// I'm able to successfuly find out the largest values in the object, however. There is a problem, and that is inability to display the associated key with that value.
var obj = {
t1: 1,
t2: 33,
t3: 10,
t4: 9,
t5: 45,
t6: 101
//...
}
// create an array
var arr = [];
// loop through the object and add values to the array
for (var p in obj) {
arr.push(obj[p]);
}
// sort the array, largest numbers to lowest
arr.sort(function(a,b){return b - a});
// grab the first 10 numbers
var firstThree = arr.slice(0, 3);
console.log(firstThree);I'm able to successfuly find out the largest values in the object, however. There is a problem, and that is inability to display the associated key with that value.
var obj = {
t1: 1,
t2: 33,
t3: 10,
t4: 9,
t5: 45,
t6: 101
//...
}
// create an array
var arr = [];
// loop through the object and add values to the array
for (var p in obj) {
arr.push(obj[p]);
}
// sort the array, largest numbers to lowest
arr.sort(function(a,b){return b - a});
// grab the first 10 numbers
var firstThree = arr.slice(0, 3);
console.log(firstThree);It can display the top values, however i'm having a hard time to display keys with that.
The result should be like
var result = {
t6: 101,
t5: 45,
t2: 33
}
-
Do you need
resultto be the same object asobj(with the other keys removed), or an entirely new object (so, you now have two)? – VLAZ Commented Jan 28, 2019 at 12:00 - Whatever is the most efficient way, is the way im after, I just want to send and object with highest values and associated keys with that – Stas Commented Jan 28, 2019 at 12:03
- Possible duplicate of Sorting JavaScript Object by property value – Ayman Safadi Commented Jan 28, 2019 at 12:07
- not really a dublicate, i think here is more straightforward and more flexible – Stas Commented Jan 28, 2019 at 12:11
8 Answers
Reset to default 8You could get the entries, sort and slice them and build a new object by assigning mapped objects to a single object.
var object = { t1: 1, t2: 33, t3: 10, t4: 9, t5: 45, t6: 101 },
result = Object.assign( // collect all objects into a single obj
...Object // spread the final array as parameters
.entries(object) // key a list of key/ value pairs
.sort(({ 1: a }, { 1: b }) => b - a) // sort DESC by index 1
.slice(0, 3) // get first three items of array
.map(([k, v]) => ({ [k]: v })) // map an object with a destructured
); // key/value pair
console.log(result);You can push objects into array instead of values, so that you can store keys
var obj = {
t1: 1,
t2: 33,
t3: 10,
t4: 9,
t5: 45,
t6: 101
}
// create an array
var arr = [];
// loop through the object and add values to the array
for (var p in obj) {
arr.push({key: p, value: obj[p]});
}
// sort the array, largest numbers to lowest
arr.sort(function(a,b){return b.value - a.value});
// grab the first 10 numbers
var firstThree = arr.slice(0, 3);
console.log(firstThree);You could push to your array an object containing the key and value association arr.push({p: obj[p]})
var obj = {
t1: 1,
t2: 33,
t3: 10,
t4: 9,
t5: 45,
t6: 101
}
// create an array
var arr = [];
// loop through the object and add values to the array
for (var p in obj) {
arr.push({
p: obj[p]
});
}
// sort the array, largest numbers to lowest
arr.sort(function(a, b) {
return b - a
});
// grab the first 10 numbers
var firstThree = arr.slice(0, 3);
console.log(firstThree);Nothing fancy but does the job :-)
you can do something like this
var obj = {
t1: 1,
t2: 33,
t3: 10,
t4: 9,
t5: 45,
t6: 101
};
// create an array
var arr = [];
// loop through the object and add values to the array
for (var p in obj) {
arr.push({ key: p, val: obj[p] });
}
// sort the array, largest numbers to lowest
arr.sort(function(a, b) {
return b.val - a.val;
});
// grab the first 10 numbers
var firstThree = arr.slice(0, 3);
console.log(...firstThree);
You can first get the values of that object and sort it. Then loop over to the key-value of that object using Object.entries() and determine the key of that value which belong to top 3 values:
var obj = {
t1: 1,
t2: 33,
t3: 10,
t4: 9,
t5: 45,
t6: 101
};
var values = Object.values(obj).sort((a,b) => b-a).slice(0,3);
let resultObj = {};
Object.entries(obj).forEach((item) => {
if(values.indexOf(item[1]) !== -1){
resultObj[item[0]] = item[1];
}
});
console.log(resultObj);Also note that the key ordering in object does not really matter when you work with object, so
{
t6: 101,
t5: 45,
t2: 33
}
is similar to
{
t2: 33,
t5: 45,
t6: 101
}
Alternative solution using lodash
_.chain(object)
.map((value, key) => ({value, key}))
.sortBy("value")
.reverse()
.take(3)
.reduce((acc, item) => ({ ...acc, [item.key]: item.value}), {})
.value();
var obj = {
t1: 1,
t2: 33,
t3: 10,
t4: 9,
t5: 45,
t6: 101
};
const sortedKey = Object.keys(obj).sort((a,b)=>{
if(obj[a] > obj[b]) return -1;
if(obj[a] < obj[b]) return 1;
return 0;
})
//print top three key -- val
for(let i=0; i< 3; i++) console.log(sortedKey[i], '--', obj[sortedKey[i]])
Objects key are not guaranteed to remain in order, so don't trust that.
It's safe though to create and use an array for that:
const top10KeyValArr = Object.entries(obj).sort(([_k1, val1], [_k2, val2])=> val2 - val1).slice(0, 3);
Then you can do what you want with it and it will remain in that order.