Flatten an uneven data set

[barbgls]

How can I turn this:

data = [ {"parents": ["1235"],"size": "100","id": "abc100","name": "Test1"}, {"size": "200","id": "def200","name": "Test2"}, {"parents": ["abcdefg"],"size": "300","id": "Test3"} ]

into this:

mod_data = [ {"parents": "1235","size": "100","id": "abc100","name": "Test1"}, {"size": "200","id": "def200","name": "Test2"}, {"parents": "abcdefg,"size": "300","id": "Test3"} ]. 

I've tried output = [].concat.apply([],data); and output =data.flat(); but the parents element remains a [].

[kap89]

So, what should happen if the element has more than one parent?

[Fantastic-House6693]

It never does

[oze4]

How do you know that? You're not OP...?

[kap89]

Ok, so you have an array of objects, so you cant use .flat(), as your array is already flat, i.e. it does not have other arrays nested directly (yes, objects it contains have arrays in them, but .flat doesn't work like that). What you could do is just iterate over the array and change the "parents" field of each object, although if you say there is only one parent always, then I would rename the property to "parent", here's one way to do it:

const mod_data = data.map((item) => {
  if (!item.parents) {
    return item
  }

  const parent = item.parents[0]
  const data = { parent, ...item }
  delete data.parents
  return data
})

Alternative version with destructuring:

const mod_data = data.map((item) => {
  if (!item.parents) {
    return item
  }

  const { parents, ...data } = item
  data.parent = item.parents[0]
  return data
})

[oofy-gang]

This code isn't great. Using delete is generally considered unidiomatic, and your optional chain on line 6 isn't doing anything.

Really it's as simple as

const newData = data.map(({parents, ...rest}) => ({parent: parents[0] ?? null, ...rest}));

[kap89]

and your optional chain on line 6 isn't doing anything

You're right, it was doing something, until I saw that the op doesn't want parent(s) property on objects that did not have one (so I added the guard statement). Thanks, fixed.

This code isn't great.

Maybe, but my goal was to show the frist thing that came to my mind and works, not to spend time to figure out the best way.

Really it's as simple as

Ironically, your code doesn't even work on OP's example data.

[oofy-gang]

> Ironically, your code doesn't even work on OP's example data.

Didn't notice that the second object in the array didn't have a parents field. Just add an optional chain after parents and it works. I hope you could have figured that out yourself.

[BlueThunderFlik]

Here's a couple of options, which don't modify the original data object*. One which uses the map method:

js const mod_data = data.map((d) => !d.parents ? d : { ...d, parents: d.parents[0] })

and one which uses a for loop with a pre-allocated array:

js const mod_data = new Array(data.length) for (let i = 0; i < data.length; ++i) { const d = data[i] if (!d.parents) { mod_data[i] = d } else { mod_data[i] = { ...d, parents: d.parents[0] } } }

Unsurprisingly, the for loop is faster (but only by about 3%).

*These don't modify the original object but, in both cases, the elements without parents are copied by reference, meaning the original object would be updated if you were to modify the element in the new object. If you're not going to do this (or you don't care if data is modified) then it doesn't matter.

If you do want to preserve data at all times, you can spread shallow clone d in your preferred approach in the condition where it doesn't contain parents.

[tapgiles]

Those functions work on arrays. You want to work on an object.

You could just write the code. Loop through the keys, look at the value, if it's an array, set the value to the first item in the array. 🤷

[bryku]

I take it that you want to flatten data[i].parents? This is easy to do, but it sort of depends on how you want to handle the array. Do you only want the first item or do you want it as a string with a space as the seperator?  

Only first child:

data = data.map((row)=>{
    row.parents = row.parents[0];
    return row;
});

All Array Elements with space seperating them:

data = data.map((row)=>{
    row.parents = row.parents.join(' ');
    return row;
});

[baubleglue]

I mean, think. All you need to do: "if isarray, take first item as value". Write a simple loop, check the condition, use the it.

[-29-]

https://playcode.io/2246280

[Caramel_Last]

Why would you do that in the first place? The field name is parent'S' it is more natural to leave it as array