How to efficiently evaluate element in javascript array by using reduce function

Question

I have an array called people which is an array of objects(person's name and his/her city's name), and i want to make a function that calculate the total number of distinct cities in that array. I used a function that use for loop but it seems to be a better way by using reduce functions in javascript. Here is the snippet

const people = [
  { name: "Jessica", city: "New York"},
  { name: "Steve",   city: "Los Angels"},
  { name: "Peter",   city: "Boston"},
  { name: "Elaine",  city: "Montreal"},
  { name: "Chris",  city: "Montreal"},
  { name: "Mike",  city: "Boston"},
  { name: "George",  city: "Vancouver"},
];
let nbre_distinct_cities = 0;

countDistinctCity(people);
console.log('Total number of distinct cities: ',nbre_distinct_cities); 

function countDistinctCity(people)
{
  for(let i = 0; i < people.length; i++)
  {
      if(i === people.length - 1)
      {
          break;
      }
      else if(people[i].city !== people[i + 1].city)
      {
        nbre_distinct_cities++
      }
  }

}

I would appreciate if someone suggest an efficient function using reduce() function

What's the point of that first if statement? What if the last city in the list is unique? — Pointy
– Pointy, Commented Jul 31, 2019 at 15:12
@Pointy It's because the logic is wrong to avoid an index out of bounds. This logic doesn't count the unique entries. Consider using a Set — Christopher Schneider
– Christopher Schneider, Commented Jul 31, 2019 at 15:14
@Pointy. i use the if just to make sure i don't compare with undefined element. i compare previous element and next element of the array — Kenzo
– Kenzo, Commented Jul 31, 2019 at 15:14
In regards to your title and the last line of your question: every() certainly does not apply here. Check out the documentation. reduce() is closer, but you could probably just .map() the cities into a new Set and get its size. — Tyler Roper
– Tyler Roper, Commented Jul 31, 2019 at 15:16
@ChristopherSchneider oh right, I see now. Yea this is just all wrong; a Set (or even a simple object since the city names are just strings) would work. — Pointy
– Pointy, Commented Jul 31, 2019 at 15:21

slider · Accepted Answer · 2019-07-31 15:17:43Z

6

You can use a Set to store all the cities from the array and, since a set only has unique entries, the final size of the set will give you the number of distinct cities:

const people = [
  { name: "Jessica", city: "New York"},
  { name: "Steve",   city: "Los Angels"},
  { name: "Peter",   city: "Boston"},
  { name: "Elaine",  city: "Montreal"},
  { name: "Chris",  city: "Montreal"},
  { name: "Mike",  city: "Boston"},
  { name: "George",  city: "Vancouver"},
];

let nbre_distinct_cities = new Set(people.map(({city}) => city)).size;
console.log('Total number of distinct cities: ', nbre_distinct_cities);

answered Jul 31, 2019 at 15:17

slider

13.1k1 gold badge29 silver badges45 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

jjbskir · Accepted Answer · 2019-07-31 15:24:09Z

1

Using reduce

Object.keys(people.reduce((acc, ppl) => (acc[ppl.city] = ppl.city, acc), {})).length

answered Jul 31, 2019 at 15:24

jjbskir

11.3k10 gold badges49 silver badges63 bronze badges

Comments

Max · Accepted Answer · 2019-07-31 15:19:14Z

0

you can solve the problem via reduce method

    const cities = people.reduce((accumulator, current) => {
       const isItNotExistInAccumulator
          = accumulator.every(city => city !== current.city);
       if (isItNotExistInAccumulator) return [current.city, ...accumulator];
       return accumulator;
    }, []);

    console.log(cities.length);

answered Jul 31, 2019 at 15:19

Max

8019 silver badges21 bronze badges

7 Comments

Tyler Roper Over a year ago

This would work but I'd suggest a simple accumulator.indexOf(current.city) < 0 instead of using .every(), personally.

ggorlen Over a year ago

accumulator.every is O(n^2), looping repeatedly over the accumulator for each and every element. Set or object is O(n), with a constant time check for adding each element (and less verbose). indexOf(current.city >= 0) suffers from the same problems.

Tyler Roper Over a year ago

@ggorlen I wasn't arguing that this answer was better than the other - in fact I suggested the very same Set implementation in the comments above. Though, indexOf is at least more efficient than .every(), as it doesn't require the invocation of a callback method. I'm simply noting that using .every() to check if a string exists in an array is never a great choice.

Tyler Roper Over a year ago

@ggorlen Again, you seem to be debating something that we agree upon. I myself suggested the Set method in the comments on OP before there were any answers, because I agree - it is the best one, given its time complexity. However, I am telling Max that you should never use .every() to determine if a string exists in an array. I don't think this is the best answer regardless, but it's still a valuable takeaway.

ziggy wiggy Over a year ago

Using .every() reimplements functionality found elsewhere. Using .indexOf would be preferable, or .includes() is better yet. Even using .some() would arguably be clearer than .every().

|

Karim · Accepted Answer · 2019-07-31 15:30:55Z

0

An alternative solution that removes the duplicates with Array.indexOf

const people = [
  { name: "Jessica", city: "New York"},
  { name: "Steve",   city: "Los Angels"},
  { name: "Peter",   city: "Boston"},
  { name: "Elaine",  city: "Montreal"},
  { name: "Chris",  city: "Montreal"},
  { name: "Mike",  city: "Boston"},
  { name: "George",  city: "Vancouver"},
];

let nbre_distinct_cities = people.map(el => el.city)
 .filter((city, idx, arr) => arr.indexOf(city) === idx).length;

console.log('Total number of distinct cities: ', nbre_distinct_cities);

edited Jul 31, 2019 at 15:30

answered Jul 31, 2019 at 15:25

Karim

8,6823 gold badges27 silver badges36 bronze badges

Collectives™ on Stack Overflow

How to efficiently evaluate element in javascript array by using reduce function

4 Answers 4

Comments

Comments

7 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

Comments

Comments

7 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related