Remove Duplicates From Arrays Using Reduce

I am trying to remove duplicates from a list of arrays. The way I was trying to do this is by using reduce to create an empty array that pushes all undefined indexes onto that arr

Solution 1:

First concatenate the two arrays, next use filter() to filter out only the unique items-

var a = [1, 2, 2, 4], b = [1, 1, 4, 5, 6];
var c = a.concat(b);
var d = c.filter(function (item, pos) {return c.indexOf(item) == pos});
console.log(d);

Solution 2:

There are a number of issues with how you're calling methods and where you return acc from:

functionnoDuplicates(arrays) {
  var arrayed = Array.prototype.slice.call(arguments);

  // reduce is a method of an array, so call it as a method// return reduce(arrayed, function(acc, cur) {return arrayed.reduce(function(acc, cur) {
  
    // Same with forEach
    cur.forEach(function(item) {
      if (acc[item] === undefined) {
        acc.push(item);
      }
       // Return acc from the reduce callback, forEach returns undefined always// return acc;
    });
    return acc;
  }, []);
}

console.log(noDuplicates([1, 2, 2, 4], [1, 1, 4, 5, 6]));

You could also call reduce directly on arguments using call:

Array.prototype.reduce.call(arguments, function(acc, curr) {
  // ...
});

The above makes your code run, but it doesn't produce the correct output as the test:

  if (acc[item] === undefined)

doesn't do what you want. What you need to do is remember each value and only push it to acc if it's not been seen before:

functionnoDuplicates(arrays) {
  var arrayed = Array.prototype.slice.call(arguments);
  var seen = {};

  return arrayed.reduce(function(acc, cur) {
    cur.forEach(function(item) {
      if (!seen[item]) {
        acc.push(item);
        seen[item] = true;
      }
    });
    return acc;
  }, []);
}

console.log(noDuplicates([1, 2, 2, 4], [1, 1, 4, 5, 6]));

Some other approaches:

// A more concise version of the OPfunctionnoDupes() {
  return [].reduce.call(arguments, function(acc, arr) {
    arr.forEach(function(value) {
      if (acc.indexOf(value) == -1) acc.push(value);
    });
    return acc;
   },[]);
}
 
console.log(noDupes([1, 2, 2, 4], [1, 1, 4, 5, 6]));

// Some ECMAScript 2017 goodnessfunctionnoDupes2(...args){
 return [].concat(...args).filter((v, i, arr) => arr.indexOf(v)==i);
}

console.log(noDupes2([1, 2, 2, 4], [1, 1, 4, 5, 6]));

Solution 3:

Any reason of using reduce? because we can do this easily by first merging these two arrays then by using Set to remove the duplicate keys.

Check this:

functionnoDuplicates(a, b){
    var k = a.concat(b);
    return [...newSet(k)];
}

console.log(noDuplicates([1,2,2,4],[1,1,4,5,6]));

Check the DOC, how Set works.

Solution 4:

My Solution is -

var numbers = [1, 1, 2, 3, 4, 4];

functionunique(array){
  returnarray.reduce(function(previous, current) {
     if(!previous.find(function(prevItem){
         return prevItem === current;
     })) {
        previous.push(current);
     }
     return previous;
 }, []);
}

unique(numbers);

Solution 5:

looking for a smoother solution for MDN's exact same problem, I've came up with that solution, I find it simple and nice. I have also just updated it in MDN and wanted to share it here (I'm really new to that stuff, so sorry if did something wrong)

let myArray = ['a', 'b', 'a', 'b', 'c', 'e', 'e', 'c', 'd', 'd', 'd', 'd'];
var myOrderedArray = myArray.reduce(function (accumulator, currentValue) {
  if (accumulator.indexOf(currentValue) === -1) {
    accumulator.push(currentValue);
  }
  return accumulator
}, [])

console.log(myOrderedArray);

(I'm really new to this, hope it'd help)