What is the most efficient way to group elements in an Array by their shared property value and push them into a new Array as a single element?

Question

What is the most efficient way to group elements in an Array by their shared property value and push them into a new Array as a single element?

Is there a way to group elements in an Array by their property value and put them into a new element, then push that new element into a new Array?

data = [
    {status: 0,name:'a'},
    {status: 0,name:'b'},
    {status: 1,name:'b'},
];

var newData = [
           {
               id: 0,
                name: 'a',
                    services: [
                        {id: 'a', name: 'a',status: 0}
                    ]
                },
                {
                    id: 1,
                    name: 'b',
                    services: [
                        {id: 'b', name: 'b',status: 0},
                        {id: 'b', name: 'b',status: 1},
                    ]
                }
            ]

javascript arrays

Answer 1

Answer №1

To organize the items into groups with the same name, you can utilize a hash table as a reference and then add the actual item to the 'services' array.

var data = [{ status: 0, name: 'a' }, { status: 0, name: 'b' }, { status: 1, name: 'b' }],
    grouped = data.reduce(function (hash) {
        return function (r, a) {
            if (!hash[a.name]) {
                hash[a.name] = { id: a.name, name: a.name, services: [] };
                r.push(hash[a.name]);
            }
            a.id = a.name;
            hash[a.name].services.push(a);
            return r;
        };
    }(Object.create(null)), []);

console.log(grouped);

.as-console-wrapper { max-height: 100% !important; top: 0; }

The id property starts counting from zero in this scenario.

var data = [{ status: 0, name: 'a' }, { status: 0, name: 'b' }, { status: 1, name: 'b' }],
    grouped = data.reduce(function (hash, id) {
        return function (r, a) {
            if (!hash[a.name]) {
                hash[a.name] = { id: id++, name: a.name, services: [] };
                r.push(hash[a.name]);
            }
            a.id = a.name;
            hash[a.name].services.push(a);
            return r;
        };
    }(Object.create(null), 0), []);

console.log(grouped);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 2

To organize the items into groups with the same name, you can utilize a hash table as a reference and then add the actual item to the 'services' array.

var data = [{ status: 0, name: 'a' }, { status: 0, name: 'b' }, { status: 1, name: 'b' }],
    grouped = data.reduce(function (hash) {
        return function (r, a) {
            if (!hash[a.name]) {
                hash[a.name] = { id: a.name, name: a.name, services: [] };
                r.push(hash[a.name]);
            }
            a.id = a.name;
            hash[a.name].services.push(a);
            return r;
        };
    }(Object.create(null)), []);

console.log(grouped);

.as-console-wrapper { max-height: 100% !important; top: 0; }

The id property starts counting from zero in this scenario.

var data = [{ status: 0, name: 'a' }, { status: 0, name: 'b' }, { status: 1, name: 'b' }],
    grouped = data.reduce(function (hash, id) {
        return function (r, a) {
            if (!hash[a.name]) {
                hash[a.name] = { id: id++, name: a.name, services: [] };
                r.push(hash[a.name]);
            }
            a.id = a.name;
            hash[a.name].services.push(a);
            return r;
        };
    }(Object.create(null), 0), []);

console.log(grouped);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 3

Answer №2

var inputData = [
             {status: 0,name:'apple'},
             {status: 0,name:'banana'},
             {status: 1,name:'banana'}
         ];
var outputData = [];
var index=0;
inputData.forEach(function(data){
                           var matchingElem = outputData.find(function(element){
                           return element.name === data.name;
             });

if(matchingElem) 
  {
    matchingElem.services.push({
                        "id":data.name,
                        "name": data.name,
                        "status": data.status
                      });
  }
  else{
           outputData.push({
                      "id":index++,
                      "name":data.name,
                      "services": [{
                                 "id":data.name,
                                 "name": data.name,
                                 "status": data.status
                               }]
                    });
     }

});
console.log(JSON.stringify( outputData ));

Answer 4

var inputData = [
             {status: 0,name:'apple'},
             {status: 0,name:'banana'},
             {status: 1,name:'banana'}
         ];
var outputData = [];
var index=0;
inputData.forEach(function(data){
                           var matchingElem = outputData.find(function(element){
                           return element.name === data.name;
             });

if(matchingElem) 
  {
    matchingElem.services.push({
                        "id":data.name,
                        "name": data.name,
                        "status": data.status
                      });
  }
  else{
           outputData.push({
                      "id":index++,
                      "name":data.name,
                      "services": [{
                                 "id":data.name,
                                 "name": data.name,
                                 "status": data.status
                               }]
                    });
     }

});
console.log(JSON.stringify( outputData ));

Answer 5

Answer №3

data = [
{status: 0,name:'a'},
{status: 0,name:'b'},
{status: 1,name:'b'}, 
];

NewData=data.reduce((newobj,obj)=>((newobj[obj.name]=newobj[obj.name]||[]).push(obj),newobj),{});

The newly transformed data structure, NewData, will now look like this:

{
    "a":[
      {status: 0,name:'a'},
    ],
    "b":[
      {status: 0,name:'b'},
      {status: 1,name:'b'}, 
    ]
}

If you want to see the code in action, check out this link.

