Finding the max value of a property in an array of objects

Question

I'm looking for a really quick, clean and efficient way to get the max "y" value in the following JSON slice:

[ { "x": "8/11/2009", "y": 0.026572007 }, { "x": "8/12/2009", "y": 0.025057454 }, { "x": "8/13/2009", "y": 0.024530916 }, { "x": "8/14/2009", "y": 0.031004457 } ]

Is a for-loop the only way to go about it? I'm keen on somehow using Math.max.

Wimanicesir · Accepted Answer · 2023-10-24 11:40:37Z

1284

To find the maximum y value of the objects in array:

Math.max.apply(Math, array.map(function(o) { return o.y; }))

or in more modern JavaScript:

Math.max(...array.map(o => o.y))

Warning:

This method is not advisable, it is better to use reduce. With a large array, Math.max will be called with a large number of arguments, which can cause stack overflow.

edited Oct 24, 2023 at 11:40

Wimanicesir

5,1592 gold badges14 silver badges34 bronze badges

answered Oct 26, 2010 at 5:04

tobyodavies

28.2k5 gold badges45 silver badges58 bronze badges

Sign up to request clarification or add additional context in comments.

16 Comments

Mike Lyons Over a year ago

Could you expand this answer to show how to return the object that the max value was found in? That would be very helpful, thanks!

mili Over a year ago

Here is the fiddle! hope this will help to someone jsfiddle.net/45c5r246

tobyodavies Over a year ago

@MikeLyons if you still care about getting the actual object: jsfiddle.net/45c5r246/34

Daniel C Over a year ago

FWIW my understanding is when you call apply on a function it executes the function with a specified value for this and a series of arguments specified as an array. The trick is that apply turns the array into a series of actual function arguments. So in this case it finally calls Math.max(0.0265, 0.0250, 0.024, 0.031) with the this of the function executed being Math. I can't see why it should be Math frankly, I don't think the function requires a valid this. Oh and here is a proper explanation: stackoverflow.com/questions/21255138/…

Fabien Over a year ago

Math.max(...array.map(o => o.y)) <3 thx to atom code formatter

|

Ooker · Accepted Answer · 2024-05-13 16:40:06Z

Find the object whose property "Y" has the greatest value in an array of objects

One way would be to use Array reduce..

const max = data.reduce(function(prev, current) { return (prev && prev.y > current.y) ? prev : current }) //returns object

If you don't need to support IE (only Edge), or can use a pre-compiler such as Babel you could use the more terse syntax.

const max = data.reduce((prev, current) => (prev && prev.y > current.y) ? prev : current)

Note that if you have multiple objects sharing the same max value, it will only return the first matched object.

This is a good answer, however, you would like to pass an initial value or you would get an error in case the data array is empty. i.e. for an autoincrement index of objects. const max = data.reduce((prev, current) => (prev.y > current.y) ? prev : current, 1)
You raise a good point, I would probably choose null over 1.
Note that this returns the object that had the max value not the max value from the object. This may or may not be what you want. In my case it was what I wanted. +1
Excellent answer! At first, I hesitated due to the reduce, but we're gonna need to iterate anyway, so why not?
Also, make sure the data type is a number and not a string. Otherwise you could get mixed results. Use parseInt() where appropriate.

Alexei - check Codidact · Accepted Answer · 2018-09-10 08:49:21Z

260

clean and simple ES6 (Babel)

const maxValueOfY = Math.max(...arrayToSearchIn.map(o => o.y), 0);

The second parameter should ensure a default value if arrayToSearchIn is empty.

edited Sep 10, 2018 at 8:49

Alexei - check Codidact

23.2k22 gold badges159 silver badges179 bronze badges

answered Sep 13, 2016 at 3:43

Vitaliy Kotov

3,4521 gold badge16 silver badges5 bronze badges

6 Comments

Juraj Over a year ago

also it's good to know that it returns -Infinity (a truthy value) for an empty array

Eugene Kulabuhov Over a year ago

This is supported in most modern browsers without Babel now.

