Fast way to get the min/max values among properties of object

Question

I have an object in javascript like this:

{ "a":4, "b":0.5 , "c":0.35, "d":5 }

Is there a fast way to get the minimum and maximum value among the properties without having to loop through them all? because the object I have is huge and I need to get the min/max value every two seconds. (The values of the object keeps changing).

@Oleg: Well, given only this, it could very well be JSON. Youssef: Parse the JSON into an object and iterate over its properties. — Felix Kling, Jun 21 '12 at 16:48
@OlegV.Volkov I'm using JSON.parse() shouldn't that make it Json? — Y2theZ, Jun 21 '12 at 16:49
@Youssef It was JSON (which is a String value) before parsing. It's an Object value after parsing. — Šime Vidas, Jun 21 '12 at 16:50
JSON is the string notation of objects. When you parse JSON to an object, it's no longer in the JSON format — altschuler, Jun 21 '12 at 16:50
@Youssef: The value you *pass* to `JSON.parse` is a string containing JSON. The result is an object. (maybe that's what you mean) — Felix Kling, Jun 21 '12 at 16:50
@FelixKling I am trying to avoid iterating over the entire data because it is huge and it need to be done every 2 seconds. wont that affect performance? — Y2theZ, Jun 21 '12 at 16:51
Does this mean every to seconds you get new JSON, containing all the data? — Felix Kling, Jun 21 '12 at 16:52
no. I have the data and I do math operations on some of the fields every two seconds and then need to get the new min/max — Y2theZ, Jun 21 '12 at 16:54
So you actually have an object with changing fields and you always want to know the min and max value. See carlosfigueira's answer. Your problem does not seems to be related to JSON at all btw. — Felix Kling, Jun 21 '12 at 16:56
I took liberty of fixing JSON -> object in your question, as comments confirm that it is what you've meant. — Oleg V. Volkov, Jun 21 '12 at 17:05
Updated my solution to make use of the fact that Objects retain their order in ES6: https://stackoverflow.com/a/53661894/4722345 — JBallin, Mar 25 '19 at 00:21

score 197 · Accepted Answer · edited Jun 20 '20 at 09:12

197

Update: ^{_{_{Modern version (ES6+)}}}

let obj = { a: 4, b: 0.5 , c: 0.35, d: 5 };

let arr = Object.values(obj);
let min = Math.min(...arr);
let max = Math.max(...arr);

console.log( `Min value: ${min}, max value: ${max}` );

Original Answer:

Try this:

let obj = { a: 4, b: 0.5 , c: 0.35, d: 5 };
var arr = Object.keys( obj ).map(function ( key ) { return obj[key]; });

and then:

var min = Math.min.apply( null, arr );
var max = Math.max.apply( null, arr );

Live demo: http://jsfiddle.net/7GCu7/1/

edited Jun 20 '20 at 09:12

Community

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answered Jun 21 '12 at 16:56

Šime Vidas

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Can also do `max = Object.keys(obj).reduce(function(m, k){ return obj[k] > m ? obj[k] : m }, -Infinity);` – levi Sep 18 '14 at 16:08
4

Can also do this now: `Math.max(...arr);` – cmac Feb 17 '17 at 22:24
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@cmac I’ve added an ES6 version. – Šime Vidas Feb 17 '17 at 22:36
@ŠimeVidas - what do the 3 dots on the Math.min & max function represent? Thanks – AME May 07 '17 at 07:01
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@AME Read here: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Spread_operator – Šime Vidas May 07 '17 at 17:24
Can we get value and key as well?? – Sangwin Gawande Dec 29 '17 at 11:10
@SangwinGawande What if more than one key have the same minimum value? – Šime Vidas Dec 29 '17 at 13:21
Get the first one, like this {'key', 'value'} – Sangwin Gawande Dec 29 '17 at 14:38
@SangwinGawande Properties in an object are not ordered, so between two properties there is no “first one.” The `Object.keys()` API does return an array, but I don’t think that order can be relied upon (the order may change between browsers or over time). – Šime Vidas Dec 30 '17 at 14:13
I would also like to see which key had the min/max values – mplungjan Jul 09 '18 at 17:22
thx, simple logic – Sathis Sakthivel Feb 20 '23 at 06:58

carlosfigueira · Answer 2 · 2020-10-27T13:39:36.367

There's no way to find the maximum / minimum in the general case without looping through all the n elements (if you go from, 1 to n-1, how do you know whether the element n isn't larger (or smaller) than the current max/min)?

