Find duplicates of some keys in array of JavaScript objects and set a related key to true

Question

I have an array of JS Objects:

  var a = [{a:"a1",b:"b",c:"c",prop:false},
           {a:"a1",b:"b",c:"c",prop:false},
           {a:"a",b:"b",c:"c",prop:false},
           {a:"a2",b:"b",c:"c",prop:false},
           {a:"a3",b:"b",c:"c",prop:false},
           {a:"a1",b:"b",c:"c",prop:false}];

I must find duplicates of the last element but WITHOUT considering the key prop: when I find them, the related prop boolean must be set to true.

I did the simplest thing in the world and it works fine but I was wandering if there is something JS can do (without 3rd party libraries), better either in "elegance", modernity or performance:

var toTest = a[a.length-1];
for(var i=0;i<a.length;i++){
      if(toTest.a==a[i].a && toTest.b==a[i].b && toTest.c==a[i].c){
         a[i].prop = true;
         a[a.length-1].prop = true;
      }
    }

Any tip is appreciated

@meowgoesthedog what's a dictionary in JS? Do you mean an object, or a Map? — evolutionxbox
– evolutionxbox, Commented Aug 29, 2017 at 9:50
You could create an aggregate key for a b and c and put them into a hash, then you just need to check if there is already an entry in the hash when going through the array. — neuhaus
– neuhaus, Commented Aug 29, 2017 at 9:50

Rajesh Dixit · Accepted Answer · 2017-08-29 10:00:18Z

1

You can use a combination of Array.forEach and Array.every. Create a list of keys to test and in every iteration, check for these keys and update object.

a[a.length-1].prop = true looks unnecessary to me as you are not removing the item, so it will be updated in the last iteration.

var a = [{a:"a1",b:"b",c:"c",prop:false},
         {a:"a1",b:"b",c:"c",prop:false},
         {a:"a",b:"b",c:"c",prop:false},
         {a:"a2",b:"b",c:"c",prop:false},
         {a:"a3",b:"b",c:"c",prop:false},
         {a:"a1",b:"b",c:"c",prop:false}];
var toTest = a[a.length -1];
var keysToTest = ['a', 'b', 'c'];

a.forEach((o)=>{
  var valid = keysToTest.every((k) => o[k] === toTest[k]);
  if(valid){
    o.prop = true;
  }
});

console.log(a)

answered Aug 29, 2017 at 10:00

Rajesh Dixit

25k5 gold badges52 silver badges86 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

user1094081 Over a year ago

Wow, very compact

T.J. Crowder · Accepted Answer · 2017-08-29 10:02:53Z

1

What you have is pretty much it. There are tweaks you can make, but it's fine as is except you want to stop one step earlier, when i < a.length - 1 (since you don't want to compare the last entry to itself and set prop incorrectly).

Some tweaks you could do:

a[a.length-1].prop = true; could be toTest.prop = true;
You can cache a[i] to a variable rather than looking it up each time
You can cache a, b, and c instead of looking them up on the object every time (but the JavaScript engine will do that if it matters anyway)
In ES2015+ you could use destructuring assignment to get a, b, and c

So for instance:

// Caching the last item
var last = a[a.length - 1];
// Caching the values to check
var checka = last.a, checkb = last.b, checkc = last.c;
// Stop at i < a.length - 1, caching that to l
for (var i = 0, l = a.length - 1; i < l; ++i) {
    // Cache a[i]
    var entry = a[i];
    if (entry.a == checka && entry.b == checkb && entry.c == checkc) {
        entry.prop = last.prop = true;
    }
}

Live Copy:

var a = [{a:"a1",b:"b",c:"c",prop:false},
           {a:"a1",b:"b",c:"c",prop:false},
           {a:"a",b:"b",c:"c",prop:false},
           {a:"a2",b:"b",c:"c",prop:false},
           {a:"a3",b:"b",c:"c",prop:false},
           {a:"a1",b:"b",c:"c",prop:false}];

