Retrieve objects in array if property matches another array

Question

I want to create a new array containing contact objects if the value property in contacts matches the values in selectedContact. Any simpler way to do this?

selectedContact: number[] = [0,2] //value
contacts: Contact[] = [{ 
  firstName:"Dan";
  lastName:"Chong";
  email:"[email protected]";
  value:0;
},
{ 
  firstName:"Mark";
  lastName:"Wong";
  email:"[email protected]";
  value:1;
},
{ 
  firstName:"Layla";
  lastName:"Sng";
  email:"[email protected]";
  value: 2;
}]

Intended final result:

newArray = [{ 
 firstName:"Dan";
 lastName:"Chong";
 email:"[email protected]";
 value:0;
},{ 
 firstName:"Layla";
 lastName:"Sng";
 email："[email protected]";
 value：2;
}];

My current solution:

const newArray: Contact[] = [];
this.selectedContact.forEach(index => {
  newArray.push(this.contacts.find(c => c.value === index));
});

Failed to mention how current solution differs from expected result or any errors encountered — charlietfl
– charlietfl, Commented Nov 12, 2018 at 4:34
There's definitely a better way, but what do you mean by "simpler", what are you looking for? — CertainPerformance
– CertainPerformance, Commented Nov 12, 2018 at 4:36
What are your criteria for "simpler"? It's difficult to conceive of a simpler method. There might be more efficient ones though, such as creating an index of contacts for reuse. — RobG
– RobG, Commented Nov 12, 2018 at 4:41
Sorry I meant a more efficient way to go about doing this, without modifications to the arrays. The current solution provides the desired output but is there a more efficient method e.g. without using a forEach loop? — iBlehhz
– iBlehhz, Commented Nov 12, 2018 at 4:45

lucascaro · Accepted Answer · 2018-11-12 05:57:16Z

2

In terms of performance, it would be better to iterate over selectedContacts rather than contacts, especially since contacts are indexed (as an array) and you are selecting through the index.

Say the length of contacts is N and the length of selectedContacts is M.

Since selectedContacts is a subset of contacts, we know M <= N. For large databases of contacts, this difference could be significant.

The code in the question:

this.selectedContact.forEach(index => {
  newArray.push(this.contacts.find(c => c.value === index));
});

Has O(M*N) since it iterates over selectedContact O(M) and on each iteration it find a value in contacts (O(N)).

The code from the accepted answer iterates over contact (O(N)) and looks for a value in selectedContact which is O(M). This makes the algorithm equivalent, with O(N*M)

In your example, you already have a cheap way of looking up contacts by number since contacts is an array and your indexes are simply the index in the array.

This means you can use code like this:

return this.selectedContact.map(index => this.contacts[index]);

Since accessing an array element by index has O(1), this would have O(M) which is the smallest of the sizes.

If you can't use the array index as a key, you can use other data structures, like a Map where the id is the key, and the contact is the value. This would have similar lookup speeds (roughly O(1)).

answered Nov 12, 2018 at 5:57

lucascaro

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1 Comment

iBlehhz Over a year ago

I see. Thanks for your detailed explanation!! :)

Mamun · Accepted Answer · 2018-11-12 05:24:34Z

2

You can use Array.prototype.filter()

The filter() method creates a new array with all elements that pass the test implemented by the provided function.

and Array.prototype.includes()

The includes() method determines whether an array includes a certain element, returning true or false as appropriate.

Working Code Example:

var selectedContact = [0,2];
var contacts = [{ 
  firstName: "Dan",
  lastName: "Chong",
  email: "[email protected]",
  value: 0
},
{ 
  firstName: "Mark",
  lastName: "Wong",
  email: "[email protected]",
  value: 1
},
{ 
  firstName: "Layla",
  lastName: "Sng",
  email: "[email protected]",
  value: 2
}]
let newArray =  contacts.filter(c => selectedContact.includes(c.value));

console.log(newArray);

edited Nov 12, 2018 at 5:24

answered Nov 12, 2018 at 4:36

Mamun

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1 Comment

RobG Over a year ago

@iBlehhz—yes, but "works" is not a measure of efficiency. ;-) This will be efficient where selectedContacts is small, but if it's not, an indexed solution may be better (assuming the cost of creating the index can be recovered).

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