I just want to return an array containing the first and second objects based on the "point" property of an object
and I have a condition below, When the condition is satisfied it will swap places.
Default:
ENGLISH[0]
ENGLISH[1]
MATH[0]
MATH[1]
HISTORY[0]
HISTORY[1]
...
if(ENGLISH[0].POINT/ENGLISH[1].POINT > MATH[0].POINT/MATH[1].POINT)
MATH[0]
MATH[1]
ENGLISH[0]
ENGLISH[1]
HISTORY[0]
HISTORY[1]
...
Given the following object:
const arr = [
{ name: "John", age: 32, group: "Math", point: 70 },
{ name: "David", age: 23, group: "Math", point: 20 },
{ name: "Justin", age: 28, group: "Math", point: 20 },
{ name: "Neymar", age: 30, group: "Math", point: 50 },
{ name: "Arnauld", age: 35, group: "History", point: 40 },
{ name: "Ivan", age: 18, group: "History", point: 50 },
{ name: "Nekko", age: 13, group: "History", point: 80 },
{ name: "Lena", age: 25, group: "English", point: 90 },
{ name: "Test", age: 45, group: "English", point: 30 },
{ name: "Ann", age: 19, group: "English", point: 38 }
]
the result I want is
const result = [
{ name: "Lena", age: 25, group: "English", point: 90 },
{ name: "Test", age: 45, group: "English", point: 30 },
{ name: "John", age: 32, group: "Math", point: 70 },
{ name: "David", age: 23, group: "Math", point: 20 },
{ name: "Arnauld", age: 35, group: "History", point: 40 },
{ name: "Ivan", age: 18, group: "History", point: 50 },
{ name: "Ann", age: 19, group: "English", point: 38 },
{ name: "Justin", age: 28, group: "Math", point: 20 },
{ name: "Neymar", age: 30, group: "Math", point: 50 },
{ name: "Nekko", age: 13, group: "History", point: 80 }
]
You could group the data with an object for the indices and an array of references.
const
order = { English: 0, Math: 1, History: 2 },
data = [{ name: "John", age: 32, group: "Math", point: 70 }, { name: "David", age: 23, group: "Math", point: 20 }, { name: "Justin", age: 28, group: "Math", point: 20 }, { name: "Neymar", age: 30, group: "Math", point: 50 }, { name: "Arnauld", age: 35, group: "History", point: 40 }, { name: "Ivan", age: 18, group: "History", point: 50 }, { name: "Nekko", age: 13, group: "History", point: 80 }, { name: "Lena", age: 25, group: "English", point: 90 }, { name: "Test", age: 45, group: "English", point: 30 }, { name: "Ann", age: 19, group: "English", point: 38 }],
references = [0, 0, 0],
result = data
.reduce((r, o) => {
const i = order[o.group];
if (r[references[i]][i].length === 2) {
references[i]++;
r[references[i]] ??= []
r[references[i]][i] = [];
}
r[references[i]][i].push(o);
return r;
}, [[[], [], []]])
.flat(Infinity);
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }
In case reactjs Run time environment has been tested.
