Create array of unique objects by property

JavascriptJqueryFilter

Javascript Problem Overview


I created an array of objects like so:

[    {        "lat": 12.123,        "lng": 13.213,        "city": "New York"    },    {        "lat": 3.123,        "lng": 2.213,        "city": "New York"    },    {        "lat": 1.513,        "lng": 1.113,        "city": "London"    }]

I'm trying to create a new array that filters the places to only contains objects that don't have the same city property (lat/lng duplicates are ok). Is there a built in JS or Jquery function to achieve this?

Javascript Solutions


Solution 1 - Javascript

I'd probably use a flags object during the filtering (edit: I wouldn't anymore, see the note at the end of the answer about ES2015's Set), like this:

var flags = {};
var newPlaces = places.filter(function(entry) {
    if (flags[entry.city]) {
        return false;
    }
    flags[entry.city] = true;
    return true;
});

That uses Array#filter from ECMAScript5 (ES5), which is one of the ES5 additions that can be shimmed (search for "es5 shim" for several options).

You can do it without filter, of course, it's just a bit more verbose:

var flags = {};
var newPlaces = [];
var index;
for (index = 0; index < places.length; ++index) {
    if (!flags[entry.city]) {
        flags[entry.city] = true;
        newPlaces.push(entry);
    }
});

Both of the above assume the first object with a given city should be kept, and all other discarded.


Note: As user2736012 points out below, my test if (flags[entry.city]) will be true for cities with names that happen to be the same as properties that exist on Object.prototype such as toString. Very unlikely in this case, but there are four ways to avoid the possibility:

  • (My usual preferred solution) Create the object without a prototype: var flags = Object.create(null);. This is a feature of ES5. Note that this cannot be shimmed for obsolete browsers like IE8 (the single-argument version of Object.create can be except when that argument's value is null).

  • Use hasOwnProperty for the test, e.g. if (flags.hasOwnProperty(entry.city))

  • Put a prefix on that you know doesn't exist for any Object.prototype property, such as xx:

      var key = "xx" + entry.city;
      if (flags[key]) {
          // ...
      }
      flags[key] = true;
    
  • As of ES2015, you could use a Set instead:

      const flags = new Set();
      const newPlaces = places.filter(entry => {
          if (flags.has(entry.city)) {
              return false;
          }
          flags.add(entry.city);
          return true;
      });
    

Solution 2 - Javascript

Shortest, but not best performance (see update bellow) solution for es6 :

function unique(array, propertyName) {
   return array.filter((e, i) => array.findIndex(a => a[propertyName] === e[propertyName]) === i);
}

performance: https://jsperf.com/compare-unique-array-by-property

Solution 3 - Javascript

https://lodash.com/docs#uniqBy

https://github.com/lodash/lodash/blob/4.13.1/lodash.js#L7711

/**
 * This method is like `_.uniq` except that it accepts `iteratee` which is
 * invoked for each element in `array` to generate the criterion by which
 * uniqueness is computed. The iteratee is invoked with one argument: (value).
 *
 * @static
 * @memberOf _
 * @since 4.0.0
 * @category Array
 * @param {Array} array The array to inspect.
 * @param {Array|Function|Object|string} [iteratee=_.identity]
 *  The iteratee invoked per element.
 * @returns {Array} Returns the new duplicate free array.
 * @example
 *
 * _.uniqBy([2.1, 1.2, 2.3], Math.floor);
 * // => [2.1, 1.2]
 *
 * // The `_.property` iteratee shorthand.
 * _.uniqBy([{ 'x': 1 }, { 'x': 2 }, { 'x': 1 }], 'x');
 * // => [{ 'x': 1 }, { 'x': 2 }]
 */

Solution 4 - Javascript

You can filter using a Set by only including elements with a property value that has not yet been added to the Set (after which it should be added to the Set). This can be accomplished in one line using the logical and operator (&&).

Below is a general function to obtain a unique array of objects based on a specific property (prop) from an array of objects (arr). Note that in the case of duplicates, only the first object with the property value will be retained.

const getUniqueBy = (arr, prop) => {
  const set = new Set;
  return arr.filter(o => !set.has(o[prop]) && set.add(o[prop]));
};

Demo:

var places = [{
  lat: 12.123,
  lng: 13.213,
  city: 'New York'
}, {
  lat: 3.123,
  lng: 2.213,
  city: 'New York'
}, {
  lat: 3.123,
  lng: 4.123,
  city: 'Some City'
}];
const getUniqueBy = (arr, prop) => {
  const set = new Set;
  return arr.filter(o => !set.has(o[prop]) && set.add(o[prop]));
};
console.log(getUniqueBy(places, 'city'));

Solution 5 - Javascript

I expanded a bit on @IgorL solution, but extended prototype and gave it a selector function instead of a property to make it a little more flexible:

Array.prototype.unique = function(selector) {
   return this.filter((e, i) => this.findIndex((a) => {
      if (selector) {
        return selector(a) === selector(e);
      }
      return a === e;
    }) === i);
};

Usage:

// with no param it uses strict equals (===) against the object
let primArr = ['one','one','two','three','one']
primArr.unique() // ['one','two','three']

let a = {foo:123}
let b = {foo:123}
let fooArr = [a,a,b]
fooArr.unique() //[a,b]

// alternatively, you can pass a selector function
fooArr.unique(item=>item.foo) //[{foo:123}] (first "unique" item returned)

Definitely NOT the most performant way to do this but as long as the selector is simple and the array isn't massive, it should work fine.

