[
{"marks": 99, "set":"A"},
{"marks": 90, "set":"B"},
{"marks":"10", "set":"A"},
{"marks":"10", "set":"B"},
]
Working with TypeOrm at the moment and looking to apply filtering on data.
Expected outcome array:
[
{"marks": 99, "set":"A"},
{"marks": 90, "set":"B"},
]
const dataset = [
{"rating": 99, "category":"A"},
{"rating": 90, "category":"B"},
{"rating": 10, "category":"A"},
{"rating": 10, "category":"B"}]
const outcome = [...new Set(dataset.map(i=>i.category))] // obtain distinct category ids
.map(c=>dataset.filter(({category})=>c===category) // explore members of each group
.sort(({rating:a},{rating:b})=>a-b).pop()) // find the one with highest rating
console.log(outcome)If your data includes ratings that are strings rather than numbers, modify a-b to (+a)-(+b) to coerce the strings into numbers.
In situations like this, my go-to choice would be utilizing the powerful capabilities of Lodash library.
function findMaxScoreGroups() {
let data = [
{"score": 99, "group":"1"},
{"score": 90, "group":"2"},
{"score": 10, "group":"1"},
{"score": 10, "group":"2"},
]
let groupedData = _.groupBy(data, "group")
let maxScoresByGroup = _.map(groupedData, group => (_.maxBy(group, 'score')));
console.dir(maxScoresByGroup)
}
If you want a more efficient solution, consider using a single loop along with an object to store the group information.
const
info = [{ score: 99, group: "1" }, { score: 90, group: "2" }, { score: 10, group: "1" }],
highestScore = Object.values(info.reduce((result, obj) => {
if (!result[obj.group] || result[obj.group].score < obj.score) result[obj.group] = obj;
return result;
}, {}));
console.log(highestScore);.as-console-wrapper { max-height: 100% !important; top: 0; }To achieve this, aggregate the collection based on the group attribute while retaining only the objects with the highest score. The approach involves utilizing a Map for accumulation and then spreading the iterator generated by Map.values() to obtain the desired outcome.
const data = [{ score: 99, group: '1' }, { score: 90, group: '2' }, { score: '10', group: '1' }, { score: '10', group: '2' },];
const aggregated = data.reduce(
(accumulator, obj) =>
!accumulator.has(obj.group) || obj.score > accumulator.get(obj.group)?.score
? accumulator.set(obj.group, { ...obj })
: accumulator,
new Map()
);
const finalResult = [...aggregated.values()];
console.log(finalResult);Just for kicks, here's an alternative version:
let numbers = [2, 4, 6, 8, 10];
let sum = numbers.reduce((acc, curr) => acc + curr, 0);
console.log(sum)This method is not only simple and easy to understand but also efficient in terms of performance.
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