Trjtdtk

There are several ways to arrive at what you want. Please see the following methods:

const obj = lst1: 300, lst2: 381, lst3: 4, lst4: 4, lst5: 49 
const values = Object.values(obj)
const lowest = Math.min.apply(null, values)

// Using Object.keys(), Object.values() and findIndex()
const keys = Object.keys(obj)
const indexOfLowest = values.findIndex(function (x) return x === lowest )
console.log(`The lowest value is $lowest from $keys[indexOfLowest]`)



// OR Using Object.keys() and filter()
const key = Object.keys(obj).filter(function (x) return obj[x] === lowest )[0]
console.log(`The lowest value is $lowest from $key`)



// OR Using Object.entries() and sort()
const [[lowestKey, lowestVal]] = Object.entries(obj).sort(function ([,valA], [,valB]) return valA - valB );
console.log(`The lowest value is $lowestVal from $lowestKey`)

There are many ways to approach your problem, but here is one simple alternative way to do it. Basically, it is a brute force way of doing things, whereby you iterate through every property to check for the value.

const obj = lst1: 300, lst2: 381, lst3: 4, lst4: 4, lst5: 49;

const getSmallest = function (object) 
 let smallestValue = Number.MAX_VALUE;
 let selectedKey = '';
 for (let key in object) 
 if (object[key] < smallestValue) 
 smallestValue = object[key];
 selectedKey = key;
 
 ; 
 console.log(selectedKey, smallestValue)
 return `The lowest value is $smallestValue from $selectedKey`;
;

getSmallest(obj);

To get only the minimum number you can use Math.min(...Object.entries(arr).map(o => o[1])):

const arr = lst1: 300, lst2: 381, lst3: 4, lst4: 4, lst5: 49;
const lowestValue = Math.min(...Object.entries(arr).map(o => o[1]));

console.log(lowestValue);

And to get the object with the minimum value you can use Object.entries(arr).reduce((a, [k, v]) => a[Object.keys(a)[0]] < v ? a : [k]: v, ):

const arr = lst1: 300, lst2: 381, lst3: 4, lst4: 4, lst5: 49;
const lowestObj = Object.entries(arr).reduce((a, [k, v]) => a[Object.keys(a)[0]] < v ? a : [k]: v, );

console.log(lowestObj);

Here we first assume that the min, key and value as first property and the value respectively. Then we iterate through all the elements and compare with minValue to find any value less than that in the object. If it is, then we update both minValue and minKey. Anyway, the arr type is an Object. it is not an array in javascript.

var arr = lst1: 300, lst2: 381, lst3: 4, lst4: 4, lst5: 49;
var minkey = Object.keys(arr)[0];
var minValue = arr[Object.keys(arr)[0]];
for (var key in arr) 
 if (arr.hasOwnProperty(key)) 
 if(arr[key] < minValue)
 minValue = arr[key];
 minKey = key;
 
 

You can try this:

obj = lst1: 300, lst2: 381, lst3: 4, lst4: 4, lst5: 49, lst9: 2; // This is an object, That;s ehy you getting infinite

var arr = Object.keys( obj ).map(function ( key ) return obj[key]; );

alert(Math.min.apply( null, arr ));

Fiddle: https://jsfiddle.net/ozvubmdr/

Easy, though not optimal, as it traverses the array twice: find the smallest value, then find the key belonging to it.

arr = lst1: 300, lst2: 381, lst3: 4, lst4: 4, lst5: 49, lst9: 2
min = Math.min(...Object.values(arr));
minkey = Object.keys(arr).find(key => arr[key] = min);
console.log(`Lowest is $min at $minkey`);

You may try this:

const arr = lst1: 300, lst2: 381, lst3: 4, lst4: 4, lst5: 49, lst9: 2
const minVal = Math.min(...Object.values(arr))
const fromKey = Object.keys(arr).find(key => arr[key] === minVal)
console.log(`The lowest value is $minVal from $fromKey`)

You can use Object.entries and use Array.reduce to reduce the resulting [key-value] pair to an object which will have lowest entry:

const data = lst1: 300, lst2: -381, lst3: 4, lst4: 4, lst5: 49, lst6: -49, lst7: 0, lst7: 78, lst7: -90;
const obj = Object.entries(data).reduce((acc, [key, value]) =>
 acc["lowest"] = acc["lowest"] ? Object.values(acc["lowest"])[0] < value ? acc["lowest"] : [key] : value : [key] : value ;
 return acc;
, );
console.log(`Lowest value is found in $JSON.stringify(obj.lowest)`);