I have an array that I want to get the most occurring elements,
First scenario
let arr1 = [\'foo\', \'foo\', \'foo\', \
You can count the items with reduce
and find the maximum occurring count. Then you can filter any keys that have that count:
let arr = ['foo', 'foo', 'foo', 'bar', 'bar', 'bar', 'baz', 'baz'];
let counts = arr.reduce((a, c) => {
a[c] = (a[c] || 0) + 1;
return a;
}, {});
let maxCount = Math.max(...Object.values(counts));
let mostFrequent = Object.keys(counts).filter(k => counts[k] === maxCount);
console.log(mostFrequent);
You can also use a for of
loop and ìn
For example
const arrayFrecuent = [3, 1, 2, 1, 3, 2, 5, 4, 2, 10];
const mostFrecuent = givenArray => {
let counts = {};
let maxValue = -1;
let maxItem = null;
for (const num of givenArray) {
if (!(num in counts)) {
counts[num] = 1;
} else {
counts[num] = counts[num] + 1;
}
if (counts[num] > maxValue) {
maxValue = counts[num];
maxItem = num;
}
}
return maxItem;
};
const mostFrecuentNumber = mostFrecuent(arrayFrecuent);
console.log("mostFrecuentNumber", mostFrecuentNumber);
You can calculate the max
for each of the values and only return those which match via grouping them with an Array.reduce:
const mostFrequent = data => data.reduce((r,c,i,a) => {
r[c] = (r[c] || 0) + 1
r.max = r[c] > r.max ? r[c] : r.max
if(i == a.length-1) {
r = Object.entries(r).filter(([k,v]) => v == r.max && k != 'max')
return r.map(x => x[0])
}
return r
}, {max: 0})
console.log(mostFrequent(['foo', 'foo', 'foo', 'bar', 'bar', 'bar', 'baz', 'baz']))
console.log(mostFrequent(['foo', 'foo', 'foo', 'bar', 'baz']))