To achieve the desired result, you need to place each object from the old array into the corresponding obj.name array of the new object. If the array does not exist yet, create a new one with ' ||[] ' method. Some formatting may still be required to tailor the output further to your needs.

Answer 6

data = [
{status: 0,name:'a'},
{status: 0,name:'b'},
{status: 1,name:'b'}, 
];

NewData=data.reduce((newobj,obj)=>((newobj[obj.name]=newobj[obj.name]||[]).push(obj),newobj),{});

The newly transformed data structure, NewData, will now look like this:

{
    "a":[
      {status: 0,name:'a'},
    ],
    "b":[
      {status: 0,name:'b'},
      {status: 1,name:'b'}, 
    ]
}

If you want to see the code in action, check out this link.

To achieve the desired result, you need to place each object from the old array into the corresponding obj.name array of the new object. If the array does not exist yet, create a new one with ' ||[] ' method. Some formatting may still be required to tailor the output further to your needs.

Answer 7

Answer №4

Try out this method

Start by creating a mapping of name with an array of corresponding statuses.
Go through the map and build the array accordingly.

EXAMPLE

var data = [
    {status: 1, name:'x'},
    {status: 0, name:'y'},
    {status: 1, name:'z'},
];

var map = {};
data.forEach( function(item){
 map[ item.name ] = map[ item.name ] || [];
 map[ item.name ].push( item );
});

var newData = Object.keys( map ).map(function(name, index){
  var arr = map[ name ];
  arr = arr.map( function(arrItem){ arrItem.refId = arrItem.status; return arrItem });
  return { refId : index, title : name, entries : arr };
});

console.log(newData);

Answer 8

Try out this method

Start by creating a mapping of name with an array of corresponding statuses.
Go through the map and build the array accordingly.

EXAMPLE

var data = [
    {status: 1, name:'x'},
    {status: 0, name:'y'},
    {status: 1, name:'z'},
];

var map = {};
data.forEach( function(item){
 map[ item.name ] = map[ item.name ] || [];
 map[ item.name ].push( item );
});

var newData = Object.keys( map ).map(function(name, index){
  var arr = map[ name ];
  arr = arr.map( function(arrItem){ arrItem.refId = arrItem.status; return arrItem });
  return { refId : index, title : name, entries : arr };
});

console.log(newData);

Answer 9

Answer №5

    var elements = [
        {status: 0, name:'a'},
        {status: 0, name:'b'},
        {status: 1, name:'b'}
    ];

    var idNumber = 0;
    var newData = elements.reduce(function(accumulator, current) {
      var item = accumulator.find(element => {
        return element.name === current.name
      });
      if (item) {
        item.services.push({
          id: current.name,
          name: current.name,
          status: current.status
        })
      } else {
        accumulator.push({
          id: idNumber++,
          name: current.name,
          services: [{
            id: current.name,
            name: current.name,
            status: current.status
          }]
        })
      }
      return accumulator;
    }, []);
    
    console.log(newData)

Answer 10

    var elements = [
        {status: 0, name:'a'},
        {status: 0, name:'b'},
        {status: 1, name:'b'}
    ];

    var idNumber = 0;
    var newData = elements.reduce(function(accumulator, current) {
      var item = accumulator.find(element => {
        return element.name === current.name
      });
      if (item) {
        item.services.push({
          id: current.name,
          name: current.name,
          status: current.status
        })
      } else {
        accumulator.push({
          id: idNumber++,
          name: current.name,
          services: [{
            id: current.name,
            name: current.name,
            status: current.status
          }]
        })
      }
      return accumulator;
    }, []);
    
    console.log(newData)

What is the most efficient way to group elements in an Array by their shared property value and push them into a new Array as a single element?

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Similar questions

Is it possible for JavaScript to be cached when it is located within the body tag of an HTML document?

Utilizing JSON data from Jade in local JavaScript: A comprehensive guide

Tips on sending form data, including a file, to Ajax using the onclick() method

Exploring Json parsing in Python with Django

Tracking mouse movement: calculating mouse coordinates in relation to parent element

Having trouble getting the JavaScript slider to animate

Is it possible to verify if a web application is currently active on a different browser tab?

Triggering form submission through JavaScript by simulating keypress Enter does not work

What is the best way to adjust the camera position in three.js while keeping it from rotating?

Enhance your Rails application by dynamically updating table cells with Ajax, eliminating the need for

Unable to use .ajax within autocomplete function

Can you explain the distinction between using <router-view/> and <router-view></router-view>?

What is the best way to automatically adjust a panel's width and height based on the user's screen resolution?

Using jQuery to create sliding animations with left and right transitions

Doing an asynchronous function in the route.js file in Node.js

Error with ThreeJs OrbitControls.js functionality not operating as expected

Why does this image slider begin with blankness?

Angular directive for resizing C3 charts

I am looking for a way to locate all meta tags that begin with either "ABC_" or "XYZ_" by using jQuery

When dealing with arrays, the parentElement.remove() function may not be defined