What is the best way to determine the amount of distinct elements in an array of objects based on a specific object property?

Question

What is the best way to determine the amount of distinct elements in an array of objects based on a specific object property?

I am working with an array called orders.

orders = [
  {table_id: 3, food_id: 5},
  {table_id: 4, food_id: 2},
  {table_id: 1, food_id: 6},
  {table_id: 3, food_id: 4},
  {table_id: 4, food_id: 6},
];

I am looking to create a function that can calculate the number of unique table_ids and food_ids in this array. For example,

For instance,
The unique table_ids in the list are 1, 3, 4 totaling 3.
The unique food_ids in the list are 2, 4, 5, 6 totaling 4

Can anyone guide me on how to accomplish this task?

javascript typescript ecmascript-6

Answer 1

Answer №1

If you're looking for a solution, here's one that might do the trick:

function findUniqueTableId(orders){
    let uniqueTableId = new Set();
    for(let order of orders){
        uniqueTableId.add(order.table_id);
    }
    return uniqueTableId.size;
}

You can also apply a similar approach to the food_id property in a separate function.

Kudos to @Felix King for coming up with a succinct one-liner using the set object and Array.prototype.map()

let uniqueTableIdCount = new Set(orders.map(x => x.table_id)).size
let uniqueFoodIdCount = new Set(orders.map(x => x.food_id)).size

The .map() method iterates through the array, adds each table_id to the set, and retains only unique values. By chaining methods, we can efficiently determine the size of the set.

Answer 2

If you're looking for a solution, here's one that might do the trick:

function findUniqueTableId(orders){
    let uniqueTableId = new Set();
    for(let order of orders){
        uniqueTableId.add(order.table_id);
    }
    return uniqueTableId.size;
}

You can also apply a similar approach to the food_id property in a separate function.

Kudos to @Felix King for coming up with a succinct one-liner using the set object and Array.prototype.map()

let uniqueTableIdCount = new Set(orders.map(x => x.table_id)).size
let uniqueFoodIdCount = new Set(orders.map(x => x.food_id)).size

The .map() method iterates through the array, adds each table_id to the set, and retains only unique values. By chaining methods, we can efficiently determine the size of the set.

Answer 3

Answer №2

I am confident that you have not attempted this yet because the solution is straightforward:

1 - Initialize the arrays

var uniqueIds = [];
var uniqueItems = [];

2 - Iterate through orders and fill the arrays

for (var i = 0; i < orders.length; i++) {
    if (uniqueIds.indexOf(orders[i].id) == -1) // if this ID is not in the unique ID list
        uniqueIds.push(orders[i].id);          // Add the ID to the unique ID list
    if (uniqueItems.indexOf(orders[i].item_id) == -1)  // if this item ID is not in the unique item ID list
        uniqueItems.push(orders[i].item_id);            // Add the item ID to the unique item ID list
}

With your current orders array, the values of uniqueIds and uniqueItems are :

uniqueIds // = [3, 4, 1];
uniqueItems // = [5, 2, 6, 4];

You can also find their lengths :

uniqueIds.length // = 3
uniqueItems.length // = 4

Answer 4

I am confident that you have not attempted this yet because the solution is straightforward:

1 - Initialize the arrays

var uniqueIds = [];
var uniqueItems = [];

2 - Iterate through orders and fill the arrays

for (var i = 0; i < orders.length; i++) {
    if (uniqueIds.indexOf(orders[i].id) == -1) // if this ID is not in the unique ID list
        uniqueIds.push(orders[i].id);          // Add the ID to the unique ID list
    if (uniqueItems.indexOf(orders[i].item_id) == -1)  // if this item ID is not in the unique item ID list
        uniqueItems.push(orders[i].item_id);            // Add the item ID to the unique item ID list
}

With your current orders array, the values of uniqueIds and uniqueItems are :

uniqueIds // = [3, 4, 1];
uniqueItems // = [5, 2, 6, 4];