juliangonzalez Over a year ago

as it returns -Infinity for an empty array, you can pass an initial value Math.max(...state.allProjects.map(o => o.id), 1);

Kamil Kiełczewski Over a year ago

to handle case for minus values change 0 to arrayToSearchIn[0].y. Time complexity comparision: stackoverflow.com/a/53654364/860099

Nabeel K Over a year ago

how to get the object here ?

|

isherwood · Accepted Answer · 2023-11-30 20:27:31Z

Comparison of three ONELINERS which handle minus numbers case (input in a array):

var maxA = a.reduce((a,b)=>a.y>b.y?a:b).y; // 30 chars time complexity: O(n) var maxB = a.sort((a,b)=>b.y-a.y)[0].y; // 27 chars time complexity: O(nlogn) var maxC = Math.max(...a.map(o=>o.y)); // 26 chars time complexity: >O(2n)

Editable example. Ideas from: maxA, maxB and maxC (side effect of maxB is that array a is changed because sort is in-place).

var a = [ {"x":"8/11/2009","y":0.026572007},{"x":"8/12/2009","y":0.025057454}, {"x":"8/14/2009","y":0.031004457},{"x":"8/13/2009","y":0.024530916} ] var maxA = a.reduce((a,b)=>a.y>b.y?a:b).y; var maxC = Math.max(...a.map(o=>o.y)); var maxB = a.sort((a,b)=>b.y-a.y)[0].y; document.body.innerHTML=`<pre>maxA: ${maxA}\nmaxB: ${maxB}\nmaxC: ${maxC}</pre>`;

For bigger arrays the Math.max... will throw exception: Maximum call stack size exceeded (Chrome 76.0.3809, Safari 12.1.2, date 2019-09-13)

let a = Array(400*400).fill({"x": "8/11/2009", "y": 0.026572007 }); // Exception: Maximum call stack size exceeded try { let max1= Math.max.apply(Math, a.map(o => o.y)); } catch(e) { console.error('Math.max.apply:', e.message) } try { let max2= Math.max(...a.map(o=>o.y)); } catch(e) { console.error('Math.max-map:', e.message) }

Benchmark for the 4 element array

one thing to point out is option B makes it much easier to get the whole object with the max y value by leaving off the .y at the end.
Im not sure to understand why maxA's Math.max() has a second parameter? It should work with only var maxA = Math.max(...a.map(o=>o.y));, wouldn't it?
For those curious about the difference (if any) between O(n) and O(2n) see stackoverflow.com/questions/25777714/…
@jHova - I not use strict big O notation - this is more "practical / intuitive" notation and meaning is following: the number of operations in O(2n) is twice as high as in O(n) (I'm not going to define it precisely)
With my lack of math expertise, I just wanted to know which is faster without having to edit the code above with timers and sample runs. According to stackoverflow.com/a/25899325/2578125, best bet is O(n), though actual usage could vary.

Community · Accepted Answer · 2017-05-23 12:02:47Z

I'd like to explain the terse accepted answer step-by-step:

var objects = [{ x: 3 }, { x: 1 }, { x: 2 }]; // array.map lets you extract an array of attribute values var xValues = objects.map(function(o) { return o.x; }); // es6 xValues = Array.from(objects, o => o.x); // function.apply lets you expand an array argument as individual arguments // So the following is equivalent to Math.max(3, 1, 2) // The first argument is "this" but since Math.max doesn't need it, null is fine var xMax = Math.max.apply(null, xValues); // es6 xMax = Math.max(...xValues); // Finally, to find the object that has the maximum x value (note that result is array): var maxXObjects = objects.filter(function(o) { return o.x === xMax; }); // Altogether xMax = Math.max.apply(null, objects.map(function(o) { return o.x; })); var maxXObject = objects.filter(function(o) { return o.x === xMax; })[0]; // es6 xMax = Math.max(...Array.from(objects, o => o.x)); maxXObject = objects.find(o => o.x === xMax); document.write('<p>objects: ' + JSON.stringify(objects) + '</p>'); document.write('<p>xValues: ' + JSON.stringify(xValues) + '</p>'); document.write('<p>xMax: ' + JSON.stringify(xMax) + '</p>'); document.write('<p>maxXObjects: ' + JSON.stringify(maxXObjects) + '</p>'); document.write('<p>maxXObject: ' + JSON.stringify(maxXObject) + '</p>');

