Compare n objects in array

Question

I want to compare an unlimited amount of objects in an array, like

{ a: true, b: false, c: "foo"}

and

{ a: false, b: false, c: "foo"}

and get a result like

[b,c]

This can be done with just two objects like this:

function compare(first, second) {
    var shared = []
    $.each(first, function(index, value){
        if(second[index] == value){
            shared.push(index)
        }
    })
    return shared
}

So my question is: How to write a (recursive) function that compares n objects and returns the properties they all have in common.

Thank you for reading this

So i could compare the first to the second one and then compare the result to the third and so on? — lukas293
– lukas293, Commented Sep 24, 2015 at 17:52
Right - From the first object you'd pick up all the properties/values, and then for each subsequent object you'd throw out the ones that don't match. At the end, you'd just collect up the property names as a final step. — Pointy
– Pointy, Commented Sep 24, 2015 at 17:54
You actually need .filter and .every Object.keys(first).filter(key => objects.every(obj => obj[key] === first[key]) ); — Yury Tarabanko
– Yury Tarabanko, Commented Sep 24, 2015 at 18:04

Bergi · Accepted Answer · 2015-09-24 18:10:43Z

2

That's pretty simple if written literally:

function allCommonProperties(objects) {
    if (!objects.length) return [];
    var first = objects[0];
    return Object.keys(first).filter(function(p) {
        return objects.every(function(o) {
             return first[p] === o[p];
        });
    });
});

No need for recursion, no need for jQuery.

answered Sep 24, 2015 at 18:10

Bergi

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Comments

Yury Tarabanko · Accepted Answer · 2015-09-24 18:11:29Z

0

An alternative approach using .filter and .every. No libs.

Object.keys(first)
      .filter(function(key){
          return objects.every(function(obj){
             return obj[key] === first[key];
          });
      });

For ES6 envs

Object.keys(first).filter(key => objects.every(obj => obj[key] === first[key]));

Where

objects - an array of all objects to compare
first = objects.shift() - the first element

answered Sep 24, 2015 at 18:11

Yury Tarabanko

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Comments

Nina Scholz · Accepted Answer · 2015-09-24 18:11:36Z

0

This solution features Array.prototype.filter in combination with Object.keys.

var setA = { a: true, b: false, c: "foo" },
    setB = { a: false, b: false, c: "foo" },
    commonProperties = Object.keys(setA).filter(function (k) {
        return setA[k] === setB[k];
    });

document.write('<pre>' + JSON.stringify(commonProperties, 0, 4) + '</pre>');

answered Sep 24, 2015 at 18:11

Nina Scholz

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Comments

Pointy · Accepted Answer · 2015-09-24 18:12:50Z

I'd do it with .reduce(), going through the array and accumulating an object containing only properties that have been found (with matching values) in all the objects so far.

I'll give my example with an "extend" function; if you're working with jQuery you could use $.extend() or you can write a simple one. The idea is to do a shallow copy of properties from one object into another, which'll be necessary for the first object in the list.

var objects = [ ... ]; // the array of objects
var resultObject = objects.reduce(function(result, obj) {
  if (result === null)
    result = extend({}, obj);
  else {
    Object.keys(result).forEach(function(key) {
      if (!(key in obj) || obj[key] !== result[key])
        delete result[key];
    });
  }
  return result;
}, null);
var resultProperties = Object.keys(resultObject);

Here's what extend() might look like:

function extend(obj) {
  var args = arguments;
  for (var i = 1; i < args.length; ++i)
    Object.keys(args[i]).forEach(function(key) {
      obj[key] = args[i][key];
    });
  return obj;
}

WriteYour NameHere · Accepted Answer · 2015-09-24 18:16:10Z

0

I think that work:

function compareObj(obj1,obj2){
    var ar =[];//you need an  array to put the diffirent property
    for(key in obj1){//getting all proo
        try{
            if (obj1[key]!=obj2[key]) //if there is diffirence 
                {
                    ar[ar.length-1]=key;
                    //key is the name of property 
                }
        }
        catch(err){
            console.log("Not the same type of objects");
            return false;
        }


    }
    return ar ;
}

answered Sep 24, 2015 at 18:16

WriteYour NameHere

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Comments

MinusFour · Accepted Answer · 2015-09-24 18:33:09Z

This tries to iterate as least as possible. It mutates an array that holds the keys of the initial object every time a property is no longer found on the next object and will also stop iterating the rest of the array elements if there's no common properties left.

function filterByPropertyValue(arr) {
  var acc = {},
    initial, objectProperties;
  if (arr.length > 0) {
    initial = arr[1];
    objectProperties = Object.keys(initial);
    for (var i = 1; i < arr.length; i++) {
      if (objectProperties.length > 0) {
        objectProperties.forEach(function(prop, index) {
          if (initial[prop] === arr[i][prop]) {
            acc[prop] = initial[prop];
          } else {
            delete acc[prop];
            delete objectProperties[index];
          }
        });
      } else {
        break;
      }
    }
  }
  return acc;
}

var arr = [{ a: true, b: false, c: "foo"}, { a: false, b: false, c: "foo"}, { a: true, b: false, c: "bar"}];
document.getElementById('results').innerHTML = JSON.stringify(filterByPropertyValue(arr), null, '\t');

<pre id="results"></pre>

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