Compare two array object for matched data and return new array object

Question

How can i check the available data from two array and return new array. For example i want to compare data from one array and check with another array and if it available then it will return with new array with count . Below is the two array and my expected result code.

        const restaurant = [
                { name: 'La mesa', cuisine: ['chiness', 'arabic'] },
                { name: 'Purnima', cuisine: ['thai'] },
                { name: 'Red Bull', cuisine: ['french', 'arabic'] },
                { name: 'Pasta', cuisine: ['indian'] },
            ];

            const cuisine = [
                { name: 'chiness' },
                { name: 'arabic' },
                { name: 'thai' },
                { name: 'french' },
                { name: 'italian' },
                { name: 'indian' },
                { name: 'mexican' },
            ];

            // Expected Output a new array like this below
            const myCuisine = [
                { name: 'chiness', restaurant: 1 },
                { name: 'arabic', restaurant: 2 },
                { name: 'thai', restaurant: 1 },
                { name: 'french', restaurant: 1 },
                { name: 'italian', restaurant: 0 },
                { name: 'indian', restaurant: 1 },
                { name: 'mexican', restaurant: 0 },
            ];

Thank you

Please add the code you've tried

adiga
– adiga

2020-09-05 11:24:16 +00:00
Commented Sep 5, 2020 at 11:24 — adiga
– adiga, Commented Sep 5, 2020 at 11:24

Ele · Accepted Answer · 2020-09-05 11:00:11Z

2

You can use the functions map, reduce, and some all together to build the desired output as follow:

const restaurant = [    { name: 'La mesa', cuisine: ['chiness', 'arabic'] },    { name: 'Purnima', cuisine: ['thai'] },    { name: 'Red Bull', cuisine: ['french', 'arabic'] },    { name: 'Pasta', cuisine: ['indian'] }],
      cuisine = [    { name: 'chiness' },    { name: 'arabic' },    { name: 'thai' },    { name: 'french' },    { name: 'italian' },    { name: 'indian' },    { name: 'mexican' }],
      myCuisine = cuisine.map(({name}) => ({name, restaurant: restaurant.reduce((r, {cuisine}) => r + cuisine.some(c => c === name) , 0)}));

console.log(myCuisine)

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

answered Sep 5, 2020 at 11:00

Ele

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Comments

DanieleAlessandra · Accepted Answer · 2020-09-05 11:03:08Z

1

With map and filter, this way:

const myCuisine = cuisine.map(
  item => {
    return {
      ...item,
      restaurant: restaurant.filter(
        res => res.cuisine.indexOf(item.name) >= 0
      ).length
    }
  }
);

answered Sep 5, 2020 at 11:03

DanieleAlessandra

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1 Comment

Ele Over a year ago

Using the function filter for counting should be avoided.

stranded · Accepted Answer · 2020-09-05 11:03:36Z

1

You can map the cuisines and filter the restaurants to get the number of restaurants

cuisine.map((cuisineObject) => {
  const numberOfRestaurants = restaurant.filter((restaurantObject) => restaurantObject.cuisine.includes(cuisineObject.name)).length
    return {
        ...cuisineObject,
        restaurant: numberOfRestaurants
    }
})

answered Sep 5, 2020 at 11:03

stranded

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1 Comment

Ele Over a year ago

Using the function filter for counting should be avoided.

Nithish · Accepted Answer · 2020-09-05 11:15:46Z

First we can format the restaurants with cuisine information into an object and then use the same object for finding out the number of restaurants serving the particular cuisine. This can be achieved using Array.reduce and Array.map.

const restaurant = [{name:'La mesa',cuisine:['chiness','arabic']},{name:'Purnima',cuisine:['thai']},{name:'Red Bull',cuisine:['french','arabic']},{name:'Pasta',cuisine:['indian']}];

const cuisine = [{name:'chiness'},{name:'arabic'},{name:'thai'},{name:'french'},{name:'italian'},{name:'indian'},{name:'mexican'}];

const getFormattedList = (cuisines, restaurants) => {
  return cuisines.map(cuisine => {
    return {
      ...cuisine,
      restaurant: restaurants[cuisine.name] || 0
    }
  })
}

const formatRestaurantCuisines = (restaurants) => {
  return restaurants.reduce((result, restaurant) => {
    restaurant.cuisine.forEach(cuisine => {
        result[cuisine] = (result[cuisine]||0) + 1;
    })
    return result;
  }, {});
}

//Formatted object to convert the restaurant with cuisine info to count
const formattedObj = formatRestaurantCuisines(restaurant);
console.log(formattedObj);
console.log(getFormattedList(cuisine, formattedObj))

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

Nick Parsons · Accepted Answer · 2020-09-05 11:18:27Z

You can build an object which stores all the frequencies of each cuisine with .reduce() and then use .map() on your cuisine array like so:

const restaurant = [ { name: 'La mesa', cuisine: ['chiness', 'arabic'] }, { name: 'Purnima', cuisine: ['thai'] }, { name: 'Red Bull', cuisine: ['french', 'arabic'] }, { name: 'Pasta', cuisine: ['indian'] }, ]; 
const cuisine = [ { name: 'chiness' }, { name: 'arabic' }, { name: 'thai' }, { name: 'french' }, { name: 'italian' }, { name: 'indian' }, { name: 'mexican' }, ];

const cusineFreq = restaurant.reduce((o, {cuisine}) => {
  cuisine.forEach(type => o[type] = (o[type] || 0) + 1);
  return o;
}, {}); 
const res = cuisine.map(o => ({...o, restaurant: (cusineFreq[o.name] || 0)}));
console.log(res);

This approach of creating an object for look-up is particularly useful if restaurant is large, as it allows for a time complexity of O(n + k) rather than O(n*k). Thus, it will allow for better overall performance compared to nested loops and is more scalable.

Ranjeet Eppakayala · Accepted Answer · 2020-09-05 11:41:00Z

0

Using map, flatMap and filter

Edited: using flatMap instead of map.flat

cuisine.map(({name}) => ({name: name,restaurant: restaurant.flatMap(v => v.cuisine).filter(v=>v === name).length}))

edited Sep 5, 2020 at 11:41

answered Sep 5, 2020 at 11:08

Ranjeet Eppakayala

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5 Comments

Ele Over a year ago

Using the function filter for counting should be avoided.

Ranjeet Eppakayala Over a year ago

@Ele what would be better?

Nick Parsons Over a year ago

You could use .flatMap() instead of .map().flat()

Ranjeet Eppakayala Over a year ago

@NickParsons thats interesting!

Ele Over a year ago

@RadicalEdward reduce is a better approach because you avoid creating an array which is the case of using the function filter.

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