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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).

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).

• javascript
• jquery

Share Improve this question Follow edited Jun 21, 2012 at 17:06 Y2theZ asked Jun 21, 2012 at 16:45 Y2theZY2theZ 10.4k3939 gold badges133133 silver badges204204 bronze badges 11

• 3 @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 Commented Jun 21, 2012 at 16:48
• @OlegV.Volkov I'm using JSON.parse() shouldn't that make it Json? – Y2theZ Commented Jun 21, 2012 at 16:49
• @Youssef It was JSON (which is a String value) before parsing. It's an Object value after parsing. – Šime Vidas Commented Jun 21, 2012 at 16:50
• 2 JSON is the string notation of objects. When you parse JSON to an object, it's no longer in the JSON format – altschuler Commented Jun 21, 2012 at 16:50
• 1 I took liberty of fixing JSON -> object in your question, as comments confirm that it is what you've meant. – Oleg V. Volkov Commented Jun 21, 2012 at 17:05

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16 Answers 16

Sorted by: Reset to default 202

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/

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;
}

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

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];
}

// 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

// 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 ]

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);

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]

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)

Using the lodash library you can write shorter

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

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));

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;
    }
}

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))

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); });

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

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)

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