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Magima Felix O
Magima Felix O

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Help with includes() in js

#filter #includes 
 ANY HELP WITH JAVASCRIPT includes() function
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const formA = [
    {id:37,name:'New',another:'things'},
    {id:31,name:'Old',another:'things'},           
    {id:3,name:'Oldest',another:'things'}
 ];

 const formB = [
    {id:37,name:'New',another:'things'},
   {id:31,name:'Old',another:'things'},          
   ];

   const difference = formA.filter(x => !formB.includes(x.name))
   console.log(difference);
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would someone kindly explain to me how includes() method works?
i was expecting____ [{id:3,name:'Oldest',another:'things'}] as output

but instead getting.

//__________output
[
{ id: 37, name: 'New', another: 'things' },
{ id: 31, name: 'Old', another: 'things' },
{ id: 3, name: 'Oldest', another: 'things' }
]

Top comments (8)

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vonheikemen profile image
Heiker • • Edited

Let's just say that Array.includes is making a "shallow comparison". What is happening is that you're comparing an object against a string. Imagine something like this.

'Oldest' === ({ id:3, name:'Oldest', another:'things' })
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If you try that it would return false.

Array.includes doesn't try to find the string inside the object it, just compares the whole item in the array against your argument.

If I were to try implement Array.includes as a function it would be like this.

function includes(array, search) {
   for(const item of array) {
      if(item === search) {
         return true;
      }
   }

   return false;
}
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magimart profile image
Magima Felix O •

i have to look into this approach, thank you very much for the response

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vonheikemen profile image
Heiker •

Actually I was just trying to explain how Array.includes works.

What would solve your problem is to transform formB into an array of names before using formA.filter. Like this.

const formBNames = formB.map((x) => x.name);

const difference = formA.filter((x) => !formBNames.includes(x.name));
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magimart profile image
Magima Felix O •

thank your very much for your kind efforts, i tried it like you stated already but got got a desirable result. i used a set method instead and worked well.
Another contribute here also commented with reducer() that also works perfect :)

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peerreynders profile image
peerreynders •

Array.prototype.includes()

The includes() method determines whether an array includes a certain value

Given that your arrays itemize objects, those arrays contain references to objects, not the actual objects.
So includes() can't help here because it can only find one-and-the-same object (because its reference is the same) but it cannot find objects that look the same.

const object = { id: 37, name: 'New', another: 'things' };
console.log(object === object); // true
console.log(object === { id: 37, name: 'New', another: 'things' }); // false
console.log(
  { id: 37, name: 'New', another: 'things' } ===
    { id: 37, name: 'New', another: 'things' }
); // false
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const object = { id: 37, name: 'New', another: 'things' };
const items = [{ id: 37, name: 'New', another: 'things' }];

console.log(items.includes(items[0])); // true
console.log(items.includes(object));  // false
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I believe what you are looking for is Array.prototype.some().

const formA = [
  { id: 37, name: 'New', another: 'things' },
  { id: 31, name: 'Old', another: 'things' },
  { id: 3, name: 'Oldest', another: 'things' },
];

const formB = [
  { id: 37, name: 'New', another: 'things' },
  { id: 31, name: 'Old', another: 'things' },
];

function containsNameOf(list, other) {
  return list.some((item) => item.name === other.name);
}

console.log(containsNameOf(formB, formA[0])); // true
console.log(containsNameOf(formB, formA[1])); // true
console.log(containsNameOf(formB, formA[2])); // false
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magimart profile image
Magima Felix O •

because i was looking at this kind of approach with includes() to get the Difference between two arrays
const arrA = ['john','matha','lisa','adam']; const arrB = ['tom','andrew','lisa','adam'];

                    const myDiff = arrA.filter(name => !arrB.includes(name))
                    console.log(myDiff)
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but this seems like it does not work with objects

Otherwise thank very much for your take, helps a lot lot...going to look into the approaches

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peerreynders profile image
peerreynders • • Edited 
const arrA = ['john','matha','lisa','adam']; 
const arrB = ['tom','andrew','lisa','adam'];

console.log(arrA.filter(name => !arrB.includes(name))); // ["john", "matha"]
console.log(arrA.filter(name => !arrB.some(other => other === name))); // ["john", "matha"]
console.log(arrA.filter(name => arrB.every(other => other !== name))); // ["john", "matha"]
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JavaScript data types and data structures

Strings are a primitive type.
Objects are a structural type.

So with objects you simply have to "dig deeper" until you can compare the primitive types (referenced by the object) that matter to you.

const formA = [
  { id: 37, name: 'New', another: 'things' },
  { id: 31, name: 'Old', another: 'things' },
  { id: 3, name: 'Oldest', another: 'things' },
];

const formB = [
  { id: 37, name: 'New', another: 'things' },
  { id: 31, name: 'Old', another: 'things' },
];

console.log(
  formA.filter((itemA) => formB.every((itemB) => itemB.name !== itemA.name))
); // [ {id: 3, name: "Oldest", another: "things" } ]

function differenceByName(itemsA, itemsB) {
  const keepNotInB = (itemA) => {
    const isNameDifferent = (itemB) => itemB.name !== itemA.name;
    return itemsB.every(isNameDifferent);
  };
  return itemsA.filter(keepNotInB);
}

console.log(differenceByName(formA, formB)); // [ {id: 3, name: "Oldest", another: "things" } ]
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Array.prototype.every()

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magimart profile image
Magima Felix O •

thank you peerreynders, i like this approach, quiet understandable too, thanks alot for the efforts i am more than ever energised :)

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