import sortBy from 'lodash/sortBy';
let testData = [
{ name: 'Lena', age: 25, group: 'English', point: 90 },
{ name: 'Test', age: 45, group: 'English', point: 30 },
{ name: 'John', age: 32, group: 'Math', point: 70 },
{ name: 'David', age: 23, group: 'Math', point: 20 },
{ name: 'Arnauld', age: 35, group: 'History', point: 40 },
{ name: 'Ivan', age: 18, group: 'History', point: 50 },
{ name: 'Ann', age: 19, group: 'English', point: 38 },
{ name: 'Justin', age: 28, group: 'Math', point: 20 },
{ name: 'Neymar', age: 30, group: 'Math', point: 50 },
{ name: 'Nekko', age: 13, group: 'History', point: 80 }
];
let sortedDataByName = sortBy(result, 'name');
let sortedDataByAge = sortBy(result, 'age');
let sortedDataByGroup = sortBy(result, 'group');
let sortedDataByPoints = sortBy(result, 'point');
Related
How can i get a part of the object and put it in a new element in that object
I have an original Array like this
[
{
stt:1,
code: 24,
name: 'n',
age: 27,
address: 'somewhere',
},
{
stt:2,
code: 82,
name: 'jo',
age: 23,
address: 'tree',
},
{
stt:3,
code: 2,
name: 'con',
age: 23,
address: 'somewhere',
},
]
array that i expected to be a result :
[
{
stt:1,
code: 24,
name: 'n',
employee: {
age: 27,
address: 'somewhere',
}
},
{
stt:2,
code: 82,
name: 'jo',
employee: {
age: 23,
address: 'tree',
}
},
{
stt:3,
code: 2,
name: 'con',
employee: {
age: 23,
address: 'somewhere',
}
},
]
i tried to get first 3 element of object but dont know how to do next
you can use Array.map for that
like this
const transform = data =>data.map(
({age, address, ...rest}) => ({...rest, employee: {age, address}})
)
const data = [
{
stt:1,
code: 24,
name: 'n',
age: 27,
address: 'somewhere',
},
{
stt:2,
code: 82,
name: 'jo',
age: 23,
address: 'tree',
},
{
stt:3,
code: 2,
name: 'con',
age: 23,
address: 'somewhere',
},
]
console.log(transform(data))
This question already has answers here:
Is there a way to sort/order keys in JavaScript objects?
(9 answers)
Closed last year.
I could add "id" to each object in the array "data":
const data = [
{ name: "John", age: 24 },
{ name: "Marry", age: 18 },
{ name: "Tom", age: 15 },
]
for(const key in data) {
data[key]['id'] = key;
}
console.log(data);
But "id" is added to the last in each object in the array "data":
[
{ name: "John", age: 24, id: "0" },
{ name: "Marry", age: 18, id: "1" },
{ name: "Tom", age: 15, id: "2" }
]
My desired result is this below adding "id" to the first in each object in the array "data":
[
{ id: "0", name: "John", age: 24 },
{ id: "1", name: 'Marry', age: 18 },
{ id: "2", name: 'Tom', age: 15 }
]
Are there any ways to do that?
Create the array "newData" with "id" and the array "data":
const data = [
{ name: "John", age: 24 },
{ name: "Marry", age: 18 },
{ name: "Tom", age: 15 },
]
const newData = [];
for(const key in data) {
const obj = {
id: key,
...data[key]
}
newData.push(obj);
}
console.log(newData);
This is the result:
[
{ id: "0", name: "John", age: 24 },
{ id: "1", name: 'Marry', age: 18 },
{ id: "2", name: 'Tom', age: 15 }
]
In addition, if you want "id" of Number type, use "Number()":
const data = [
{ name: "John", age: 24 },
{ name: "Marry", age: 18 },
{ name: "Tom", age: 15 },
]
const newData = [];
for(const key in data) {
const obj = {
id: Number(key), // Here
...data[key]
}
newData.push(obj);
}
console.log(newData);
This is the result:
[
{ id: 0, name: "John", age: 24 },
{ id: 1, name: 'Marry', age: 18 },
{ id: 2, name: 'Tom', age: 15 }
]
I'm looking for a simple methodology to merge the Array of the nested object into an array of the same object
Input:
const data=[
{ id: 123, name: "dave", age: 23 , address:{city:"chennai", zipcode:600001}},
{ id: 456, name: "chris", age: 23, address:{city:"cbe", zipcode:600002} },
{ id: 789, name: "bob", age: 23, address:{city:"tiruppur", zipcode:600003}},
{ id: 101, name: "tom", age: 23, address:{city:"erode", zipcode:600004} },
{ id: 102, name: "tim", age: 23, address:{city:"selam", zipcode:600005} }
]
Needed Output:
{ id: 123, name: "dave", age: 23, city:"chennai", zipcode:600001},
{ id: 456, name: "chris", age: 23, city:"cbe", zipcode:600002},
{ id: 789, name: "bob", age: 23, city:"tiruppur", zipcode:600003},
{ id: 101, name: "tom", age: 23, city:"erode", zipcode:600004},
{ id: 102, name: "tim", age: 23, city:"selam", zipcode:600005}
]
Use a combination of destructuring and spread syntax:
const data=[{ id: 123, name: "dave", age: 23 , address:{city:"chennai", zipcode:600001}},{ id: 456, name: "chris", age: 23, address:{city:"cbe", zipcode:600002} },{ id: 789, name: "bob", age: 23, address:{city:"tiruppur", zipcode:600003}},{ id: 101, name: "tom", age: 23, address:{city:"erode", zipcode:600004} },{ id: 102, name: "tim", age: 23, address:{city:"selam", zipcode:600005} }];
const result = data.map(({address, ...rest}) => ({...rest, ...address}));
console.log(result);
You can use Object.entries to extract each property and its value in the address property and assign it to the parent object, then delete the address property.
