在对象列表上使用.reduce（）

Question

I have this sample data set:

let list = [
  {'first': 'morgan', 'id': 1},
  {'first': 'eric', 'id': 1},
  {'first': 'brian', 'id': 2 },
  {'first': 'derek', 'id' : 2},
  {'first': 'courtney', 'id': 3},
  {'first': 'eric', 'id': 4},
  {'first': 'jon', 'id':4},
]

I am trying to end up with this:

[[1, [morgan, eric]], [2, [brian, derek]], [3, [courtney]], [4, [eric, jon]]

I am using the .reduce() function to map over the list. However, I'm somewhat stuck.

I got this working:

let b = list.reduce((final_list, new_item) => {
  x = final_list.concat([[new_item.id, [new_item.first]]])
  return x
}, [])

However, that flattens out the list into a list of lists of lists, but doesn't combine names that share a similar id.

I tried using .map() the code below does not work

I tried to map over the final_list (which is a list of [id, [names]] looking to see if the id of the new_item exists in the smaller_list and then add new_item.first to the smaller_list[1] (which should be the list of names).

Is this the right approach?

let b = list.reduce((final_list, new_item) => {
  final_list.map((smaller_list) => {
    if (smaller_list.indexOf((new_item.id)) >=0) {
      smaller_list[1].concat(new_item.first)
      // not sure what to do here...
    } else {
        // not sure what to do here either...
    }

  })
  x = final_list.concat([[item.id, [item.first]]])
  return x
}, [])

Answer 1

In my approach, i'm using reduce to create a sparse array with the index as the list id. I then use Object.values to squash it down.

 let list = [ {'first': 'morgan', 'id': 1}, {'first': 'eric', 'id': 1}, {'first': 'brian', 'id': 2 }, {'first': 'derek', 'id' : 2}, {'first': 'courtney', 'id': 3}, {'first': 'eric', 'id': 4}, {'first': 'jon', 'id':4} ]; let result = list.reduce( (acc, item, index) => { if(acc[item.id]){ acc[item.id][1].push(item.first); } else { acc[item.id] = [item.id, [item.first]]; } return acc; }, []); console.log(Object.values(result)); 

Answer 2

My solution was to first map over your list and combine the ones with similar ID and then loop over that result in order to change the format to the one requested.

 let list = [ {'first': 'morgan', 'id': 1}, {'first': 'eric', 'id': 1}, {'first': 'brian', 'id': 2 }, {'first': 'derek', 'id' : 2}, {'first': 'courtney', 'id': 3}, {'first': 'eric', 'id': 4}, {'first': 'jon', 'id':4}, ] let temp = [], result = []; list.map(entry => { if(!temp[entry.id]) temp[entry.id] = [entry.first] else temp[entry.id].push(entry.first) }) temp.forEach((names, id) => { result.push([id, names]) }) console.log(JSON.stringify(result)) 

EDIT: I used an array as temp because the ID is an integer and for the convenience of being able to use a forEach loop. If the key was something else then you would use a Hash table.

Answer 3

Try first creating a map of IDs to names, then map that into your final array, eg

 let list = [ {'first': 'morgan', 'id': 1}, {'first': 'eric', 'id': 1}, {'first': 'brian', 'id': 2 }, {'first': 'derek', 'id' : 2}, {'first': 'courtney', 'id': 3}, {'first': 'eric', 'id': 4}, {'first': 'jon', 'id':4}, ] const idMap = list.reduce((map, item) => { if (Array.isArray(map[item.id])) { map[item.id].push(item.first) } else { map[item.id] = [item.first] } return map }, {}) const b = Object.keys(idMap).map(id => [parseInt(id), idMap[id]]) console.info(JSON.stringify(b)) 

