Eliminate any repetitive items from the array while simultaneously boosting their quantity

Question

Eliminate any repetitive items from the array while simultaneously boosting their quantity

Here is a list that I have:

const myList = [
  { name: 'AA', quantity: 1 },
  { name: 'AA', quantity: 1 },
  { name: 'BB', quantity: 1 },
  { name: 'CC', quantity: 1 },
  { name: 'CC', quantity: 2 },
]

I want the output to be like this:

const updatedList = [
  { name: 'AA', quantity: 2 },
  { name: 'BB', quantity: 1 },
  { name: 'CC', quantity: 3 },
]

I am trying to remove duplicates and increase the quantity when a duplicate is found, but struggling with the logic.

const uniqueSet = new Set()
const finalResult = myList.reduce((acc, item) => {
  if (!uniqueSet.has(item.name)) {
    uniqueSet.add(item.name)
    acc.push({ name: item.name, quantity: item.quantity })
  }

  // for (const iterator of myList) {
  //   if (uniqueSet.has(item.name)) {
  //     console.log('🚀 ~ file: Untitled-1 ~ line 15 ~ iterator', iterator)
  //   }
  // }

  console.log()
  return acc
}, [])

javascript arrays object

Answer 1

Answer №1

To achieve this, you can utilize the reduce method and then search for the element in the resulting array. If it does not exist, use push to add it to the array. If it does exist, simply add the current element's quantity to the existing element with the same name in the result array.

Here is the code snippet:

const list=[{name:"AA",quantity:1},{name:"AA",quantity:1},{name:"BB",quantity:1},{name:"CC",quantity:1},{name:"CC",quantity:2}];

const res = list.reduce((acc, e) => {
  const found = acc.find(x => e.name === x.name)
  found ? found.quantity += e.quantity : acc.push(e)
  return acc
}, [])

console.log(res)

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

Answer 2

To achieve this, you can utilize the reduce method and then search for the element in the resulting array. If it does not exist, use push to add it to the array. If it does exist, simply add the current element's quantity to the existing element with the same name in the result array.

Here is the code snippet:

const list=[{name:"AA",quantity:1},{name:"AA",quantity:1},{name:"BB",quantity:1},{name:"CC",quantity:1},{name:"CC",quantity:2}];

const res = list.reduce((acc, e) => {
  const found = acc.find(x => e.name === x.name)
  found ? found.quantity += e.quantity : acc.push(e)
  return acc
}, [])

console.log(res)

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

Answer 3

Answer №2

Using a unique method to manipulate objects and return an array with Object.values()

const items = [
  { item: 'AA', count: 1 },
  { item: 'AA', count: 1 },
  { item: 'BB', count: 1 },
  { item: 'CC', count: 1 },
  { item: 'CC', count: 2 },
]

const groupedItems = {};

items.forEach(element => {    
   const obj =  groupedItems[element.item] = groupedItems[element.item] || {...element, count: 0}
   obj.count += element.count
})

const result = Object.values(groupedItems)
console.log(result)

Answer 4

Using a unique method to manipulate objects and return an array with Object.values()

const items = [
  { item: 'AA', count: 1 },
  { item: 'AA', count: 1 },
  { item: 'BB', count: 1 },
  { item: 'CC', count: 1 },
  { item: 'CC', count: 2 },
]

const groupedItems = {};

items.forEach(element => {    
   const obj =  groupedItems[element.item] = groupedItems[element.item] || {...element, count: 0}
   obj.count += element.count
})

const result = Object.values(groupedItems)
console.log(result)

Answer 5

Answer №3

To efficiently combine and sum up values based on a specific property, you can use a loop like the following:

let mergedList = [];
list.forEach((item) => {
   let existingItem = mergedList.find(i => i.name == item.name);
   if (existingItem) {
       existingItem.quantity += item.quantity;
   } else {
       mergedList.push(item);
   }
});

Answer 6

To efficiently combine and sum up values based on a specific property, you can use a loop like the following:

let mergedList = [];
list.forEach((item) => {
   let existingItem = mergedList.find(i => i.name == item.name);
   if (existingItem) {
       existingItem.quantity += item.quantity;
   } else {
       mergedList.push(item);
   }
});

Answer 7

Answer №4

The best approach to this problem is utilizing an object structure, as it allows for the creation of a unique entity with distinct keys.

const itemList = [
  { itemName: 'AA', itemQuantity: 1 },
  { itemName: 'AA', itemQuantity: 1 },
  { itemName: 'BB', itemQuantity: 1 },
  { itemName: 'CC', itemQuantity: 1 },
  { itemName: 'CC', itemQuantity: 2 },
];

const updatedList = itemList.reduce((accumulator, currentItem) => {
  if (!accumulator[currentItem.itemName]) accumulator[currentItem.itemName] = 0;
  accumulator[currentItem.itemName] += currentItem.itemQuantity;
  return accumulator;
}, {});

Answer 8

The best approach to this problem is utilizing an object structure, as it allows for the creation of a unique entity with distinct keys.

const itemList = [
  { itemName: 'AA', itemQuantity: 1 },
  { itemName: 'AA', itemQuantity: 1 },
  { itemName: 'BB', itemQuantity: 1 },
  { itemName: 'CC', itemQuantity: 1 },
  { itemName: 'CC', itemQuantity: 2 },
];

const updatedList = itemList.reduce((accumulator, currentItem) => {
  if (!accumulator[currentItem.itemName]) accumulator[currentItem.itemName] = 0;
  accumulator[currentItem.itemName] += currentItem.itemQuantity;
  return accumulator;
}, {});

Eliminate any repetitive items from the array while simultaneously boosting their quantity

Answer №1

Answer №2

Answer №3

Answer №4

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