const data=[{id:123,name:"dave",age:23,address:{city:"chennai",zipcode:600001}},{id:456,name:"chris",age:23,address:{city:"cbe",zipcode:600002}},{id:789,name:"bob",age:23,address:{city:"tiruppur",zipcode:600003}},{id:101,name:"tom",age:23,address:{city:"erode",zipcode:600004}},{id:102,name:"tim",age:23,address:{city:"selam",zipcode:600005}}];
const result = data.map(e => (Object.entries(e.address).forEach(f => e[f[0]] = f[1]), delete e.address, e))
console.log(result)
let data = [
{ id: 123, name: "dave", age: 23, address: { city: "chennai", zipcode: 600001 } },
{ id: 456, name: "chris", age: 23, address: { city: "cbe", zipcode: 600002 } },
{ id: 789, name: "bob", age: 23, address: { city: "tiruppur", zipcode: 600003 } },
{ id: 101, name: "tom", age: 23, address: { city: "erode", zipcode: 600004 } },
{ id: 102, name: "tim", age: 23, address: { city: "selam", zipcode: 600005 } }
]
data = data.map((node) => ({ id: node.id, name: node.name, age: node.age, city: node.address.city, zipcode: node.address.zipcode }))
console.log(data)
const data=[
{ id: 123, name: "dave", age: 23 , address:{city:"chennai", zipcode:600001}},
{ id: 456, name: "chris", age: 23, address:{city:"cbe", zipcode:600002} },
{ id: 789, name: "bob", age: 23, address:{city:"tiruppur", zipcode:600003}},
{ id: 101, name: "tom", age: 23, address:{city:"erode", zipcode:600004} },
{ id: 102, name: "tim", age: 23, address:{city:"selam", zipcode:600005} }
];
const result = data.map(item => ({id: item.id, name: item.name, age: item.age, ...item.address}));
console.log(result);
I have an object so defined:
{
_id: "5d406a171ed43384972f04b5",
index: 0,
age: 28,
eyeColor: "brown",
name: {
first: "Myra",
last: "Navarro"
},
company: "SUSTENZA",
email: "myra.navarro#sustenza.net"
}
I would need to be able to do the search on all keys, even in compound keys like name.
For example I have a search string name.first and the like, it could be even deeper and compound objects.
I managed to do this, but on name.first for example I am failing to get it to work.
Can you give me a hand?
let datatable = [
{
_id: "5d406a171ed43384972f04b5",
index: 0,
age: 28,
eyeColor: "brown",
name: {
first: "Myra",
last: "Navarro"
},
company: "SUSTENZA",
email: "myra.navarro#sustenza.net"
},
{
_id: "5d406a170db0f4b04d9a9acf",
index: 1,
age: 23,
eyeColor: "blue",
name: {
first: "Harriett",
last: "Tanner"
},
company: "VALPREAL",
email: "harriett.tanner#valpreal.com"
},
{
_id: "5d406a17e95da8ff80a759c5",
index: 2,
age: 39,
eyeColor: "blue",
name: {
first: "Vega",
last: "Hanson"
},
company: "BEDLAM",
email: "vega.hanson#bedlam.tv"
},
{
_id: "5d406a175505da190e6875ec",
index: 3,
age: 31,
eyeColor: "blue",
name: {
first: "Rosemary",
last: "Fields"
},
company: "QUAILCOM",
email: "rosemary.fields#quailcom.me"
},
{
_id: "5d406a17ea96044c027f4e50",
index: 4,
age: 27,
eyeColor: "brown",
name: {
first: "Dale",
last: "Wilkinson"
},
company: "QIAO",
email: "dale.wilkinson#qiao.org"
},
{
_id: "5d406a17c5fff1ff6653a555",
index: 5,
age: 25,
eyeColor: "blue",
name: {
first: "Beatrice",
last: "Contreras"
},
company: "ZENOLUX",
