split an Js object into array

Question

I have an object and I need to split the values into an array like below

Ex: {firstName1: "Tom", lastName1: "Jerry", firstName2: "Stuart", lastName2: "Little"}

I want the output to be like

[{fName: "Tom", lName: "Jerry"},
{fName: "Stuart", lName: "Little"}]

I can do this like

      let persons= new Array<Person>(); //Assume Person is a Typescript class
      let p1 = new Person(); 
      p1.fName= data.firstName1;
      p1.lName= data.lastName1;
      persons.push(p1);
      let p2= new Person(); 
      p2.fName= data.firstName2;
      p2.lName= data.lastName2;
      persons.push(p2);

I'm looking for an optimized/recommended way of achieving this.

What exactly do you mean by optimising it. What you are doing here seems fine. Recommended will be to put new person bit in a loop. — Atul Kumar
– Atul Kumar, Commented Oct 1, 2020 at 11:26
Optimizing in the sense like I feel this is not the only option to achieve this if possible to reduce number of lines of code as well. I would have done for loop part but the object many more additional properties which are not related to Person object. — Syed Abbas
– Syed Abbas, Commented Oct 1, 2020 at 11:34
I am assuming firstname and lastname remains same for all, the suffix (number) changes!? you still have to traverse through the object, but need to condition it according to the string parameter of the key i,e my assumption — Atul Kumar
– Atul Kumar, Commented Oct 1, 2020 at 11:43

Dharman · Accepted Answer · 2020-10-01 11:59:30Z

2

let requiredObject  = [];
let firstNames = []; 
let lastNames = [];
let inputObject = {firstName1: "Tom", lastName1: "Jerry", firstName2: "Stuart", lastName2: "Little"}
firstNames = Object.keys(inputObject).filter(each => each.includes('firstName'));
lastNames = Object.keys(inputObject).filter(each => each.includes('lastName'));
for(var i = 0; (firstNames.length > i &&  lastNames.length > i); i++) {
  var data = {};
  data.fName = inputObject[firstNames[i]];
  data.lName = inputObject[lastNames[i]]
  requiredObject.push(data);
}

console.log(requiredObject);

edited Oct 1, 2020 at 11:59

Dharman♦

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answered Oct 1, 2020 at 11:54

Surya Kavutarapu

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Comments

Chintan Dhandhusariya · Accepted Answer · 2020-10-01 11:31:43Z

1

How about this?

let persons= new Array<Person>();
let person = {}; //Not familiar with typescript much. Not sure if it would be {} or new Person();

Object.keys(data).forEach((key,index)=>{
    if(index%2 == 0) {
        //firstName
        person.fName = data[key];
    } else {
        //lastName
        person.lName = data[key];
        persons.push(person);
        person = {}; // or new Person();
    }
});

answered Oct 1, 2020 at 11:31

Chintan Dhandhusariya

416 bronze badges

2 Comments

Syed Abbas Over a year ago

Thanks for your input. The object contain many more properties which are not related to Person class. so looping through properties and plucking using Index may not help.

Chintan Dhandhusariya Over a year ago

You can use a regex instead of index%2 to determine the required key names

pilchard · Accepted Answer · 2020-10-01 13:56:40Z

Update Reduce time complexity by avoiding nested iterations

const combined = {firstName1: "Tom", lastName1: "Jerry", age1: 101, firstName2: "Stuart", lastName2: "Little", age2: 15, petName2: "dog"};

const personMap = Object.entries(combined).reduce((acc, [k, v]) => {
   const pNum = k.match(/[0-9]+$/)[0];
   const normalKey = k.slice(0, (k.length - pNum.length));
   
   const match = acc.get(pNum),
          attr = {[normalKey]: v};
        match ?
          Object.assign(match, attr) :
          acc.set(pNum, {...attr});
          
   return acc;
}, new Map());

const refactored = [...personMap.values()].map(p => ({...p}));

console.log(refactored);

You can use a .reduce() on the keys of the initial object and generate the lastName key from the firstName to avoid ordering problems.

const combined = {firstName1: "Tom", lastName1: "Jerry", firstName2: "Stuart", lastName2: "Little"};

const refactored = Object.keys(combined).reduce((a, k) => {
    if (k.startsWith('firstName')){
      a.push({fName: combined[k], lName: combined[k.replace('firstName', 'lastName')]});
    }
   return a;
  }, []);
  
  console.log(refactored);

You can generalize this for any number of attributes and pass an array of attributes to map to the new array.

const combined = {firstName1: "Tom", lastName1: "Jerry", age1: 101, firstName2: "Stuart", lastName2: "Little", age2: 15};
const attributes = ['firstName', 'lastName', 'age'];

const refactored = Object.keys(combined).reduce((a, k) => {
    if (k.startsWith('firstName')){
      const p = attributes.reduce((b, attr) => (b[attr] = combined[k.replace('firstName', attr)], b), {});
      a.push({...p});
    }
   return a;
  }, []);
  
  console.log(refactored);

Or for an arbitrary number of attributes .filter the keys of the object by personNumber.

const combined = {firstName1: "Tom", lastName1: "Jerry", age1: 101, firstName2: "Stuart", lastName2: "Little", age2: 15, petName2: "dog"};

const refactored = Object.keys(combined).reduce((a, k) => {
    if (k.startsWith('firstName')){
      const pNum = k.replace('firstName', '');
      const p = Object.keys(combined)
          .filter(k => k.endsWith(pNum))
          .reduce((b, attr) => (b[attr.replace(pNum, '')] = combined[attr], b), {});
      a.push({...p});
    }
   return a;
  }, []);
  
  console.log(refactored);

Eka Cipta · Accepted Answer · 2020-10-01 12:24:30Z

0

let obj = {firstName1: "Tom", lastName1: "Jerry", firstName2: "Stuart", lastName2: "Little"}
let keysArr = Object.keys(obj)
let valueArr = Object.values(obj)

let result = []

keysArr.map((x, i) => {
  if(x.indexOf('first') != -1){
    result.push({ fName: valueArr[i], lName: valueArr[i+1] })
  }
})

console.log(result)

answered Oct 1, 2020 at 12:24

Eka Cipta

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