What is the process for verifying the existence of a key value pair within one array in comparison to another array?

Question

What is the process for verifying the existence of a key value pair within one array in comparison to another array?

I have two different sets of data, represented by arrays:

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 response retrieved in JSON format via a GET request and may be larger than the result array. For example, the invoices array includes Bob while he does not appear in the result array.

How can I identify the id of objects whose Client does not match any entry in the Person field of the result array? I attempted to accomplish this using nested for loops without success.

javascript arrays typescript loops

Answer 1

Answer №1

By creating an object with keys representing the Person values found in the result array, we can efficiently filter the values in the invoices array and then use map to extract only the id values of the filtered results:

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);

Answer 2

By creating an object with keys representing the Person values found in the result array, we can efficiently filter the values in the invoices array and then use map to extract only the id values of the filtered results:

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);

Answer 3

Answer №2

To apply a filter on the invoices Array, utilize the result Array as the argument and search for any Client in it (refer to MDN for Array.filter). Remember, the callback function used with filter should be a standard function (not an arrow function) so that you can access the result-Array for thisArg.

console.log( `Ids not present 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; }

Answer 4

To apply a filter on the invoices Array, utilize the result Array as the argument and search for any Client in it (refer to MDN for Array.filter). Remember, the callback function used with filter should be a standard function (not an arrow function) so that you can access the result-Array for thisArg.

console.log( `Ids not present 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; }

Answer 5

Answer №3

Utilizing Set and flatMap provides a concise and effective solution. Set enables quick member checking in an array. As for flatMap, it allows filtering and mapping simultaneously.

// 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);

Answer 6

Utilizing Set and flatMap provides a concise and effective solution. Set enables quick member checking in an array. As for flatMap, it allows filtering and mapping simultaneously.

// 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);

Answer 7

Answer №4

To achieve this, you can create a Map of persons.

An advantage of this approach is that you can easily access person data by their name.

Note that I adjusted the invoices by 1 row to improve the result view.

// Enter data 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 without corresponding results
const orphans = invoices.filter((i)=>!persons.has(i.Client));
// ID of invoices without corresponding results
const orphansId = orphans.map((o)=>o.id)

console.log(orphansId);

Resulting in: [ 90, 92 ]

Answer 8

To achieve this, you can create a Map of persons.

An advantage of this approach is that you can easily access person data by their name.

Note that I adjusted the invoices by 1 row to improve the result view.

// Enter data 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 without corresponding results
const orphans = invoices.filter((i)=>!persons.has(i.Client));
// ID of invoices without corresponding results
const orphansId = orphans.map((o)=>o.id)

console.log(orphansId);

Resulting in: [ 90, 92 ]

What is the process for verifying the existence of a key value pair within one array in comparison to another array?

Answer №1

Answer №2

Answer №3

Answer №4

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