In Typescript

Array.prototype.unique = function<T>(this: T[], selector?: (item: T) => object): T[] {
   return this.filter((e, i) => this.findIndex((a) => {
      if (selector) {
        return selector(a) === selector(e);
      }
      return a === e;
    }) === i);
};

Solution 6 - Javascript

My suggestion :

Array.prototype.uniqueCity = function() {
	var processed = [];
	for (var i=this.length-1; i>=0; i--){
		if (processed.indexOf(this[i].city)<0) {
			processed.push(this[i].city);
		} else {
			this.splice(i, 1);
		}
	}
}

in use :

places.uniqueCity();

or

Array.prototype.uniqueObjectArray = function(field) {
	var processed = [];
	for (var i=this.length-1; i>=0; i--) {
		if (this[i].hasOwnProperty(field)) {
			if (processed.indexOf(this[i][field])<0) {
				processed.push(this[i][field]);
			} else {
				this.splice(i, 1);
			}
		}
	}
}

places.uniqueObjectArray('city');

With the above you can sort the array by any of the fields in the objects, even if they are not present for some of the objects.

or

function uniqueCity(array) {
    var processed = [];
    for (var i=array.length-1; i>=0; i--){
        if (processed.indexOf(array[i].city)<0) {
            processed.push(array[i].city);
        } else {
            array.splice(i, 1);
        }
    }
	return array;
}

places = uniqueCity(places);

Solution 7 - Javascript

You could use a Map so the entries with the same key property (in your case 'city') only appear once

module.exports = (array, prop) => {
   const keyValueArray = array.map(entry => [entry[prop], entry]);
   const map = new Map(keyValueArray);
   return Array.from(map.values());
};

More info about Map and array objects here

Basic example on Codepen

Solution 8 - Javascript

Another option:

const uniqueBy = prop => list => {
    const uniques = {}
    return list.reduce(
        (result, item) => {
            if (uniques[item[prop]]) return result
            uniques[item[prop]] = item
            return [...result, item]
        },
        [],
    )
}

const uniqueById = uniqueBy('id')

uniqueById([
    { id: 1, name: 'one' },
    { id: 2, name: 'two' },
    { id: 1, name: 'one' },
    { id: 3, name: 'three' }
])

You can paste it on your console to see it working. It should work for the scenario presented and a few others.

Solution 9 - Javascript

We can create the list of unique objects by any property using JavaScript Map.

For example :

var places = [{ 'lat': 12.123, 'lng': 13.213, 'city': "New York"},
                { 'lat': 3.123, 'lng': 2.213, 'city': "New York"},
                { 'lat': 43.123, 'lng': 12.213, 'city': "London"}];
                
var cityMap = new Map();
places.forEach(p=> cityMap.set(p.city, p));

console.log([...cityMap.values()]);

Execute code snippet to see the result.

Solution 10 - Javascript

As pointed out in the comments, you could use an object as a map, which will allow you to avoid duplicates, you can then enumerate the properties of the object.

working fiddle: http://jsfiddle.net/gPRPQ/1/

var places = [];
var a = {};
a.lat = 12.123;
a.lng = 13.213;
a.city = "New York";

places.push(a);

var b = {};
b.lat = 3.123;
b.lng = 2.213;
b.city = "New York";

places.push(b);

var unique = {}

for (var i = 0; i < places.length; i++) {
    var place = places[i];
    unique[place.city] = place;
}

for (var name in unique) {
    var place = unique[name];
    console.log(place);
}

Solution 11 - Javascript

var places = [];
var a = {};
a.lat = 12.123;
a.lng = 13.213;
a.city = "New York";

places.push(a);

var b = {};
b.lat = 3.123;
b.lng = 2.213;
b.city = "New York";

places.push(b);

getUniqAR(places,'city'); //Return Uniq Array by property

function getUniqAR(Data,filter){
var uniar =[];
Data.forEach(function(item,ind,arr){
    var dupi=false;
    if(!uniar.length) uniar.push(item) //push first obj into uniq array 
    uniar.forEach(function(item2, ind2,arr){
    if(item2[filter] == item[filter]){  //check each obj prop of uniq array 
      dupi=true; //if values are same put duplicate is true
        }     
    })
if(!dupi){  uniar.push(item)} //if no duplicate insert to uniq

