CODEWITHSUNDEEP
javascriptduplicates
user1094081
user1094081

Reputation:

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

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

Upvotes: 4

Views: 455

Answers (4)

Batman25663
Batman25663

Reputation: 272

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);

Upvotes: 1

neuhaus
neuhaus

Reputation: 4094

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.

Upvotes: 1

T.J. Crowder
T.J. Crowder

Reputation: 1074295

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.

Upvotes: 1

Rajesh
Rajesh

Reputation: 24915

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)

Upvotes: 1

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