How can I remove a specific object from my array based on object values

Question

I have the following array of objects in javascript.

[{
     's': 'ETHBTC',
     'q': 123456
},
{
     's': 'XMLBTC',
     'q': 545454
},
{
     's': 'ETHBTC',
     'q': 123451
}]

I want to remove the duplicate objects based on condition that if all the objects have same value of the key s ,I want to keep the only one that has the highest value of key q . In the above example, I would want to only keep

[{
     's': 'ETHBTC',
     'q': 123456
},
{
     's': 'XMLBTC',
     'q': 545454
}]

because it has highest value of key q for key s : ETHBTC and s : XMLBTC was missing so I pushed it into the array. What can be the best approach for it?

Answer 1

I'd reduce into an object indexed by s , checking whether the current object at that index (if any) has a higher or lower q , and then getting the Object.values of the result to turn it back into an array:

 const input = [{ 's': 'ETHBTC', 'q': 123456 }, { 's': 'XMLBTC', 'q': 545454 }, { 's': 'ETHBTC', 'q': 123451 }]; console.log( Object.values( input.reduce((a, item) => { const { s, q } = item; if (!a[s] || a[s].q < q) a[s] = item; return a; }, {}) ) ); 

ES5 compatible solution:

 var input = [{ 's': 'ETHBTC', 'q': 123456 }, { 's': 'XMLBTC', 'q': 545454 }, { 's': 'ETHBTC', 'q': 123451 }]; var outputObj = input.reduce(function (a, item) { var s = item.s, q = item.q; if (!a[s] || a[s].q < q) a[s] = item; return a; }, {}); console.log( Object.keys(outputObj).map(key => outputObj[key]) );