Jrobm2k9
Jrobm2k9

Reputation: 73

Filtering a nested array - javascript

I have a nested array and each array has a string and an integer, the strings in some of the array are the same but i want to filter the array so that it only contains nested arrays with unique names and that have the highest values. Here's an example of what i have and what i want:

[['a', 1],['a', 2],['a', 3],['b',2],['b',5]]

what i want to do is filter so that it contains this:

[['a', 3],['b', 5]]

I originally tried doing this with a for loop and an if statement, then a for loop and a while statement, when i looked at filtering but i'm not sure how to implement it where it will keep the string with the highest value, please help!!!!

Upvotes: 1

Views: 221

Answers (5)

Osama
Osama

Reputation: 3040

const data = [['a', 1],['a', 2],['a', 3],['b',2],['b',5]];
var result = data.sort(function(a,b){
    return Math.max(b[1]-a[1]);
});
var new_data= result.slice(0,2);
console.log(new_data.reverse());

Upvotes: 0

Nina Scholz
Nina Scholz

Reputation: 386519

You could take a Map for grouping by the first element and get the maximum values by checking stored values.

var array = [['a', 1], ['a', 2], ['a', 3], ['b', 2], ['b', 5]],
    result = Array.from(
        array.reduce((m, [k, v]) => m.set(k, m.has(k) ? Math.max(v, m.get(k)) : v), new Map)
    );
    
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }

If you like to keep the original arrays, you could store the array instead of the value and take later only the values of the map.

var array = [['a', 1], ['a', 2], ['a', 3, 'bar'], ['b', 2], ['b', 5, 'foo']],
    result = Array.from(array
        .reduce(
            (m, a) => m.has(a[0]) && m.get(a[0])[1] > a[1] ? m : m.set(a[0], a),
            new Map
        )
        .values()
    );
    
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }

Upvotes: 3

Vladimir Trifonov
Vladimir Trifonov

Reputation: 1385

const obj = [['a', 1],['a', 2],['a', 3],['b',2],['b',5]].reduce((res, arr) => {
    res[arr[0]] = !res[arr[0]] ? arr[1] : Math.max(res[arr[0]], arr[1])
    return res
}, {})
const result = Object.keys(obj).map((key) => [key, obj[key]])

Upvotes: 0

wang
wang

Reputation: 1780

const data = [['a', 1],['a', 2],['a', 3],['b',2],['b',5]]
const buf = {}
data.map(arr => {
    if(!buf[arr[0]] || buf[arr[0]] < arr[1])
        buf[arr[0]] = arr[1]
})
const result = Object.keys(buf).map(k => [k, buf[k]]);
console.log(result)

Upvotes: 0

kockburn
kockburn

Reputation: 17606

Here is a solution using reduce and Object.keys and map.

const data = [['a', 1],['a', 2],['a', 3],['b',2],['b',5]];

//in reducer we work with object instead of array
//efficient since we avoid doing an extra loop
const result = data.reduce((acc, cur)=>{

  //create variables from current array (ex: ['a', 1])
  const letter = cur[0];
  const value = cur[1];

  //Acc (accumulator) holds the highest values (ex {a: 1, b: 2} )
  //If the letter doesn't yet exist or if the cur value is higher we update the acc
  if(!acc[letter] || acc[letter] < value){
    acc[letter] = value;
  }
  
  return acc;
}, {});

//Not in the correct format, so we transform the result into the requested format
const final = Object.keys(result).map(key=>[key, result[key]]);

console.log(final);

Upvotes: 0

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