Further information:

Great explanation! It might be a bit easier to read if it wasn't in code comments, but still - great work

Guffa · Accepted Answer · 2010-10-26 05:34:11Z

Well, first you should parse the JSON string, so that you can easily access it's members:

var arr = $.parseJSON(str);

Use the map method to extract the values:

arr = $.map(arr, function(o){ return o.y; });

Then you can use the array in the max method:

var highest = Math.max.apply(this,arr);

Or as a one-liner:

var highest = Math.max.apply(this,$.map($.parseJSON(str), function(o){ return o.y; }));

Abdelsalam Shahlol · Accepted Answer · 2020-05-23 16:06:14Z

18

Here is the shortest solution (One Liner) ES6:

Math.max(...values.map(o => o.y));

edited May 23, 2020 at 16:06

Abdelsalam Shahlol

1,7891 gold badge25 silver badges33 bronze badges

answered May 9, 2020 at 8:33

Subodh Singh

3493 silver badges4 bronze badges

1 Comment

Sam Watkins Over a year ago

again, this is unsafe for large arrays as it will cause a stack overflow crash

VLAZ · Accepted Answer · 2020-06-03 12:16:17Z

if you (or, someone here) are free to use lodash utility library, it has a maxBy function which would be very handy in your case.

hence you can use as such:

_.maxBy(jsonSlice, 'y');

Ooki Koi · Accepted Answer · 2018-01-10 10:45:04Z

10

Or a simple sort! Keeping it real :)

array.sort((a,b)=>a.y<b.y)[0].y

answered Jan 10, 2018 at 10:45

Ooki Koi

3304 silver badges4 bronze badges

5 Comments

Kamil Kiełczewski Over a year ago

Nice idea +1 (shortest code), but there is small bug - change a.y<a.y to b.y-a.y. Time complexity comparison here: stackoverflow.com/a/53654364/860099

Wildhammer Over a year ago

Finding the max is O(n). This is O(nlogn). Writing simple code is good as long as efficiency is not sacrificed.

Kamil Kiełczewski Over a year ago

@Wildhammer - actually Micro-optimisation is worth it when you have evidence that you're optimising a bottleneck.. In most cases, the simple code is better choice than high-efficient code.

Wildhammer Over a year ago

@KamilKiełczewski Both array comparisons in that article have the same time complexity the difference is in their coefficient. For instance, one takes n time units to find the solution while the other one is 7n. In time complexity theory, both of these are O(n). What we are talking about in the problem of finding max is the comparison of O(n) with O(n logn). Now if you can guarantee n doesn't exceed 10 then you can use your solution otherwise O(n) algorithm is always the winner and performance(user experience) is always prior to developer experience(ask industry people, they tell you that!).

Kamil Kiełczewski Over a year ago

@Wildhammer nope - even if your array have n=10000 elements user will not see difference - proof HERE. Performance optimisation is good only for app bottleneck (eg. you need to process large arrays) - but in most case a performance-focus is wrong approach and time(=money) wasting. This is well known code approach mistake - read more: "micro-optimisation"

Carlos · Accepted Answer · 2023-03-09 18:55:01Z

Using one function find the max obj.value return obj

 let List= [{votes:4},{votes:8},{votes:7}] let objMax = List.reduce((max, curren) => max.votes > curren.votes ? max : curren); console.log(objMax)

Or using two functions return max value

 Math.max(...List.map(el => el.votes)))

This answer is far from what is being questioning here. This will return the object, not the value.