You can also find their lengths :

uniqueIds.length // = 3
uniqueItems.length // = 4

Answer 5

Answer №3

To put it simply:

let uniqueResults = {};

items.forEach(function(order) {
  for (let prop in order) {
    uniqueResults[prop] = uniqueResults[prop] || [];
    if (uniqueResults[prop].indexOf(order[prop]) == -1)
      uniqueResults[prop].push(order[prop]);
  }
})

for (let prop in uniqueResults) {
  console.log("The unique "+ prop +" are "+ uniqueResults[prop] +" with a total count of "+ uniqueResults[prop].length);
}

With only a few lines of code, no need for additional data structures besides the uniqueResults object, no alterations to your current code, and, most importantly, no prior knowledge of the keys in the orders objects required

Answer 6

To put it simply:

let uniqueResults = {};

items.forEach(function(order) {
  for (let prop in order) {
    uniqueResults[prop] = uniqueResults[prop] || [];
    if (uniqueResults[prop].indexOf(order[prop]) == -1)
      uniqueResults[prop].push(order[prop]);
  }
})

for (let prop in uniqueResults) {
  console.log("The unique "+ prop +" are "+ uniqueResults[prop] +" with a total count of "+ uniqueResults[prop].length);
}

With only a few lines of code, no need for additional data structures besides the uniqueResults object, no alterations to your current code, and, most importantly, no prior knowledge of the keys in the orders objects required

Answer 7

Answer №4

One way to find the total number of unique values is by adding them to a Set and checking the size of the set.

For those looking for an ES5 compatible solution, it is possible to create a custom set class to achieve the same result.

Answer 8

One way to find the total number of unique values is by adding them to a Set and checking the size of the set.

For those looking for an ES5 compatible solution, it is possible to create a custom set class to achieve the same result.

Answer 9

Answer №5

If you need to generate a list of unique table and food IDs and calculate their occurrences simultaneously, the following piece of code will help you accomplish that task. By executing this code snippet, you will create two objects to store this information...

var orders = [
  {table_id: 6, food_id: 8},
  {table_id: 2, food_id: 3},
  {table_id: 4, food_id: 7},
  {table_id: 6, food_id: 5},
  {table_id: 2, food_id: 7},
];

var tables = {};
var foods = {};

for (var i in orders) {
  var order = orders[i];
  var tableId = "table_id_" + order.table_id;
  var foodId = "food_id_" + order.food_id;

if (tables.hasOwnProperty(tableId)) {
  tables[tableId]++;
  }
  else {
  tables[tableId] = 1;
  }
  
  if (foods.hasOwnProperty(foodId)) {
  foods[foodId]++;
  }
  else {
  foods[foodId] = 1;
  }
}

console.log(tables);
console.log(foods);

To those bothered by my choice of the variable "foods," I must apologize—it's just a bad habit of mine.

Answer 10

If you need to generate a list of unique table and food IDs and calculate their occurrences simultaneously, the following piece of code will help you accomplish that task. By executing this code snippet, you will create two objects to store this information...

var orders = [
  {table_id: 6, food_id: 8},
  {table_id: 2, food_id: 3},
  {table_id: 4, food_id: 7},
  {table_id: 6, food_id: 5},
  {table_id: 2, food_id: 7},
];

var tables = {};
var foods = {};

for (var i in orders) {
  var order = orders[i];
  var tableId = "table_id_" + order.table_id;
  var foodId = "food_id_" + order.food_id;

if (tables.hasOwnProperty(tableId)) {
  tables[tableId]++;
  }
  else {
  tables[tableId] = 1;
  }
  
  if (foods.hasOwnProperty(foodId)) {
  foods[foodId]++;
  }
  else {
  foods[foodId] = 1;
  }
}

console.log(tables);
console.log(foods);

To those bothered by my choice of the variable "foods," I must apologize—it's just a bad habit of mine.

