JS Boolean Logic Operation With Object

Question

...but with objects.

for example :

one = {a: false, c: false, e: false};
two = {a: true, b: false, c: false, d:false};

result = somethingJavascriptHas.ObjectAnd(one, two);

console.log(result);

would return :

{a: false, c: false};

(I'm interested in the key, not the rest, so I discard the second object's value in favor of the first's here. In my case the two object's values will always be the same simply some of their entries will be missing or not and I want a final object that includes only key's (with the first object key's value) that are present in both objects)

is this possible?

For completeness' sake OR, XOR and NOT. (again assume priority of values for the first passed object)

Yes it's possible. You'd just have to write some code to do it. — Pointy
– Pointy, Commented Mar 2, 2018 at 14:59
not somtheing Object.something or lodash would already have then? — tatsu
– tatsu, Commented Mar 2, 2018 at 15:02
@tatsu Asking for library suggestions is one of the reasons for closing questions at Stack Overflow — Ruan Mendes
– Ruan Mendes, Commented Mar 2, 2018 at 15:23

Zenoo · Accepted Answer · 2018-03-02 15:04:09Z

2

You could use the Array#reduce function on one of your objects keys and check if the properties of both objects are not undefined :

var one = {a: false, c: false, e: false};
var two = {a: true, b: false, c: false, d:false};

var three = Object.keys(one).reduce((acc, curr) => {
  if(two[curr] !== undefined){
    acc[curr] = one[curr] && two[curr];
  }
  return acc;
}, {});

console.log(three);

answered Mar 2, 2018 at 15:04

Zenoo

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Comments

messerbill · Accepted Answer · 2018-03-02 15:03:19Z

2

just loop through your objects:

var one = {a: false, c: false, e: false};
var two = {a: true, b: false, c: false, d:false};

var res = {}

for (var o in one) {
   for (var t in two) {
     if (t == o) res[t] = one[t]
   }
}

console.log(res)

answered Mar 2, 2018 at 15:03

messerbill

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Comments

th3n3wguy · Accepted Answer · 2018-03-02 15:06:39Z

0

// Your example:
/*one = {a: false, c: false, e: false};
two = {a: true, b: false, c: false, d:false};

result = somethingJavascriptHas.ObjectAnd(a, b);

console.log(result);
// would return :

{a: false, c: false};*/

// Working code:
const one = {a: false, c: false, e: false};
const two = {a: true, b: false, c: false, d:false};

const oneTwoIntersection = Object.keys(one)
  .reduce((currentObj, nextKey) => {
    return !two.hasOwnProperty(nextKey) ? currentObj : Object.assign({}, currentObj, { 
      [nextKey]: one[nextKey] 
    });
  }, {});
  
console.log(oneTwoIntersection);

answered Mar 2, 2018 at 15:06

th3n3wguy

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Ruan Mendes · Accepted Answer · 2018-03-02 15:21:27Z

0

Since you said you wanted just keys, you can combine Object.keys, Array#filter and Object#hasOwnProperty

const one = {a: false, c: false, e: false};
const two = {a: true, b: false, c: false, d:false};

const intersection = (a, b) => 
    Object.keys(a).filter(aKey => b.hasOwnProperty(aKey));


console.log(intersection(one, two));

edited Mar 2, 2018 at 15:21

answered Mar 2, 2018 at 15:15

Ruan Mendes

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1 Comment

tatsu Over a year ago

okay I'll admit my wording was misleading but to me "not being interested in" doesn't by definition promote itself to "time to delete". I should have said "I'm interested in the key, not the rest, for this part" or "for now". the distinction being that the second array's object's were not being taken into consideration by the operation. interesting possibility though.

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