Diego Santa Cruz Mendezú · Accepted Answer · 2018-07-12 13:22:03Z

8

Each array and get max value with Math.

data.reduce((max, b) => Math.max(max, b.costo), data[0].costo);

answered Jul 12, 2018 at 13:22

Diego Santa Cruz Mendezú

4,6771 gold badge42 silver badges28 bronze badges

2 Comments

Pilaf T. Pilafian Over a year ago

+1 but use y: data.reduce((max, point) => Math.max(max, point.y), data[0].y); Many of the other answers create unnecessary temporary arrays or do expensive sorting. Using reduce() plus Math.max() is memory and CPU efficient, and it's more readable.

john nowlin Over a year ago

As mentioned above, can replace initial value with -Infinity.

isherwood · Accepted Answer · 2023-11-30 20:29:20Z

It returns the object simplified @andy polhill answare

var data = [{ y: 90 }, { y: 9 }, { y: 8 } ] const max = data.reduce((prev, current) => ((prev.y > current.y) ? prev : current), 0) //returns object console.log(max)

Mephisto07 · Accepted Answer · 2014-08-26 12:39:15Z

var max = 0; jQuery.map(arr, function (obj) { if (obj.attr > max) max = obj.attr; });

ndey96 · Accepted Answer · 2017-09-28 21:19:20Z

ES6 solution

Math.max(...array.map(function(o){return o.y;}))

For more details see https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Math/max

Ryan · Accepted Answer · 2020-11-18 21:57:01Z

// Here is very simple way to go: // Your DataSet. let numberArray = [ { "x": "8/11/2009", "y": 0.026572007 }, { "x": "8/12/2009", "y": 0.025057454 }, { "x": "8/13/2009", "y": 0.024530916 }, { "x": "8/14/2009", "y": 0.031004457 } ] // 1. First create Array, containing all the value of Y let result = numberArray.map((y) => y) console.log(result) // >> [0.026572007,0.025057454,0.024530916,0.031004457] // 2. let maxValue = Math.max.apply(null, result) console.log(maxValue) // >> 0.031004457

trinalbadger587 · Accepted Answer · 2021-04-07 00:46:06Z

Quick and dirty:

Object.defineProperty(Array.prototype, 'min', { value: function(f) { f = f || (v => v); return this.reduce((a, b) => (f(a) < f(b)) ? a : b); } }); Object.defineProperty(Array.prototype, 'max', { value: function(f) { f = f || (v => v); return this.reduce((a, b) => (f(a) > f(b)) ? a : b); } }); console.log([1,2,3].max()); console.log([1,2,3].max(x => x*(4-x))); console.log([1,2,3].min()); console.log([1,2,3].min(x => x*(4-x)));

Shivam Jha · Accepted Answer · 2021-10-14 11:31:06Z

Explanation for accepted answer and a more generalized way

If someone's here to find the max value among all such keys (a generalized way):

