: 13, level: 1, code: '50', value: 'lorem' }, { seq: 15, level: 1, code: '60', value: 'lorem' }, ]; const newTable = array1.filter(i => i.level === 1); const level2 = array1.filter(i => i.level === 2); level2.forEach(item => { const index = newTable.findIndex(i => i.code === item.code); newTable[index].connectedByCode = level2.filter(i => i.code === item.code); });

I don't like that code so much to be honest. Is there any better way to do this?

Answer 1

https://playcode.io/963588/

const array1 = [{
    seq: 0,
    level: 1,
    code: '20',
    value: 'lorem'
  },
  {
    seq: 2,
    level: 1,
    code: '30A',
    value: 'lorem'
  },
  {
    seq: 3,
    level: 2,
    code: '30A',
    value: 'lorem'
  },
  {
    seq: 4,
    level: 2,
    code: '30A',
    value: 'lorem'
  },
  {
    seq: 5,
    level: 2,
    code: '30A',
    value: 'lorem'
  },
  {
    seq: 8,
    level: 1,
    code: '40',
    value: 'lorem'
  },
  {
    seq: 13,
    level: 1,
    code: '50',
    value: 'lorem'
  },
  {
    seq: 15,
    level: 1,
    code: '60',
    value: 'lorem'
  },
];



const newTable = []
const level2 = []
const newTableCodes = []
const level2Codes = []

for(let [index, {level, code}]  of array1.entries()){
    if(level === 1) {
        newTable.push({...array1[index], connectedByCode:[]})
        newTableCodes.push(array1[index]['code'])
    }
    else {
        level2.push(array1[index]) 
        level2Codes.push(array1[index]['code'])
    }
}
// in order to save some iterations. Will only work if newTable code values are unique

const setLevel2 = new Set(level2Codes) // u can ignore this 

for(let [indexNew, valueNew] of newTableCodes.entries()){
    if(!setLevel2.has(valueNew)) continue // this also
    else setLevel2.delete(valueNew) // and this too
    for(let [index2, value2] of level2Codes.entries()){
        if(valueNew === value2) newTable[indexNew]['connectedByCode'].push(level2[index2])
    }
    
}


console.log(newTable)

Not sure if this is a much better approach. Atleast is another way to make it, hope it helps. (Set is a hashlist with no duplicated values)

Answer 2

It depends on what you don't like or what you want to change in your code.

Do you want to make it faster or more generic? Or is it just code style?

From what you shared, the code looks fine for me. Anyway, below is an example of how you can reformat your code a bit, see if you like it more or not.

An example:

const isRootItem = item => item.level === 1;
const isConnectedItem = item => item.level === 2;

const allConnectedItems = array1.filter(isConnectedItem);
const getConnectedItems = item => allConnectedItems.filter(connectedItem => connectedItem.code === item.code);

const items = array1
  .filter(isRootItem)
  .map(item => ({ ...item, connectedByCode: getConnectedItems(item) }));

Snippet:

 const array1 = [{ seq: 0, level: 1, code: '20', value: 'lorem' }, { seq: 2, level: 1, code: '30A', value: 'lorem' }, { seq: 3, level: 2, code: '30A', value: 'lorem' }, { seq: 4, level: 2, code: '30A', value: 'lorem' }, { seq: 5, level: 2, code: '30A', value: 'lorem' }, { seq: 8, level: 1, code: '40', value: 'lorem' }, { seq: 13, level: 1, code: '50', value: 'lorem' }, { seq: 15, level: 1, code: '60', value: 'lorem' }, ]; const isRootItem = item => item.level === 1; const isConnectedItem = item => item.level === 2; const allConnectedItems = array1.filter(isConnectedItem); const getConnectedItems = item => allConnectedItems.filter(connectedItem => connectedItem.code === item.code); const items = array1.filter(isRootItem).map(item => ({...item, connectedByCode: getConnectedItems(item) })); console.log(items);