You mentioned that the values change every couple of seconds. If you know exactly which values change, you can start with your previous max/min values, and only compare with the new ones, but even in this case, if one of the values which were modified was your old max/min, you may need to loop through them again.

Another alternative - again, only if the number of values which change are small - would be to store the values in a structure such as a tree or a heap, and as the new values arrive you'd insert (or update) them appropriately. But whether you can do that is not clear based on your question.

If you want to get the maximum / minimum element of a given list while looping through all elements, then you can use something like the snippet below, but you will not be able to do that without going through all of them

var list = { "a":4, "b":0.5 , "c":0.35, "d":5 };
var keys = Object.keys(list);
var min = list[keys[0]]; // ignoring case of empty list for conciseness
var max = list[keys[0]];
var i;

for (i = 1; i < keys.length; i++) {
    var value = list[keys[i]];
    if (value < min) min = value;
    if (value > max) max = value;
}

This does not describe how to get the min/max values of the properties of an object. — FistOfFury, Apr 04 '20 at 00:50
You are iterating over an object, not a list. `min` and `max` are undefined. Did you mean to use a `for in` loop instead? — tonix, Oct 26 '20 at 21:42

score 14 · Answer 3 · answered Feb 05 '19 at 16:34

14

You could try:

const obj = { a: 4, b: 0.5 , c: 0.35, d: 5 };
const max = Math.max.apply(null, Object.values(obj));
console.log(max) // 5

answered Feb 05 '19 at 16:34

Dave Kalu

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score 12 · Answer 4 · answered Jun 21 '12 at 16:49

12

min and max have to loop through the input array anyway - how else would they find the biggest or smallest element?

So just a quick for..in loop will work just fine.

var min = Infinity, max = -Infinity, x;
for( x in input) {
    if( input[x] < min) min = input[x];
    if( input[x] > max) max = input[x];
}

answered Jun 21 '12 at 16:49

Niet the Dark Absol

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This is great for IE7/8. Cheers @Niet the Dark Absol – ojhawkins Oct 27 '13 at 23:35
It's not necessarily true that min and max loop through the array to get their values. It's more feasible that they quicksort the array and select min and max values based on that result – goonerify Mar 16 '15 at 14:16
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@goonerify The fastest sort is `O(n log n)`, which is inherently slower than the `O(n)` that just scanning through once would be... – Niet the Dark Absol Mar 16 '15 at 15:11

Andrey Kudriavtsev · Answer 5 · 2017-07-12T14:16:41.370

5

// 1. iterate through object values and get them
// 2. sort that array of values ascending or descending and take first, 
//    which is min or max accordingly
let obj = { 'a': 4, 'b': 0.5, 'c': 0.35, 'd': 5 }
let min = Object.values(obj).sort((prev, next) => prev - next)[0] // 0.35
let max = Object.values(obj).sort((prev, next) => next - prev)[0] // 5

edited Jul 12 '17 at 14:16

answered Jul 11 '17 at 11:11

Andrey Kudriavtsev

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score 4 · Answer 6 · answered Aug 17 '19 at 18:03

4

You can also try with Object.values

const points = { Neel: 100, Veer: 89, Shubham: 78, Vikash: 67 };

const vals = Object.values(points);
const max = Math.max(...vals);
const min = Math.min(...vals);
console.log(max);
console.log(min);

answered Aug 17 '19 at 18:03

Neel Rathod

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score 4 · Answer 7 · answered Dec 22 '20 at 13:06

4

// Sorted
let Sorted = Object.entries({ "a":4, "b":0.5 , "c":0.35, "d":5 }).sort((prev, next) => prev[1] - next[1])
>> [ [ 'c', 0.35 ], [ 'b', 0.5 ], [ 'a', 4 ], [ 'd', 5 ] ]


//Min:
Sorted.shift()
>> [ 'c', 0.35 ]

// Max:
Sorted.pop()
>> [ 'd', 5 ]

answered Dec 22 '20 at 13:06

Timothy Bushell

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score 4 · Answer 8 · answered Nov 17 '21 at 07:02

You can use a reduce() function.

