How to get the sum of a key of consecutive array object and remove the duplicates?

Question

I have a main array -

const arr = [
    {  description: 'Senior', amount: 50 },
    {  description: 'Senior', amount: 50 },
    {  description: 'Adult', amount: 75 },
    {  description: 'Adult', amount: 35 },
    {  description: 'Infant', amount: 25 },
    {  description: 'Senior', amount: 150 }
]

I want help with an es6 operation which will add the amount based on the key(description) and remove the duplicates.

Result array will somewhat look like -

const newArr = [
        {  description: 'Senior', amount: 100 },
        {  description: 'Adult', amount: 110 },
        {  description: 'Infant', amount: 25 },
        {  description: 'Senior', amount: 150 }
]

Please note, the grouping will be done only on consecutive objects with similar key.

I have been using the reduce operator to achieve this using the solution, but that removes the non-consecutive objects as well.

It would be really helpful if someone can help me with some es6 operators to perform the same operation.

Mister Jojo · Accepted Answer · 2021-12-21 18:24:26Z

4

this way

const arr = 
  [ { description: 'Senior', amount:  50 } 
  , { description: 'Senior', amount:  50 } 
  , { description: 'Adult',  amount:  75 } 
  , { description: 'Adult',  amount:  35 } 
  , { description: 'Infant', amount:  25 } 
  , { description: 'Senior', amount: 150 } 
  ] 

const newArr = 
  arr.reduce((res,{description,amount},i,{[i-1]:prev})=>
    {
    if (!prev || description != prev.description )
      res.push({description,amount})
    else
      res[res.length-1].amount += amount
    return res
    },[])

console.log( newArr )

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

edited Dec 21, 2021 at 18:24

answered Dec 21, 2021 at 18:09

Mister Jojo

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2 Comments

Aritra Ganguly Over a year ago

This solution removes the 6th element of the array as well. I want to remove/group only the consecutive objects.

Mister Jojo Over a year ago

@AritraGanguly updated...

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