Given a hashmap (or a "Dictionary") as a JavaScript object, I'd like to get the Key who has the maximum Value - assuming all Values are integers.

In case there's more than one, I don't care which one of them.


var b = { '1': 9, '2': 7, '3': 7, '4': 9, '5': 3 };

I can extract the desired Key by:

parseInt(_(b).chain().pairs().max(function(p){return p[1];}).value()[0])

which returns 1.

How can this be achieved more elegantly?

I tried with _.invert as well but couldn't make it look better.

  • \$\begingroup\$ It looks like you're seeking the key with the largest value. Note that the keys are strings while the values are integers. \$\endgroup\$ May 18, 2014 at 11:24
  • \$\begingroup\$ You're right, my apologies. I modified my question, but this still does not answer. \$\endgroup\$
    – user40171
    May 18, 2014 at 12:30
  • \$\begingroup\$ There's something ambiguous here, since there are 2 keys with a value of 9: b['1'] and b['4']. You say you want the "first" key ('1'), but JS objects are technically unordered, so you could get the '4' key instead. So do you want the lowest key with the highest value, or simply any key with the highest value? \$\endgroup\$
    – Flambino
    May 18, 2014 at 13:23
  • \$\begingroup\$ "In case there's more than one, I don't care which one of them." - so I want simply any key with the highest value. \$\endgroup\$
    – user40171
    May 18, 2014 at 13:41
  • 1
    \$\begingroup\$ If you use parseInt, you should always specify a base: parseInt(…, 10). \$\endgroup\$
    – Ingo Bürk
    May 18, 2014 at 16:59

3 Answers 3


I don't know about elegant, but you can omit the initial _(), use + to convert to number instead of parseInt and just pull out the iterator.

var iterator = function(p){return p[1];}
var maxKey = +_.chain(b).pairs().max(iterator).value()[0];
// Removed the | 0 since the keys are integers to begin with

Here's an alternate way to do it using ES5 Array.prototype.reduce and Object.keys. You can also use underscore's _.reduce and _.keys for compatibility. Note that this runs through the entire array, so any duplicates found will use the latest value.

var max = Object.keys(b).reduce(function(max,key){
  return (max === undefined || b[key] > b[max]) ? +key : max;

Making it more clear what we're doing, in case you aren't familiar of reduce, is to use Array.prototype.forEach (or _.each):

var max;
  max = (b[key] > b[max]) ? +key : max;
  • 1
    \$\begingroup\$ I think | 0 is an anti-pattern as it is completely unclear what it's doing to anyone who hasn't seen it before. What's wrong with parseInt? \$\endgroup\$
    – Ingo Bürk
    May 18, 2014 at 16:58
  • \$\begingroup\$ I have to agree with @IngoBürk, |0 may be good if you're trying to Golf the code, but here we seek readability... However, I loved the idea of using an iterator function rather than using an anoynmous function. I just hoped someone will enlighten me, supplying a completely different approch for this task, avoiding the pairing and un-pairing etc. \$\endgroup\$
    – user40171
    May 18, 2014 at 17:48
  • 1
    \$\begingroup\$ @IngoBürk most of this underscore stuff should be completely unclear to anyone who hasn't seen it before. Does that make it an antipattern also? \$\endgroup\$
    – Dagg
    May 18, 2014 at 19:42
  • 1
    \$\begingroup\$ @user40171 "here we seek readability" ... just looking at your line of code from the POV of someone who didn't write it, is it obvious that it is meant to return the key of the highest value in an object's properties? IOW is it really that readable to you? I usually find these densely-packed functional sorts of snippets to be difficult to read at a glance after they're written, sort of like regex. I think I'd be able to discern the purpose of an eqivalent for loop much faster. I'm curious what others think, though. \$\endgroup\$
    – Dagg
    May 18, 2014 at 19:48
  • 1
    \$\begingroup\$ You're missing typeof for max === 'undefined' and the final expression will not find a key if all values are negative. Use var max = null and max === null || ... \$\endgroup\$ May 18, 2014 at 20:03

The way I would do this with the current iteration of underscore (1.6) would be through reduce. Note use _.reduceRight if you want to favour items to the left

_.reduce(b, function(max, current, key) {
    return max && max.value > current ? max : {
        value: current,
        key: key

However, in lodash you can use findKey which would be more intuitive. _.findKey will be available in underscore 2.0 if they accept my pull request #1587

var maxValue = _.max(b);
_.findKey(b, function(x) {
    return x === maxValue;

Ok I think I found what I was looking for. Thanks anyway for all help.

parseInt( _(b).invert()[_(b).max()] )

This returns 4, which is just good as 1 according to requirements. It looks more elegant and it is way shorter than the pairing stuff.

  • 3
    \$\begingroup\$ This returns the max value, but the question asked how to get the key of the max value. Not a big deal, though. :) \$\endgroup\$ May 19, 2014 at 15:11

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