email: "beatrice.contreras#zenolux.us"
},
{
_id: "5d406a17a199efcba25e1f26",
index: 6,
age: 34,
eyeColor: "blue",
name: {
first: "Hancock",
last: "Wynn"
},
company: "PLASMOS",
email: "hancock.wynn#plasmos.co.uk"
},
{
_id: "5d406a17019a2a4544a4f134",
index: 7,
age: 40,
eyeColor: "blue",
name: {
first: "Brown",
last: "Stanton"
},
company: "SNACKTION",
email: "brown.stanton#snacktion.name"
},
{
_id: "5d406a17e516dd71af8210d4",
index: 8,
age: 39,
eyeColor: "blue",
name: {
first: "Barnes",
last: "Dunn"
},
company: "PORTALINE",
email: "barnes.dunn#portaline.ca"
},
{
_id: "5d406a17516936a025b73c33",
index: 9,
age: 34,
eyeColor: "green",
name: {
first: "Blanche",
last: "Cherry"
},
company: "ISOSWITCH",
email: "blanche.cherry#isoswitch.io"
},
{
_id: "5d406a17527a4d2c6a7897dd",
index: 10,
age: 33,
eyeColor: "blue",
name: {
first: "Gilliam",
last: "Farley"
},
company: "AMTAS",
email: "gilliam.farley#amtas.biz"
},
{
_id: "5d406a175ff11478c416c30b",
index: 11,
age: 26,
eyeColor: "brown",
name: {
first: "Laura",
last: "Short"
},
company: "FISHLAND",
email: "laura.short#fishland.info"
},
{
_id: "5d406a1738181b471847339a",
index: 12,
age: 20,
eyeColor: "brown",
name: {
first: "Moreno",
last: "Barber"
},
company: "KEENGEN",
email: "moreno.barber#keengen.net"
},
{
_id: "5d406a17a6bcae6fe3ad1735",
index: 13,
age: 30,
eyeColor: "brown",
name: {
first: "Fischer",
last: "French"
},
company: "INCUBUS",
email: "fischer.french#incubus.com"
},
{
_id: "5d406a17600ca53e8f63f263",
index: 14,
age: 30,
eyeColor: "brown",
name: {
first: "Donaldson",
last: "Carr"
},
company: "SUNCLIPSE",
email: "donaldson.carr#sunclipse.tv"
},
{
_id: "5d406a17530655789a27174f",
index: 15,
age: 35,
eyeColor: "green",
name: {
first: "Sophia",
last: "Payne"
},
company: "PRISMATIC",
email: "sophia.payne#prismatic.me"
},
{
_id: "5d406a175dbc687b4c7669d8",
index: 16,
age: 34,
eyeColor: "green",
name: {
first: "Simone",
last: "Pollard"
},
company: "DIGIGEN",
email: "simone.pollard#digigen.org"
},
{
_id: "5d406a179f35ed326a6a5567",
index: 17,
age: 28,
eyeColor: "green",
name: {
first: "Yvette",
last: "Daugherty"
},
company: "CHILLIUM",
email: "yvette.daugherty#chillium.us"
}
];
const sortAsc = true;
const sortField = "name.first";//index,eyeColor
var result = datatable.sort((a, b) => {
let [x, z] = sortAsc ? [a, b] : [b, a];
//console.log(x[sortField], z[sortField], x[sortField] > z[sortField] ? 1 : -1);
return x[sortField] > z[sortField] ? 1 : -1;
});
console.log(result);
Edit:
It seems to work like this, but I would like to write it in a more compact form and possibly reduce the computation time.
Do you think it is possible?
var result = datatable.sort((a, b) => {
let [x, z] = sortAsc ? [a, b] : [b, a];
const arraySplit = sortField.split('.');
var vX = arraySplit.reduce((a, b) => a[b], x);
var vZ = arraySplit.reduce((a, b) => a[b], z);
return vX > vZ ? 1 : -1;
});
First issue is that javascript doesn't automatically access children using dots.
If you want to implement that, here are some solutions:
Access object child properties using a dot notation string
Next, once you have the value you want to compare with.