Answer 11

Answer №6

To implement a unique filter on an array of objects in TypeScript, you can use a combination of `Array.prototype.sort` and `Array.prototype.filter` methods:

type Order = {
    table_id: number;
    food_id: number;
}

let orders = [
  {table_id: 3, food_id: 5},
  {table_id: 4, food_id: 2},
  {table_id: 1, food_id: 6},
  {table_id: 3, food_id: 4},
  {table_id: 4, food_id: 6},
] as Order[];

let uniqueOrders = orders.sort((a, b) => a.table_id - b.table_id)
    .filter((value, index, array) => index === array.length - 1 || array[index + 1].table_id !== value.table_id);

console.log(uniqueOrders.length); // 3

(code in TypeScript playground)

Answer 12

To implement a unique filter on an array of objects in TypeScript, you can use a combination of `Array.prototype.sort` and `Array.prototype.filter` methods:

type Order = {
    table_id: number;
    food_id: number;
}

let orders = [
  {table_id: 3, food_id: 5},
  {table_id: 4, food_id: 2},
  {table_id: 1, food_id: 6},
  {table_id: 3, food_id: 4},
  {table_id: 4, food_id: 6},
] as Order[];

let uniqueOrders = orders.sort((a, b) => a.table_id - b.table_id)
    .filter((value, index, array) => index === array.length - 1 || array[index + 1].table_id !== value.table_id);

console.log(uniqueOrders.length); // 3

(code in TypeScript playground)

Answer 13

Answer №7

Task completed!

orders = [
  {table_id: 3, food_id: 5},
  {table_id: 4, food_id: 2},
  {table_id: 1, food_id: 6},
  {table_id: 3, food_id: 4},
  {table_id: 4, food_id: 6},
];

var uniqueTables=[];
var uniqueFoods=[];

for(var i=0; i<orders.length; i++){
  var current=orders[i];
  var tableId = current.table_id;
  var pushTable=true;
  for(var j=0; j<uniqueTables.length;j++){
    if(tableId==uniqueTables[j]){
      pushTable = false;
      break;
    }
  }
  if(pushTable){uniqueTables.push(tableId);}
  
  var foodId = current.food_id;
  var pushFood=true;
  for(var j=0; j<uniqueFoods.length;j++){
    if(foodId==uniqueFoods[j]){
      pushFood = false;
      break;
    }
  }
  if(pushFood){uniqueFoods.push(foodId);}
}
console.log("Unique Tables: " + uniqueTables.length + " Unique Foods: " + uniqueFoods.length);

Answer 14

Task completed!

orders = [
  {table_id: 3, food_id: 5},
  {table_id: 4, food_id: 2},
  {table_id: 1, food_id: 6},
  {table_id: 3, food_id: 4},
  {table_id: 4, food_id: 6},
];

var uniqueTables=[];
var uniqueFoods=[];

for(var i=0; i<orders.length; i++){
  var current=orders[i];
  var tableId = current.table_id;
  var pushTable=true;
  for(var j=0; j<uniqueTables.length;j++){
    if(tableId==uniqueTables[j]){
      pushTable = false;
      break;
    }
  }
  if(pushTable){uniqueTables.push(tableId);}
  
  var foodId = current.food_id;
  var pushFood=true;
  for(var j=0; j<uniqueFoods.length;j++){
    if(foodId==uniqueFoods[j]){
      pushFood = false;
      break;
    }
  }
  if(pushFood){uniqueFoods.push(foodId);}
}
console.log("Unique Tables: " + uniqueTables.length + " Unique Foods: " + uniqueFoods.length);

Answer 15

Answer №8

To extract specific elements, consider utilizing the Set method combined with mapping.

var data = [{ category_id: 3, product_id: 5 }, { category_id: 4, product_id: 2 }, { category_id: 1, product_id: 6 }, { category_id: 3, product_id: 4 }, { category_id: 4, product_id: 6 }], 
    filterUnique = key => [...new Set(data.map(d => d[key]))];
  
console.log(filterUnique('category_id'));
console.log(filterUnique('product_id'));

