Tips for iterating through an array of object literals and combining values with matching IDs

Question

Tips for iterating through an array of object literals and combining values with matching IDs

Looking for a more efficient way to iterate through an array of object literals and combine values in objects with duplicate IDs.

Is there a smarter approach than utilizing multiple nested for loops?

Here is the sample data provided:

{ "theList": [
    {
        "id": 101,
        "name": "Bubbles' Cat Farm",
        "number": "123"
    },
    {
        "id": 102,
        "name": "Sunnyvale Park",
        "number": "456"
    },
    {
        "id": 101,
        "name": "Bubbles' Cat Farm",
        "number": "789"
]};

The expected output should be:

{ "theList": [
    {
        "id": 101,
        "name": "Bubbles' Cat Farm",
        "number": "123, 789"
    }, 
    {
        "id": 102,
        "name": "Sunnyvale Park",
        "number": "456"
]}

javascript arrays javascript-objects

Answer 1

Answer №1

To consolidate and manipulate data in JavaScript arrays, you can utilize the reduce method.

let items = [{ "id": 101,"name": "Bubbles' Cat Farm","number": "123"},{"id": 102,"name": "Sunnyvale Park","number": "456"},{"id": 101,"name": "Bubbles' Cat Farm","number": "789"}];

let output = items.reduce((result, item) => {
  if(result[item.id]){
    result[item.id].number += ', ' + item.number;
  } else {
    result[item.id] = item
  }
  return result
},{});

console.log(Object.values(output))

Answer 2

To consolidate and manipulate data in JavaScript arrays, you can utilize the reduce method.

let items = [{ "id": 101,"name": "Bubbles' Cat Farm","number": "123"},{"id": 102,"name": "Sunnyvale Park","number": "456"},{"id": 101,"name": "Bubbles' Cat Farm","number": "789"}];

let output = items.reduce((result, item) => {
  if(result[item.id]){
    result[item.id].number += ', ' + item.number;
  } else {
    result[item.id] = item
  }
  return result
},{});

console.log(Object.values(output))

Answer 3

Answer №2

By utilizing the reduce method in conjunction with the findIndex function, you can accomplish your desired outcome.

const items = [{
    "id": 101,
    "name": "Bubble's Cat Farm",
    "number": "123"
  },
  {
    "id": 102,
    "name": "Sunnyvale Park",
    "number": "456"
  },
  {
    "id": 101,
    "name": "Bubble's Cat Farm",
    "number": "789"
  }
]

const updatedItems = items.reduce((accumulator, currentValue) => {
  const index = accumulator.findIndex(item => item.id === currentValue.id);
  if (index === -1) {
    accumulator.push(currentValue);
  } else {
    accumulator[index].number += ", " + currentValue.number;
  }
  return accumulator;
}, [])

console.log(updatedItems)

Answer 4

By utilizing the reduce method in conjunction with the findIndex function, you can accomplish your desired outcome.

const items = [{
    "id": 101,
    "name": "Bubble's Cat Farm",
    "number": "123"
  },
  {
    "id": 102,
    "name": "Sunnyvale Park",
    "number": "456"
  },
  {
    "id": 101,
    "name": "Bubble's Cat Farm",
    "number": "789"
  }
]

const updatedItems = items.reduce((accumulator, currentValue) => {
  const index = accumulator.findIndex(item => item.id === currentValue.id);
  if (index === -1) {
    accumulator.push(currentValue);
  } else {
    accumulator[index].number += ", " + currentValue.number;
  }
  return accumulator;
}, [])

console.log(updatedItems)

Answer 5

Answer №3

In the case where number is the sole key with unique values, you can achieve the desired outcome by utilizing the reduce function:

const data = {"items":[{"id":101,"name":"Bubbles' Cat Farm","number":"123"},{"id":102,"name":"Sunnyvale Park","number":"456"},{"id":101,"name":"Bubbles' Cat Farm","number":"789"}]}

const result = data.items.reduce((acc, {id,name,number}) =>{
  acc[id] 
    ? acc[id]["number"] += ", " + number 
    : acc[id] = {id,name,number};
    
  return acc
},{})

const finalResult = { "items": Object.values(result) }
console.log(finalResult)

Build an accumulator object with each unique id as a key and the required object for the final array as its value. Concatenate the number values if an id already exists in the accumulator, otherwise, add a new key to the accumulator.

{
  "101": {
    "id": 101,
    "name": "Bubbles' Cat Farm",
    "number": "123, 789"
  }
}

Finally, utilize Object.values to extract only the values of the items property into an array.

Answer 6

In the case where number is the sole key with unique values, you can achieve the desired outcome by utilizing the reduce function:

const data = {"items":[{"id":101,"name":"Bubbles' Cat Farm","number":"123"},{"id":102,"name":"Sunnyvale Park","number":"456"},{"id":101,"name":"Bubbles' Cat Farm","number":"789"}]}

const result = data.items.reduce((acc, {id,name,number}) =>{
  acc[id] 
    ? acc[id]["number"] += ", " + number 
    : acc[id] = {id,name,number};
    
  return acc
},{})

const finalResult = { "items": Object.values(result) }
console.log(finalResult)

Build an accumulator object with each unique id as a key and the required object for the final array as its value. Concatenate the number values if an id already exists in the accumulator, otherwise, add a new key to the accumulator.

{
  "101": {
    "id": 101,
    "name": "Bubbles' Cat Farm",
    "number": "123, 789"
  }
}

Finally, utilize Object.values to extract only the values of the items property into an array.

Answer 7

Answer №4

One approach is to transform your container into a dictionary, using the ID as the key and an empty list as the value. As you iterate through the data, check if the key exists. If it doesn't, create a new list with that item. If the key already exists, simply add the value to the existing list.

Answer 8

One approach is to transform your container into a dictionary, using the ID as the key and an empty list as the value. As you iterate through the data, check if the key exists. If it doesn't, create a new list with that item. If the key already exists, simply add the value to the existing list.

Tips for iterating through an array of object literals and combining values with matching IDs

Answer №1

Answer №2

Answer №3

Answer №4

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