Here's an O(n) solution using reduce and Object.assign
const joinById = ( ...lists ) =>
Object.values(
lists.reduce(
( idx, list ) => {
list.forEach( ( record ) => {
if( idx[ record.id ] )
idx[ record.id ] = Object.assign( idx[ record.id ], record)
else
idx[ record.id ] = record
} )
return idx
},
{}
)
)
To use this function for the OP's case, pass in the arrays you want to join to joinById (notice lists is a rest parameter).
let joined = joinById(list1, list2)
Each list gets reduced to a single object where the keys are ids and the values are the objects. If there's a value at the given key already, it gets object.assign called on it and the current record.
Here's the generic O(n*m) solution, where n is the number of records and m is the number of keys. This will only work for valid object keys. You can convert any value to base64 and use that if you need to.
const join = ( keys, ...lists ) =>
lists.reduce(
( res, list ) => {
list.forEach( ( record ) => {
let hasNode = keys.reduce(
( idx, key ) => idx && idx[ record[ key ] ],
res[ 0 ].tree
)
if( hasNode ) {
const i = hasNode.i
Object.assign( res[ i ].value, record )
res[ i ].found++
} else {
let node = keys.reduce( ( idx, key ) => {
if( idx[ record[ key ] ] )
return idx[ record[ key ] ]
else
idx[ record[ key ] ] = {}
return idx[ record[ key ] ]
}, res[ 0 ].tree )
node.i = res[ 0 ].i++
res[ node.i ] = {
found: 1,
value: record
}
}
} )
return res
},
[ { i: 1, tree: {} } ]
)
.slice( 1 )
.filter( node => node.found === lists.length )
.map( n => n.value )
This is essentially the same as the joinById method, except that it keeps an index object to identify records to join. The records are stored in an array and the index stores the position of the record for the given key set and the number of lists it's been found in.
Each time the same key set is encountered, it finds the node in the tree, updates the element at it's index, and the number of times it's been found is incremented.
After joining, the idx object is removed from the array with the slice and any elements that weren't found in each set are removed. This makes it an inner join, you could remove this filter and have a full outer join.
Finally each element is mapped to it's value, and you have the joined arrays.