const temp1 = [ { "name": "Month 8 . Week 1", "CATEGORY, Id 0": null, "CATEGORY, Id 1": 30.666666666666668, "CATEGORY, Id 2": 17.333333333333332, "CATEGORY, Id 3": 12.333333333333334, "TASK, Id 1": 30.666666666666668, "TASK, Id 2": 12.333333333333334, "TASK, Id 3": null, "TASK, Id 4": 5, "TASK, Id 5": null, "TASK, Id 6": null, "TASK, Id 7": null, "TASK, Id 8": null, "TASK, Id 9": null, "TASK, Id 10": null, "TASK, Id 12": null, "TASK, Id 14": null, "TASK, Id 16": null, "TASK, Id 17": null, "TASK, Id 26": 12.333333333333334 }, { "name": "Month 8 . Week 2", "CATEGORY, Id 0": 38, "CATEGORY, Id 1": null, "CATEGORY, Id 2": 12, "CATEGORY, Id 3": null, "TASK, Id 1": null, "TASK, Id 2": 15, "TASK, Id 3": null, "TASK, Id 4": null, "TASK, Id 5": null, "TASK, Id 6": 5, "TASK, Id 7": 5, "TASK, Id 8": 5, "TASK, Id 9": 5, "TASK, Id 10": null, "TASK, Id 12": null, "TASK, Id 14": null, "TASK, Id 16": null, "TASK, Id 17": null, "TASK, Id 26": 15 }, { "name": "Month 8 . Week 3", "CATEGORY, Id 0": 7, "CATEGORY, Id 1": 12.333333333333334, "CATEGORY, Id 2": null, "CATEGORY, Id 3": null, "TASK, Id 1": null, "TASK, Id 2": null, "TASK, Id 3": 12.333333333333334, "TASK, Id 4": null, "TASK, Id 5": null, "TASK, Id 6": null, "TASK, Id 7": null, "TASK, Id 8": null, "TASK, Id 9": null, "TASK, Id 10": null, "TASK, Id 12": null, "TASK, Id 14": 7, "TASK, Id 16": null, "TASK, Id 17": null, "TASK, Id 26": null }, { "name": "Month 8 . Week 4", "CATEGORY, Id 0": null, "CATEGORY, Id 1": null, "CATEGORY, Id 2": 10, "CATEGORY, Id 3": 5, "TASK, Id 1": null, "TASK, Id 2": null, "TASK, Id 3": null, "TASK, Id 4": null, "TASK, Id 5": 5, "TASK, Id 6": null, "TASK, Id 7": null, "TASK, Id 8": null, "TASK, Id 9": null, "TASK, Id 10": 5, "TASK, Id 12": 5, "TASK, Id 14": null, "TASK, Id 16": null, "TASK, Id 17": null, "TASK, Id 26": null }, { "name": "Month 8 . Week 5", "CATEGORY, Id 0": 5, "CATEGORY, Id 1": null, "CATEGORY, Id 2": 7, "CATEGORY, Id 3": null, "TASK, Id 1": null, "TASK, Id 2": null, "TASK, Id 3": null, "TASK, Id 4": null, "TASK, Id 5": null, "TASK, Id 6": null, "TASK, Id 7": null, "TASK, Id 8": null, "TASK, Id 9": null, "TASK, Id 10": null, "TASK, Id 12": null, "TASK, Id 14": null, "TASK, Id 16": 7, "TASK, Id 17": 5, "TASK, Id 26": null }, { "name": "Month 9 . Week 1", "CATEGORY, Id 0": 13.333333333333334, "CATEGORY, Id 1": 13.333333333333334, "CATEGORY, Id 3": null, "TASK, Id 11": null, "TASK, Id 14": 6.333333333333333, "TASK, Id 17": null, "TASK, Id 18": 7, "TASK, Id 19": null, "TASK, Id 20": null, "TASK, Id 26": 13.333333333333334 }, { "name": "Month 9 . Week 2", "CATEGORY, Id 0": null, "CATEGORY, Id 1": null, "CATEGORY, Id 3": 13.333333333333334, "TASK, Id 11": 5, "TASK, Id 14": null, "TASK, Id 17": 8.333333333333334, "TASK, Id 18": null, "TASK, Id 19": null, "TASK, Id 20": null, "TASK, Id 26": null }, { "name": "Month 9 . Week 3", "CATEGORY, Id 0": null, "CATEGORY, Id 1": 14, "CATEGORY, Id 3": null, "TASK, Id 11": null, "TASK, Id 14": null, "TASK, Id 17": null, "TASK, Id 18": null, "TASK, Id 19": 7, "TASK, Id 20": 7, "TASK, Id 26": null } ] console.log(Math.max(...[].concat([], ...temp1.map(i => Object.values(i))).filter(v => typeof v === 'number')))

one thing to note that Math.max(1, 2, 3) returns 3. So does Math.max(...[1, 2, 3]), since Spread syntax can be used when all elements from an object or array need to be included in a list of some kind.

We will take advantage of this!