})
console.log(uniar)
return uniar;
}

Solution 12 - Javascript

In simple Javascript code to remove duplicate cities from places array list is

var places = [{ 'lat': 12.123, 'lng': 13.213, 'city': "New York"},
				{ 'lat': 3.123, 'lng': 2.213, 'city': "New York"},
				{ 'lat': 43.123, 'lng': 12.213, 'city': "London"}];
var unique = [];
var tempArr = [];
places.forEach((value, index) => {
	if (unique.indexOf(value.city) === -1) {
		unique.push(value.city);
	} else {
		tempArr.push(index);    
	}
});
tempArr.reverse();
tempArr.forEach(ele => {
	places.splice(ele, 1);
});
console.log(places);

Solution 13 - Javascript

Generic Typescript answer based on https://stackoverflow.com/a/18773857/49564 above:

export function isDistinct<T>(mapper: (value: T) => string): (value: T) => boolean {
  const keys: { [index: string]: boolean } = {};

  return (entry: T) => {
    const key = mapper(entry);

    if (keys[key] !== undefined) {
      return false;
    }

    return keys[key] = true;
  };
}

// Usage example:
const items = [ { id: 1 }, { id: 2 }, { id: 3 }, { id: 1 } ];
const unique = items.filter(isDistinct(i => i.id));

Solution 14 - Javascript

I think you want this,

NOTE: No library is required.

let array = [{ id: 1}, {id: 2}, {id: 3}];

function addUniqeObj(data) {
  let index = -1;

  for(let i = 0, i < array.length; i++) {
    if(array[i].id === data.id) {
      index = i;
    }
  }

  if(index > -1) {
    array[index] = data;
  } else {
    array.push(data)
  }

}

Solution 15 - Javascript

Another variation of the rafaelbiten approach:

const dedupExample = [
    {id: 1, c: 'whatever'},
    {id: 1, c: '1whatever'},
    {id: 2, c: '2whatever'},
    {id: 2, c: '2whatever'},
    {id: 3, c: '2whatever'},
]

const getUniqueBy = (prop, list) => {
    const objUniq = list.reduce((res, item) => ({ ...res, [item[prop]]: item }), {})
    return Object.keys(objUniq).map(item => objUniq[item])
}

const uniq = getUniqueBy('id', dedupExample)

console.info('info', { uniq })
   /* [
    {id: 1, c: 'whatever'},
    {id: 2, c: '2whatever'},
    {id: 3, c: '2whatever'},
  ] */

Solution 16 - Javascript

const distinctArrayByCity= [
    ...new Map(array.map((item) => [item.city, item])).values(),
];

Solution 17 - Javascript

This thread may be old but thought I should share it. It is based on Pure JavaScript and removes Duplicate Objects based on the Properties Specified.

function removeDuplicates(originalArray, properties) {
  var newArray = [];
  var index = 0;
  var lookupObject = {};
  var totalProperties = properties.length;

  for (var i = 0; i < originalArray.length; i++) {
    var exists = false;

    for (var a = 0; a < newArray.length; a++) {
      var propsFound = 0;
      for (var b = 0; b < totalProperties; b++) {
        if (originalArray[i][properties[b]] == newArray[a][properties[b]]) {
          propsFound++;
        }
      }

      //If there is a match then break the for loop
      if (propsFound == totalProperties) {
        exists = true;
        break;
      }
    } //End of New Array

    if (!exists) {
      newArray[index] = originalArray[i];
      index++;
    }
  } //End of originalArray

  return newArray;
}

You can view the fiddle here

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestiontheblueoneView Question on Stackoverflow
Solution 1 - JavascriptT.J. CrowderView Answer on Stackoverflow
Solution 2 - JavascriptIgorLView Answer on Stackoverflow
Solution 3 - JavascriptAlec PerkeyView Answer on Stackoverflow
Solution 4 - JavascriptUnmitigatedView Answer on Stackoverflow
Solution 5 - JavascriptNSjonasView Answer on Stackoverflow
Solution 6 - JavascriptdavidkonradView Answer on Stackoverflow
Solution 7 - JavascriptTamo MaesView Answer on Stackoverflow
Solution 8 - JavascriptrafaelbitenView Answer on Stackoverflow
Solution 9 - JavascriptAnkit PrajapatiView Answer on Stackoverflow
Solution 10 - JavascriptRobert ByrneView Answer on Stackoverflow
Solution 11 - JavascriptPandi_SnklView Answer on Stackoverflow
Solution 12 - JavascriptShridhar SagariView Answer on Stackoverflow
Solution 13 - JavascriptMichaƂ TatarynowiczView Answer on Stackoverflow
Solution 14 - JavascriptRohit NishadView Answer on Stackoverflow
Solution 15 - JavascriptRomanView Answer on Stackoverflow
Solution 16 - JavascriptSafi HabhabView Answer on Stackoverflow
Solution 17 - JavascriptprogrammerboyView Answer on Stackoverflow