Answer 4

you want to do it with reduce, and doing it with a single reduce would be nice. here are 3 ways

verbose, more loops

const tuples1 = list.reduce( (tuples, obj) => {
  if( tuples.some( ( [id, _] ) => id === obj.id ) ){
    tuples.find( ( [id, _] ) => id === obj.id )[1].push(obj.first)
  }else{
    tuples.push([obj.id, [obj.first]])
  }
  return tuples
}, [])

same as tuples1, terse

const tuples2 = list.reduce( (tuples, obj) =>
  ( tuples.some( ( [id, _] ) => id === obj.id ) ) ?
    ( tuples.find( ( [id, _] ) => id === obj.id )[1].push(obj.first), tuples ) :
    ( tuples.push([obj.id, [obj.first]]), tuples ), [])

more verbose, less loops

const tuples3 = list.reduce( (tuples, obj) => {
  const existingId = tuples.find( ( [id, _] ) => id === obj.id )
  if( existingId ){
    existingId[1].push(obj.first)
  }else{
    tuples.push([obj.id, [obj.first]])
  }
  return tuples
}, [])

each produce the same result

[ [ 1, [ 'morgan', 'eric' ] ],
  [ 2, [ 'brian', 'derek' ] ],
  [ 3, [ 'courtney' ] ],
  [ 4, [ 'eric', 'jon' ] ] ]

Answer 5

Just in case you're also looking for a non reduce solution.

Here is one. It is a bit long winded but that is because I'm trying to decompose the problem into a few parts.

1. I first go through the list and sort the elements into their respective Id .

2. Then For each id in the sorted object. I loop through it again by first instantiating an outer element first.

3. Pushed the id key into the element first.
4. Loop through all the associated element and extract their first property and add it into the nested element.
5. Once the nesting element is been built. I just merge them up into the parent element and attach it into the result array.

Hopfully it is not too complicated for you but I suspect it will be easier to see the code in action through an IDE.

 var list = [ {'first': 'morgan', 'id': 1}, {'first': 'eric', 'id': 1}, {'first': 'brian', 'id': 2 }, {'first': 'derek', 'id' : 2}, {'first': 'courtney', 'id': 3}, {'first': 'eric', 'id': 4}, {'first': 'jon', 'id':4}, ] var result = []; var keyById = {}; // Sort the list of items to be keyed by: Id => [ item, item ] list.forEach(function(item) { if (!keyById.hasOwnProperty(item.id)) { keyById[item.id] = []; } keyById[item.id].push(item); }); // Build the data structure for (var id in keyById) { var item = []; // The outer element var nestedItem = []; // The nested element belonging to the outer item.push(id); keyById[id].forEach(function(element) { nestedItem.push(element.first); }); item.push(nestedItem); result.push(item); } console.log(JSON.stringify(result, '', 2)); 

Answer 6

Not very sexy, but it works

• Iterate over list and build an object using the id s as key s with value s of array s of the first names.
• Empty list (if you plan to overwrite it).
• Walk through the resultant object making array s for each id with its next sibling element being an array of the names from the value associated with that id , and push them into list (or a fresh array if maintaining the original list ).

 var list = [ {'first': 'morgan', 'id': 1}, {'first': 'eric', 'id': 1}, {'first': 'brian', 'id': 2 }, {'first': 'derek', 'id' : 2}, {'first': 'courtney', 'id': 3}, {'first': 'eric', 'id': 4}, {'first': 'jon', 'id':4}, ], o = {}, p; list.forEach( ( v ) => { if ( !o[ v.id ] ) { o[ v.id ] = []; } o[ v.id ].push( v.first ); } ); list = []; // if overwriting for ( p in o ) { if ( o.hasOwnProperty( p ) ) { list.push( [ parseInt( p ), o[ p ] ] ); } } console.log( JSON.stringify( list ).replace( /,/g, ", " ) ); 

Edited to add parseInt() (noticed in another answer) since the question's example expected output does indeed show numerals as integers, not strings.
And changed the demo output to stringify ed JSON (also noticed (easier to read) in other answers).

Answer 7

It's probably easiest to write a plain loop for this:

let final_list = []
for (k = 0; k < list.length; k += 2) {
    let item1 = list[k]
    let item2 = list[k + 1]
    final_list.push([item1.id, [item1.first, item2.first]])
}