How to check if a key value pair (in an array) exists in another array?

Question

I have two arrays, for example:

invoices: [
    {id: 90, Client: 'Bob', paid: false, total: 900},
    {id: 91, Client: 'Sarah', paid: false, total: 400}
]

and:

result: [{km: 200, hours: 20, Person: 'Sarah'}]

The invoices array is a JSON response from a GET and will sometimes be larger than result, so I in this case invoices has Bob while it doesnt exist in result.

How can I get the id of any objects whose Client does not appear as Person in result? I tried to do it with a double for loop but it didn't quite work out.

Nick · Accepted Answer · 2020-05-08 00:11:03Z

1

By generating an object with keys which are the Person values from result, we can use that to efficiently filter the values in invoices, and then use map to return only the id values of the filtered result:

const invoices =  [
    {id: 90, Client: 'Bob', paid: false, total: 900},
    {id: 91, Client: 'Sarah', paid: false, total: 400}
]

const result =  [{km: 200, hours: 20, Person: 'Sarah'}]

const resindexes = result.reduce((c, o, i) =>
  (c[o.Person] = i, c), {});

ids = invoices.filter(o => resindexes[o.Client] === undefined).map(o => o.id);

console.log(ids);

edited May 8, 2020 at 0:11

answered May 8, 2020 at 0:02

Nick

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2 Comments

Y_Lakdime Over a year ago

=== undefined? what do you mean

Nick Over a year ago

@user3660293 only Person names that are in result will be a property in the resindexes object, so resindexes[o.Client] will be undefined for any Client name which was not in result. This allows us to filter invoices to only return the ones whose Client isn't in result.

KooiInc · Accepted Answer · 2020-05-08 07:08:53Z

1

You can filter the invoices Array, using the result Array as this-argument and check for any Client in that (see MDN for Array.filter). Note, the filter-callback should be a normal function (i.e. not an arrow function) to be able to use the result-Array for thisArg.

console.log( `Id(s) not in result\n`, 
  [ {id: 90, Client: 'Bob', paid: false, total: 900}, 
    {id: 91, Client: 'Sarah', paid: false, total: 400},
    {id: 92, Client: 'Mia', paid: false, total: 200} ]
  .filter( 
    function(v) { return !this.find(n => n.Person === v.Client); }, 
    [{km: 200, hours: 20, Person: 'Sarah'}] // <= thisArg
   )
   .map(v => v.id) // <= fetch the found id's
   .join()
);

.as-console-wrapper { top: 0; max-height: 100% !important; }

edited May 8, 2020 at 7:08

answered May 7, 2020 at 23:27

KooiInc

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3 Comments

kmoser Over a year ago

Change filter(...) to filter(...).map( (invoice) => invoice.id )

Y_Lakdime Over a year ago

map( (invoice) => invoice.id ) does not exist in the answer

kmoser Over a year ago

That's why I suggested making the change, which will give you a list of just the ids you want.

Trevor Dixon · Accepted Answer · 2020-05-08 06:39:36Z

0

Using Set and flatMap makes for an expressive, efficient solution. Set consumes an array and lets you check for members in constant time. flatMap lets you filter and map at the same time, in essence.

// Data
const invoices = [
  {id: 90, Client: 'Bob', paid: false, total: 900},
  {id: 91, Client: 'Sarah', paid: false, total: 400}
];
const personsArr = [{km: 200, hours: 20, Person: 'Sarah'}];

// Query
const persons = new Set(personsArr.map(r => r.Person));
const result = invoices.flatMap(i => !persons.has(i.Client) ? i.id : []);
console.log(result);

answered May 8, 2020 at 6:39

Trevor Dixon

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Comments

Larest · Accepted Answer · 2020-05-08 07:22:13Z

Or you can do it by make persons Map.

Plus of that attempt is you have access to person data by client name.

Notice I incremented invoices for 1 row for better result view.

// Data enter code here
const invoices = [
    {id: 90, Client: 'Bob', paid: false, total: 900},
    {id: 91, Client: 'Sarah', paid: false, total: 400},
    {id: 92, Client: 'Joe', paid: false, total: 400}
];
const personsArr = [{km: 200, hours: 20, Person: 'Sarah'}];

// Query
const persons = new Map(
    personsArr.map(r =>{ 
        return [r.Person, r]
    }
  )
);
// invoices which do not have results
const orphans = invoices.filter((i)=>!persons.has(i.Client));
// id of invoices which do not have results
const orphansId = orphans.map((o)=>o.id)

console.log(orphansId);

it returns: [ 90, 92 ]

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