Sorting array with another array's objects properties

Question

I have a confusing problem right now. The situation is this:

let example = [
  {
    name: "First",
    id: 1,
    rank: 1,
    childs: [5,3] // [3,5] -> wrong, because the element which have the id: 3 
//(First's second child), have the rank 2, but with id: 5 (First's first child) have the rank 1.
  },
  {
    name: "First's first child",
    id: 5,
    rank: 1,
    childs: [4,2]
  },
  {
    name: "First's second child",
    id: 3,
    rank: 2,
    childs: [6]
  },
  {
    name: "Second's first child",
    id: 4,
    rank: 1,
    childs: []
  },
  {
    name: "Second's second child",
    id: 2,
    rank: 2,
    childs: []
  },
  {
    name: "Third's first child",
    id: 6,
    rank: 1,
    childs: []
  }
]

The childs property is an array which shows the element's childs ids (the names of the items shows the relationship between the elements). The childs array's elements order must be equal the elements ranks which has the same id the childs has. The problem here is, that the childs array's sorting is not always correct, which generates a lot of problems for me. What i need to do is get the element's childs (the childs items of the array means the item's id in example), search each of them in the example array, then sort the childs array in order of it's ranks in example. So i want to sort example[i].childs, depending on the same id'd item's rank in example. How is this possible?

The expected (ideal) output is the example above. The bad output is in the comment. The truth is in my real code i can't really control how the childs are sorted. But it must sorted depending on it's rank. — Gergő Horváth
– Gergő Horváth, Commented May 19, 2018 at 18:20
Need to explain what "sorted depending on it's rank" even means. That's why expected output would help — charlietfl
– charlietfl, Commented May 19, 2018 at 18:22
So each childs array should maintain the same order as is found in the example array? Or they should be sorted ascending by rank? — user2437417
– user2437417, Commented May 19, 2018 at 18:33

Jason Cust · Accepted Answer · 2018-05-19 18:44:53Z

1

If I understand the OP correctly then the output is what the OP example is and the comment was the input. The ordering of the childs property is based on the rank of the corresponding element's ranks from the IDs in the childs array of the "parent". If that is correct then you can use find to search the array for the element matching the child IDs, sort them based on their rank, and then replace the child property with the new sorted value.

// First map over the entire example
const sorted = example.map(({childs, ...p}) => ({
  // Copy all of the original element's properties except `childs`
  ...p,
  // Replace `childs` with a sorted array
  childs: childs
    // Find the child element based on the ID
    .map((pid) => example.find(({id: cid}) => pid === cid))
    // Sort the elements based on their rank
    .sort(({rank: a}, {rank: b}) => a - b)
    // Pick out only the `id` value
    .map(({id} = {}) => id)
}));

And a working example:

const example = [{
  name: "First",
  id: 1,
  rank: 1,
  childs: [3, 2]
},
{
  name: "First's first child",
  id: 2,
  rank: 1,
  childs: [4, 5]
},
{
  name: "First's second child",
  id: 3,
  rank: 2,
  childs: [6]
},
{
  name: "Second's first child",
  id: 4,
  rank: 1,
  childs: []
},
{
  name: "Second's second child",
  id: 5,
  rank: 2,
  childs: []
},
{
  name: "Third's first child",
  id: 6,
  rank: 1,
  childs: []
}
];

const sorted = example.map(({childs, ...p}) => ({
  ...p,
  childs: childs
    .map((pid) => example.find(({id: cid}) => pid === cid))
    .sort(({rank: a}, {rank: b}) => a - b)
    .map(({id} = {}) => id)
}));
console.log(sorted);

edited May 19, 2018 at 18:44

answered May 19, 2018 at 18:36

Jason Cust

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2 Comments

Gergő Horváth Over a year ago

Amazing... Sorry, i can't upvote yet, but it helped me a lot. Thanks for the detailed explanation!

Jason Cust Over a year ago

@DorLask No worries. I'm glad I could help. Happy coding. 👍

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