You need to use a special string compare function if the value is a string (use typeof to check). Or use valA - valB so that you get 1, 0, and -1 as the return values to sort with.
I hope this is what you need. The child function makes what you need. You have the explanation in the comments.
let datatable = [
{
_id: "5d406a171ed43384972f04b5",
index: 0,
age: 28,
eyeColor: "brown",
name: {
first: "Myra",
last: "Navarro"
},
company: "SUSTENZA",
email: "myra.navarro#sustenza.net"
},
{
_id: "5d406a170db0f4b04d9a9acf",
index: 1,
age: 23,
eyeColor: "blue",
name: {
first: "Harriett",
last: "Tanner"
},
company: "VALPREAL",
email: "harriett.tanner#valpreal.com"
},
{
_id: "5d406a17e95da8ff80a759c5",
index: 2,
age: 39,
eyeColor: "blue",
name: {
first: "Vega",
last: "Hanson"
},
company: "BEDLAM",
email: "vega.hanson#bedlam.tv"
},
{
_id: "5d406a175505da190e6875ec",
index: 3,
age: 31,
eyeColor: "blue",
name: {
first: "Rosemary",
last: "Fields"
},
company: "QUAILCOM",
email: "rosemary.fields#quailcom.me"
},
{
_id: "5d406a17ea96044c027f4e50",
index: 4,
age: 27,
eyeColor: "brown",
name: {
first: "Dale",
last: "Wilkinson"
},
company: "QIAO",
email: "dale.wilkinson#qiao.org"
},
{
_id: "5d406a17c5fff1ff6653a555",
index: 5,
age: 25,
eyeColor: "blue",
name: {
first: "Beatrice",
last: "Contreras"
},
company: "ZENOLUX",
email: "beatrice.contreras#zenolux.us"
},
{
_id: "5d406a17a199efcba25e1f26",
index: 6,
age: 34,
eyeColor: "blue",
name: {
first: "Hancock",
last: "Wynn"
},
company: "PLASMOS",
email: "hancock.wynn#plasmos.co.uk"
},
{
_id: "5d406a17019a2a4544a4f134",
index: 7,
age: 40,
eyeColor: "blue",
name: {
first: "Brown",
last: "Stanton"
},
company: "SNACKTION",
email: "brown.stanton#snacktion.name"
},
{
_id: "5d406a17e516dd71af8210d4",
index: 8,
age: 39,
eyeColor: "blue",
name: {
first: "Barnes",
last: "Dunn"
},
company: "PORTALINE",
email: "barnes.dunn#portaline.ca"
},
{
_id: "5d406a17516936a025b73c33",
index: 9,
age: 34,
eyeColor: "green",
name: {
first: "Blanche",
last: "Cherry"
},
company: "ISOSWITCH",
email: "blanche.cherry#isoswitch.io"
},
{
_id: "5d406a17527a4d2c6a7897dd",
index: 10,
age: 33,
eyeColor: "blue",
name: {
first: "Gilliam",
last: "Farley"
},
company: "AMTAS",
email: "gilliam.farley#amtas.biz"
},
{
_id: "5d406a175ff11478c416c30b",
index: 11,
age: 26,
eyeColor: "brown",
name: {
first: "Laura",
last: "Short"
},
company: "FISHLAND",
email: "laura.short#fishland.info"
},
{
_id: "5d406a1738181b471847339a",
index: 12,
age: 20,
eyeColor: "brown",
name: {
first: "Moreno",
last: "Barber"
},
company: "KEENGEN",
email: "moreno.barber#keengen.net"
},
{
_id: "5d406a17a6bcae6fe3ad1735",
index: 13,
age: 30,
eyeColor: "brown",
name: {
first: "Fischer",
last: "French"
},
company: "INCUBUS",
email: "fischer.french#incubus.com"
},
{
_id: "5d406a17600ca53e8f63f263",
index: 14,
age: 30,
eyeColor: "brown",
name: {
first: "Donaldson",
last: "Carr"
},
company: "SUNCLIPSE",
email: "donaldson.carr#sunclipse.tv"
},
{
_id: "5d406a17530655789a27174f",
index: 15,
age: 35,
eyeColor: "green",
name: {
first: "Sophia",
last: "Payne"
},
company: "PRISMATIC",
email: "sophia.payne#prismatic.me"
},
{
_id: "5d406a175dbc687b4c7669d8",
index: 16,
age: 34,
eyeColor: "green",
name: {
first: "Simone",
last: "Pollard"
},
company: "DIGIGEN",
email: "simone.pollard#digigen.org"
},
{
_id: "5d406a179f35ed326a6a5567",
index: 17,
age: 28,
eyeColor: "green",
name: {
first: "Yvette",
last: "Daugherty"
},
company: "CHILLIUM",
email: "yvette.daugherty#chillium.us"
