Identify the key in an array of objects that corresponds to the highest volume value, and return both the key and its

I am working with an array of objects that include a baseAsset key and Volume for each object, with the volume value varying between objects.

My goal is to find the object with the highest volume value based on a match with the baseAsset key. Efficiency and speed are crucial due to the large size of the array containing over 3000 objects

let tickerA = [{
    pair: 'AUDUSD',
    baseAsset: 'AUD',
    lastPriceUSD: 0.74,
    volume: 1000
}, {
    pair: 'AUDUSD',
    baseAsset: 'AUD',
    lastPriceUSD: 0.76,
    volume: 2000
}, {
    pair: 'USDEUR',
    baseAsset: 'USD',
    lastPriceUSD: 1.25,
    volume: 1200
}, {
    pair: 'USDEUR',
    baseAsset: 'USD',
    lastPriceUSD: 1.19,
    volume: 1500
}]

The expected result when running a function

tickerB = [{
    baseAsset: 'AUD',
    lastPriceUSD: 0.76,
    volume: 2000
}, {
    baseAsset: 'USD',
    lastPriceUSD: 1.25,
    volume: 1500
}]
javascriptarraysobjectecmascript-6

Answer №1

To efficiently achieve this task in linear time complexity (O(n)), iterate through the list and store the largest item in an object. Once completed, the values will be accessible through the Object.values method of the specified groups:

const tickerA = [{pair: 'AUDUSD',baseAsset: 'AUD',lastPriceUSD: 0.74,volume: 1000}, {pair: 'AUDUSD',baseAsset: 'AUD',lastPriceUSD: 0.76,volume: 2000}, {pair: 'USDEUR',baseAsset: 'USD',lastPriceUSD: 1.25,volume: 1200}, {pair: 'USDEUR',baseAsset: 'USD',lastPriceUSD: 1.19,volume: 1500}]

const groups = tickerA.reduce((largest, {baseAsset, lastPriceUSD, volume}) => {
    /* 
     * Check if it's a new baseAsset or larger than the previous one, then update the group under the baseAsset key 
    */
    if (!largest[baseAsset] || largest[baseAsset]['volume'] < volume ) {
        largest[baseAsset] = {baseAsset, lastPriceUSD, volume}
    }

    return largest
}, {})

const TickerB = Object.values(groups);
console.log(TickerB);

Answer №2

To efficiently compare and select items, we can iterate through the values of tickerA and associate them with the corresponding baseAsset keys based on the condition that the volume value is higher than the current item's volume for that particular baseAsset key:

let tickerA = [{
    pair: 'AUDUSD',
    baseAsset: 'AUD',
    lastPriceUSD: 0.74,
    volume: 1000
}, {
    pair: 'AUDUSD',
    baseAsset: 'AUD',
    lastPriceUSD: 0.76,
    volume: 2000
}, {
    pair: 'USDEUR',
    baseAsset: 'USD',
    lastPriceUSD: 1.25,
    volume: 1200
}, {
    pair: 'USDEUR',
    baseAsset: 'USD',
    lastPriceUSD: 1.19,
    volume: 1500
}];

/* Initializing a map to store max volume values per baseAsset */
const map = new Map()

/* Iterating through tickerA items to identify highest volume per baseAsset class */
for(const item of tickerA) {
  
  const assetMatch = map.get(item.baseAsset);
  
  if(assetMatch && item.volume < assetMatch.volume) {
    /* Discarding current item if matching item (by asset) has greater volume */
    continue;
  }
  else {
    /* Updating map entry if this tickerA item represents first or has higher volume in its asset class */
    map.set(item.baseAsset, item);
  }  
}
 
/* Converting map values to an array */
const tickerB = Array.from(map.values());

console.log(tickerB);

Answer №3

This unique approach combines the baseAsset elements and then extracts the combined values at the end.

It operates with a time complexity of O(n).

let tickerA = [{pair: 'AUDUSD',baseAsset: 'AUD',lastPriceUSD: 0.74,volume: 1000}, {pair: 'AUDUSD',baseAsset: 'AUD',lastPriceUSD: 0.76,volume: 2000}, {pair: 'USDEUR',baseAsset: 'USD',lastPriceUSD: 1.25,volume: 1200}, {pair: 'USDEUR',baseAsset: 'USD',lastPriceUSD: 1.19,volume: 1500}];
let result = Object.values(tickerA.reduce((a, {baseAsset, lastPriceUSD, volume}) => {
    let {volume: current} = a[baseAsset] || {volume: Number.MAX_SAFE_INTEGER};
    
    if (current < volume) a[baseAsset].volume = volume;
    else a[baseAsset] = {baseAsset, lastPriceUSD, volume};
    
    return a;
}, Object.create(null)));

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

Answer №4

Try utilizing the reduce function along with Object.values:

let tickerA = [{pair: 'AUDUSD',baseAsset: 'AUD',lastPriceUSD: 0.74,volume: 1000}, {pair: 'AUDUSD',baseAsset: 'AUD',lastPriceUSD: 0.76,volume: 2000}, {pair: 'USDEUR',baseAsset: 'USD',lastPriceUSD: 1.25,volume: 1200}, {pair: 'USDEUR',baseAsset: 'USD',lastPriceUSD: 1.19,volume: 1500}];
const res = Object.values(tickerA.reduce((acc, { baseAsset, lastPriceUSD, volume }) => {
  acc[baseAsset] = (!acc[baseAsset] || acc[baseAsset].volume < volume) ? { baseAsset, lastPriceUSD, volume } : acc[baseAsset];
  return acc;
}, {}));
console.log(res);
.as-console-wrapper { max-height: 100% !important; top: auto; }

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