Let's assume an array which looks like:

var a = [{a: 1, b: 2}, {foo: 12, bar: 141}]

And the goal is to find the max (among any attribute), (here it is bar(141))

so to use Math.max() we need values in one array (so we can do ...arr)

First let's just separate all the numbers We can each that each item of the array a is an object. While iterating through each of them, Object.values(item) will give us all the values of that item in an array form, and we can use map to generate a new array with only values

So,

var p = a.map(item => Object.values(item)) // [ [1, 2], [12, 141] ]

Also,using concat,

[].concat([], ...arr), or just [].concat(...arr) on arr, [ [1, 2], [12, 141] ] flattens it to [1, 2, 12, 141]

So,

var f = [].concat(...p) // [1, 2, 12, 141]

since we have now an array of just numbers we do Math.max(...f):

var m = Math.max(...f) // 141

isherwood · Accepted Answer · 2023-11-30 20:30:29Z

This typescript function can be called to search for the largest value that can exist in a field of the objects of an array:

function getHighestField(objArray: any[], fieldName: string) { return Number( Math.max.apply( Math, objArray?.map(o => o[fieldName] || 0), ) || 0, ); }

With this values as an example:

const scoreBoard = [ { name: 'player1', score: 4 }, { name: 'player2', score: 9 }, { name: 'player3', score: 7 } ]

You can call the function something this way:

const myHighestVariable = `This is the highest: ${getHighestField(scoreBoard, "score")}`;

The result will be something like this:

console.log(myHighestVariable);

This is the highest: 9

Rock Dial · Accepted Answer · 2021-01-04 19:02:57Z

It's very simple

 const array1 = [ {id: 1, val: 60}, {id: 2, val: 2}, {id: 3, val: 89}, {id: 4, val: 78} ]; const array2 = [1,6,8,79,45,21,65,85,32,654]; const max = array1.reduce((acc, item) => acc = acc > item.val ? acc : item.val, 0); const max2 = array2.reduce((acc, item) => acc = acc > item ? acc : item, 0); console.log(max); console.log(max2);

Tasos K. · Accepted Answer · 2022-05-24 11:45:52Z

0

const getMaxFromListByField = (list, field) => { return list[list.map(it => it[field]).indexOf(Math.max(...list.map(it => it[field])))] }

edited May 24, 2022 at 11:45

Tasos K.

8,1057 gold badges44 silver badges68 bronze badges

answered Sep 8, 2021 at 14:02

Ahmed Mohameden

1

1 Comment

Community Over a year ago

Please provide additional details in your answer. As it's currently written, it's hard to understand your solution.

Prageeth godage · Accepted Answer · 2022-12-01 18:48:24Z

carefull on null and empty and property not in array and empty array

if ((value && value.length > 0)) { var maxObj = (value && value.length > 0) value.reduce(function (prev, current) { return ((parseInt(prev["y"]) || 0) > (parseInt(current["y"]) || 0)) ? prev : current }) } { // else logic here }

Guilherme Oliveira · Accepted Answer · 2023-06-12 21:06:26Z

Just a grain of salt on Andy's answer: One can use a filter before reducing to avoid passing empty arrays, which could mess everything.

const max = data .filter(item=>item.id != undefined) // --> here .reduce(function(prev, current) {return (prev.y > current.y) ? prev : current }) //returns object

kawauso · Accepted Answer · 2024-02-22 11:04:09Z

If you want to get only the max "y",

const maxY = data.reduce((y, current) => (y == undefined || current.y > y) ? current.y : y, undefined)

.

Collectives™ on Stack Overflow

Finding the max value of a property in an array of objects

23 Answers 23

Warning:

16 Comments

Find the object whose property "Y" has the greatest value in an array of objects

11 Comments

6 Comments

5 Comments

1 Comment

1 Comment

1 Comment

Comments

5 Comments

1 Comment

2 Comments

Comments

Comments

Comments

Comments

Comments

Explanation for accepted answer and a more generalized way

Comments

Comments

Comments

1 Comment

Comments

Comments

Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

23 Answers 23

Warning:

16 Comments

Find the object whose property "Y" has the greatest value in an array of objects

11 Comments

6 Comments

5 Comments

1 Comment

1 Comment

1 Comment

Comments

5 Comments

1 Comment

2 Comments

Comments

Comments

Comments

Comments

Comments

Explanation for accepted answer and a more generalized way

Comments

Comments

Comments

1 Comment

Comments

Comments

Comments

Linked

Related