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

Answer 16

To extract specific elements, consider utilizing the Set method combined with mapping.

var data = [{ category_id: 3, product_id: 5 }, { category_id: 4, product_id: 2 }, { category_id: 1, product_id: 6 }, { category_id: 3, product_id: 4 }, { category_id: 4, product_id: 6 }], 
    filterUnique = key => [...new Set(data.map(d => d[key]))];
  
console.log(filterUnique('category_id'));
console.log(filterUnique('product_id'));

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

Answer 17

Answer №9

let ordersList = [
  {tableNumber: 3, foodNumber: 5},
  {tableNumber: 4, foodNumber: 2},
  {tableNumber: 1, foodNumber: 6},
  {tableNumber: 3, foodNumber: 4},
  {tableNumber: 4, foodNumber: 6},
  {tableNumber: 4, foodNumber: 6},
  {tableNumber: 4, foodNumber: 6}
];
  
let tableObject = {}, foodObject = {};
ordersList.forEach(function(order){
  tableObject[order.tableNumber] = null;
  foodObject[order.foodNumber] = null;
});

let uniqueTableNumbers =  Object.keys(tableObject);
let uniqueFoodNumbers = Object.keys(foodObject);

console.log("Unique table numbers: ", uniqueTableNumbers);
console.log("Unique food numbers: ", uniqueFoodNumbers);

Answer 18

let ordersList = [
  {tableNumber: 3, foodNumber: 5},
  {tableNumber: 4, foodNumber: 2},
  {tableNumber: 1, foodNumber: 6},
  {tableNumber: 3, foodNumber: 4},
  {tableNumber: 4, foodNumber: 6},
  {tableNumber: 4, foodNumber: 6},
  {tableNumber: 4, foodNumber: 6}
];
  
let tableObject = {}, foodObject = {};
ordersList.forEach(function(order){
  tableObject[order.tableNumber] = null;
  foodObject[order.foodNumber] = null;
});

let uniqueTableNumbers =  Object.keys(tableObject);
let uniqueFoodNumbers = Object.keys(foodObject);

console.log("Unique table numbers: ", uniqueTableNumbers);
console.log("Unique food numbers: ", uniqueFoodNumbers);

Answer 19

Answer №10

data = [
  {user_id: 3, item_id: 5},
  {user_id: 4, item_id: 2},
  {user_id: 1, item_id: 6},
  {user_id: 3, item_id: 4},
  {user_id: 4, item_id: 6},
];
  
function countUniqueItems(arr, key) {
    result = []; 
    arr.map(obj => obj[key])
        .filter(n => result.indexOf(n) <= -1 && result.push(n))

    return result.length;
}

print(countUniqueItems(data, 'user_id'));
print(countUniqueItems(data, 'item_id'));

By the way, the use of && here might seem a bit hacky...

Answer 20

data = [
  {user_id: 3, item_id: 5},
  {user_id: 4, item_id: 2},
  {user_id: 1, item_id: 6},
  {user_id: 3, item_id: 4},
  {user_id: 4, item_id: 6},
];
  
function countUniqueItems(arr, key) {
    result = []; 
    arr.map(obj => obj[key])
        .filter(n => result.indexOf(n) <= -1 && result.push(n))

    return result.length;
}

print(countUniqueItems(data, 'user_id'));
print(countUniqueItems(data, 'item_id'));

By the way, the use of && here might seem a bit hacky...

What is the best way to determine the amount of distinct elements in an array of objects based on a specific object property?

Answer №1

Answer №2

1 - Initialize the arrays

2 - Iterate through orders and fill the arrays

Answer №3

Answer №4

Answer №5

Answer №6

Answer №7

Answer №8

Answer №9

Answer №10

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