// Caching the last item
var last = a[a.length - 1];
// Caching the values to check
var checka = last.a, checkb = last.b, checkc = last.c;
// Stop at i < a.length - 1, caching that to l
for (var i = 0, l = a.length - 1; i < l; ++i) {
    // Cache a[i]
    var entry = a[i];
    if (entry.a == checka && entry.b == checkb && entry.c == checkc) {
        entry.prop = last.prop = true;
    }
}
console.log(a);

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

Here's an ES2015+ version, just for reference:

const last = a[a.length - 1];
const {a: checka, b: checkb, c: checkc} = last;
for (let i = 0, l = a.length - 1; i < l; ++i) {
    const {a: thisa, b: thisb, c: thisc} = a[i];
    if (thisa == checka && thisb == checkb && thisc == checkc) {
        a[i].prop = last.prop = true;
    }
}

Live Copy:

const a = [{a:"a1",b:"b",c:"c",prop:false},
           {a:"a1",b:"b",c:"c",prop:false},
           {a:"a",b:"b",c:"c",prop:false},
           {a:"a2",b:"b",c:"c",prop:false},
           {a:"a3",b:"b",c:"c",prop:false},
           {a:"a1",b:"b",c:"c",prop:false}];

const last = a[a.length - 1];
const {a: checka, b: checkb, c: checkc} = last;
for (let i = 0, l = a.length - 1; i < l; ++i) {
    const {a: thisa, b: thisb, c: thisc} = a[i];
    if (thisa == checka && thisb == checkb && thisc == checkc) {
        a[i].prop = last.prop = true;
    }
}
console.log(a);

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

Whether that meets any of your criteria is up to you.

edited Aug 29, 2017 at 10:02

answered Aug 29, 2017 at 9:57

T.J. Crowder

1.1m201 gold badges2k silver badges2k bronze badges

3 Comments

Logar Over a year ago

I won't make an answer for that, but if OP "must find duplicates of the last element but WITHOUT considering the key prop", wouldn't it be better if we loop on object properties instead of manually listing them ?

T.J. Crowder Over a year ago

@3000: I've updated to include an ES5 and earlier version

T.J. Crowder Over a year ago

@Logar: If the properties were variable, sure, we'd get the list in advance. From the OP's code I had the impression they were static.

neuhaus · Accepted Answer · 2017-08-29 10:07:03Z

1

Use Array.filter() with a function that compares only a, b and c.

Edit: I completely replaced my previous answer. I think Array.filter() is the way to go to search for duplicates of the last entry in the array as asked for.

edited Aug 29, 2017 at 10:07

answered Aug 29, 2017 at 9:55

neuhaus

4,1741 gold badge13 silver badges29 bronze badges

2 Comments

T.J. Crowder Over a year ago

I don't see how generating a lot of temporary strings and doing a lot of unnecessary string concatenation is an improvement over using &&.

neuhaus Over a year ago

OK I somewhat misread the question, I thought he had to find all duplicates in the array. If you just want to check one value it's better to iterate through it in the simple fashion you suggested.

Batman25663 · Accepted Answer · 2017-08-29 10:13:21Z

1

You can use the normal forEach Loop

var a = [{a:"a1",b:"b",c:"c",prop:false}, {a:"a1",b:"b",c:"c",prop:false}, {a:"a",b:"b",c:"c",prop:false}, {a:"a2",b:"b",c:"c",prop:false},{a:"a1",b:"b",c:"c",prop:false},{a:"a1",b:"b",c:"c",prop:false}];     
var toTest = a[a.length-1];              
a.forEach(function(item){
  item.prop = (item.a == toTest.a && item.b == toTest.b && item.c == toTest.c) ? true : false;
});
console.log(a);

edited Aug 29, 2017 at 10:13

answered Aug 29, 2017 at 10:07

Batman25663

2721 gold badge3 silver badges12 bronze badges

Collectives™ on Stack Overflow

Find duplicates of some keys in array of JavaScript objects and set a related key to true

4 Answers 4

1 Comment

3 Comments

2 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

1 Comment

3 Comments

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related