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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

}

• javascript
• jquery
• arrays

Share Improve this question Follow edited Jan 30, 2019 at 8:55 Alexandre Elshobokshy 10.9k66 gold badges3434 silver badges6060 bronze badges asked Jan 28, 2019 at 11:57 StasStas 69566 silver badges2222 bronze badges 4

• Do you need result to be the same object as obj (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

Add a ment  | 

8 Answers 8

Sorted by: Reset to default 8

You 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.

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

}

• javascript
• jquery
• arrays

Share Improve this question Follow edited Jan 30, 2019 at 8:55 Alexandre Elshobokshy 10.9k66 gold badges3434 silver badges6060 bronze badges asked Jan 28, 2019 at 11:57 StasStas 69566 silver badges2222 bronze badges 4

• Do you need result to be the same object as obj (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

Add a ment  | 

8 Answers 8

Sorted by: Reset to default 8

You 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.

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