}
];
const sortAsc = true;
const sortField = "name.first";//index,eyeColor
var result = datatable.sort((a, b) => {
let [x, z] = sortAsc ? [a, b] : [b, a];
//console.log(x[sortField], z[sortField], x[sortField] > z[sortField] ? 1 : -1);
//return x[sortField] > z[sortField] ? 1 : -1;
return child(x, sortField) > child(z, sortField) ? 1: -1;
});
console.log(result);
function child(obj, str){
//take dot-separated chain of properties' names and split them into an array
const props = str.split('.');
let child = obj;
//looop through all properties' names and get the access to the following [Object] children sequentially
props.forEach((prop)=> child = child[prop]);
return child;
}
Option 1:
const objRoute = sortField.split(".");
const getValue = obj => objRoute.reduce((v, c) => v[c], obj);
const result = datatable.sort((a, b) =>
sortAsc
? getValue(a).localeCompare() - getValue(b).localeCompare()
: getValue(b).localeCompare() - getValue(a).localeCompare()
);
Option 2:
Use eval, eval() evaluates the expression.
It is recommended to use this function with caution:
const result = datatable.sort((a, b, i) => eval(`a.${sortField}`) - eval(`b.${sortField}`));
I have a set of data where i need to format and render in HTML.
This is the data :
var data = [{
name: 'john',
age: 20,
department: 'financial',
group: 'groupA'
},
{
name: 'john',
age: 20,
department: 'healthcare',
group: 'groupA'
},
{
name: 'john',
age: 20,
department: 'insurance',
group: 'groupA'
},
{
name: 'john',
age: 20,
department: 'Realestate',
group: 'groupB'
}, {
name: 'Mark',
age: 20,
department: 'financial',
group: 'groupA'
},
{
name: 'Mark',
age: 21,
department: 'insurance',
group: 'groupA'
},
{
name: 'Mark',
age: 21,
department: 'financial',
group: 'groupB'
}
];
Here i have multiple users where he belongs to different groups and department.I want to format in such a way the if he belongs to same group then get the department which he belongs to in array and for other department show that as a new object.
The response should look like this:
var data_formatted = [
[{
name: 'john',
age: 20,
department: ['financial', 'healthcare', 'insurance'],
group: 'groupA'
},
{
name: 'john',
age: 20,
department: 'Realestate',
group: 'groupB'
}
],
[{
name: 'Mark',
age: 20,
department: ['financial', 'insurance'],
group: 'groupA'
},
{
name: 'Mark',
age: 21,
department: 'financial',
group: 'groupB'
}
]
];
Any suggestions Please!
I think this may help you.
var data = [{
name: 'john',
age: 20,
department: 'financial',
group: 'groupA'
},
{
name: 'john',
age: 20,
department: 'healthcare',
group: 'groupA'
},
{
name: 'john',
age: 20,
department: 'insurance',
group: 'groupA'
},
{
name: 'john',
age: 20,
department: 'Realestate',
group: 'groupB'
}, {
name: 'Mark',
age: 20,
department: 'financial',
group: 'groupA'
},
{
name: 'Mark',
age: 21,
department: 'insurance',
group: 'groupA'
},
{
name: 'Mark',
age: 21,
department: 'financial',
group: 'groupB'
}
];
var data_formatted = [];
var byname = _.groupBy(data, function(element){return element.name});
_.forEach(byname, function(element){
var bygroup = _.groupBy(element, function(element){return element.group});
_.forEach(bygroup, function(element, key, groups){
var departments = _.map(element, 'department');
element[0]['department'] = departments;
groups[key] = element[0];
})
data_formatted.push(_.values(bygroup));
})
console.log(data_formatted);
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.4/lodash.min.js"></script>