Example:

let obj = { "a": 4, "b": 0.5, "c": 0.35, "d": 5 }

let max = Object.entries(obj).reduce((max, entry) => entry[1] >= max[1] ? entry : max, [0, -Infinity])
let min = Object.entries(obj).reduce((min, entry) => entry[1] <= min[1] ? entry : min, [0, +Infinity])

console.log(max) // ["d", 5]
console.log(min) // ["c", 0.35]

score 4 · Answer 9 · answered Dec 23 '21 at 08:16

To get the keys for max and min

var list = { "a":4, "b":0.5 , "c":0.35, "d":5 };
var keys = Object.keys(list);
var min = keys[0]; // ignoring case of empty list for conciseness
var max = keys[0];
var i;

for (i = 1; i < keys.length; i++) {
    var value = keys[i];
    if (list[value] < list[min]) min = value;
    if (list[value] > list[max]) max = value;
}

console.log(min, '-----', max)

JBallin · Answer 10 · 2019-03-25T00:19:22.157

Here's a solution that allows you to return the key as well and only does one loop. It sorts the Object's entries (by val) and then returns the first and last one.

Additionally, it returns the sorted Object which can replace the existing Object so that future sorts will be faster because it will already be semi-sorted = better than O(n). It's important to note that Objects retain their order in ES6.

const maxMinVal = (obj) => {
  const sortedEntriesByVal = Object.entries(obj).sort(([, v1], [, v2]) => v1 - v2);

  return {
    min: sortedEntriesByVal[0],
    max: sortedEntriesByVal[sortedEntriesByVal.length - 1],
    sortedObjByVal: sortedEntriesByVal.reduce((r, [k, v]) => ({ ...r, [k]: v }), {}),
  };
};

const obj = {
  a: 4, b: 0.5, c: 0.35, d: 5
};

console.log(maxMinVal(obj));

score 2 · Answer 11 · answered Jun 25 '16 at 21:07

For nested structures of different depth, i.e. {node: {leaf: 4}, leaf: 1}, this will work (using lodash or underscore):

function getMaxValue(d){
    if(typeof d === "number") {
        return d;
    } else if(typeof d === "object") {
        return _.max(_.map(_.keys(d), function(key) {
            return getMaxValue(d[key]);
        }));
    } else {
        return false;
    }
}

score 2 · Answer 12 · answered Jun 09 '18 at 14:49

2

Using the lodash library you can write shorter

_({ "a":4, "b":0.5 , "c":0.35, "d":5 }).values().max();

answered Jun 09 '18 at 14:49

Sergey Zhigalov

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can you expand `_()` as to me, `_` is not callable in lodash – Greg Wozniak Sep 23 '22 at 09:56

score 2 · Answer 13 · answered Jan 16 '20 at 09:53

2

var newObj = { a: 4, b: 0.5 , c: 0.35, d: 5 };
var maxValue = Math.max(...Object.values(newObj))
var minValue = Math.min(...Object.values(newObj))

answered Jan 16 '20 at 09:53

user12723650

21
1

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When answering an old question, your answer would be much more useful to other StackOverflow users if you included some context to explain how your answer helps, particularly for a question that already has an accepted answer. See: [How do I write a good answer](https://stackoverflow.com/help/how-to-answer). – David Buck Jan 16 '20 at 10:09

score 1 · Answer 14 · answered Dec 05 '21 at 11:01

1

   obj.prototype.getMaxinObjArr = function (arr,propName) {
        var _arr = arr.map(obj => obj[propName]);
        return Math.max(..._arr);
    }

answered Dec 05 '21 at 11:01

AHMED RABEE

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score 0 · Answer 15 · edited May 30 '17 at 09:54

0

This works for me:

var object = { a: 4, b: 0.5 , c: 0.35, d: 5 };
// Take all value from the object into list
var valueList = $.map(object,function(v){
     return v;
});
var max = valueList.reduce(function(a, b) { return Math.max(a, b); });
var min = valueList.reduce(function(a, b) { return Math.min(a, b); });

edited May 30 '17 at 09:54

lennyklb

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answered May 26 '17 at 12:46

jaydip jadhav

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score 0 · Answer 16 · answered Nov 30 '22 at 03:29

If we are sorting date time value then follow the below described procedure

const Obj = {
    "TRADE::Trade1": {
        "dateTime": "2022-11-27T20:17:05.980Z",
    },
    "TRADE::Trade2": {
        "dateTime": "2022-11-27T20:36:10.659Z",
    },
    "TRADE::Trade3": {
        "dateTime": "2022-11-27T20:28:10.659Z",
    }
} 


const result = Object.entries(Obj).sort((prev, next) => new Date(prev[1].dateTime) - new Date(next[1].dateTime))

console.log(result)

Fast way to get the min/max values among properties of object

16 Answers16

Update: ^{_{_{Modern version (ES6+)}}}

Original Answer:

Linked

Related

Fast way to get the min/max values among properties of object

16 Answers16

Update: Modern version (ES6+)

Original Answer:

Linked

Related

Update: ^{_{_{Modern version (ES6+)}}}