Get multiple maximum values from an array of values

Question

For example in underscore/lowdash you can use _.max(list, [iterator], [context]) function to receive one maximum value. But I want to have it returned multiple maximum values if they are all equal.

var stooges = [{name: 'moe', age: 40}, {name: 'larry', age: 50}, {name: 'curly', age: 50}];

_.max(stooges, function(stooge){ return stooge.age; });

=> {name: 'curly', age: 50};

I want to have somthing like this:

var stooges = [{name: 'moe', age: 40}, {name: 'larry', age: 50}, {name: 'curly', age: 50}];

_.multiplemax(stooges, function(stooge){ return stooge.age; });

=> [{name: 'curly', age: 50},  {name: 'larry', age: 50 ];

Using underscore is ok.

Can it always return an array?

megawac
– megawac

2013-11-23 00:31:46 +00:00
Commented Nov 23, 2013 at 0:31 — megawac
– megawac, Commented Nov 23, 2013 at 0:31
Obviously... result should be an array as well.

Pol
– Pol

2013-11-23 00:32:21 +00:00
Commented Nov 23, 2013 at 0:32 — Pol
– Pol, Commented Nov 23, 2013 at 0:32

Tan Nguyen · Accepted Answer · 2013-11-23 00:55:29Z

5

Is there any special requirements like you cant not combine multiple functions to do multiplemax. If no, I have 2 solutions in my mind

The simplest solution would be to use _.max to find the max age of the array, then use _.filter to filter all values that are equal to max age

Another solution is to use _.groupBy to group the array by age and then get the group with max age

Something like this

function multiplemax(arr, compare) {
  var groups = _.groupBy(arr, compare);
  var keys = _.keys(groups);
  var max = _.max(keys);
  return groups[max];
}

more "underscore"

_.mixin({
  multiplemax: function(arr, fn) {
    var groups = _.groupBy(arr, fn);
    var keys = _.keys(groups);
    var max = _.max(keys);
    return groups[max];
  }
})

Or using max + filter

function multiplemax(arr, compare) {
  var max = _.max(arr, function(v){return v[compare]});
  return _.filter(arr, function(v){return v[compare]==max[compare]});
}

edited Nov 23, 2013 at 0:55

answered Nov 23, 2013 at 0:38

Tan Nguyen

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1 Comment

alnorth29 Over a year ago

Note that the keys will be strings - so if you take the more "underscore" approach you may want to go for var max = _.max(keys, function (d) { return parseInt(d, 10); });

megawac · Accepted Answer · 2013-11-23 00:43:48Z

1

Something like this should do the trick.

_.mixin({
    multiplymax: function(items, comparitor) {
        comparitor = comparitor || _.identity;
        var max = comparitor(items.pop());
        var found = [max];
        _.each(items, function(item) {
            var val = comparitor(item);
            if(val > max) {
                found = [item];//empty
                max = val;
            } else if (val === max) {
                found.push(item);
            }
        });

        return found;
    }
})

Update fixed the broken code ;)

_.multiplymax([{age: 1}, {age:5}, {age:7}, {age:7}, {age:3}], _.property("age")); // [{age:7}, {age:7}]

edited Nov 23, 2013 at 0:43

answered Nov 23, 2013 at 0:35

megawac

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Comments

Gruff Bunny · Accepted Answer · 2013-11-23 00:43:59Z

1

This should do the trick:

var stooges = [{name: 'moe', age: 40}, {name: 'larry', age: 50}, {name: 'curly', age: 50}];

_.mixin( { multiplemax: function(list, field){

    var max = _.max(list, function(item){
        return item[field];
    });

    return _.filter(list, function(item){
        return item[field] === max[field];
    });
}});    

var oldStooges = _.multiplemax(stooges, 'age');

answered Nov 23, 2013 at 0:43

Gruff Bunny

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Comments

Nicoowr · Accepted Answer · 2020-09-04 13:27:27Z

1

Here is a modern version using TypeScript & ES6:

const multipleMaxBy = <T>(list: T[], iteratee: (element: T) => number) => {
  const maxScore = Math.max(...list.map(iteratee));
  return list.filter((element) => iteratee(element) === maxScore);
};

edited Sep 4, 2020 at 13:27

answered Sep 4, 2020 at 10:13

Nicoowr

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Comments

aivarastee · Accepted Answer · 2024-05-10 11:33:03Z

0

Sort it from highest to lowest and take amount of values:

export const multipleMax = (list: any[], amount: number = 1) => {
  return _.take(_.reverse(_.orderBy(list)), amount)
}

answered May 10, 2024 at 11:33

aivarastee

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