Finding the max value of an attribute in an array of objects

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&开发者_如何学Cquot;: "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.

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))

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.y > current.y) ? prev : current
}) //returns object

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/Reduce http://caniuse.com/#search=reduce (IE9 and above)

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.y > current.y) ? prev : current)

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.

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 here. 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

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:

• Math.max.apply()
• Array.prototype.map()
• Array.from()
• Function.prototype.apply()
• Array.prototype.filter()
• Array.prototype.find()

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; }));

Here is the shortest solution (One Liner) ES6:

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

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');

Or a simple sort! Keeping it real :)

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

Each array and get max value with Math.

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

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)

let List= [{votes:4},{votes:8},{votes:7}]

let objMax = List.reduce((max, curren) => max.votes > curren.votes ? max : curren);

console.log(objMax)

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

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

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)));

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

Thanks for the answers I found here, I hope it can be useful to someone. 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

// 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

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);

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

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
}