How to find difference between two same array in JS?

Question

I want to find difference between two arrays. I tried below code from one of the solution given on SF community, but not working from me.

My below code is not working if there is number repeated EVEN NUMBER of time(like "7" repeated twice in both the arrays a & b and in result I am getting "7" as difference),

whereas code is working for number repeated ODD NUMBER of time(like "11" repeated thrice in both the arrays a & b and in result it cancel all the "11" and return nothing for this).

I want the solution to find difference or cancels the number repeated ANY NUMBER of times, I mean if number like "7" repeated twice in both the arrays then it should cancels both the number and return NOTHING.

And given two arrays are, you can find more examples like "7" and "11" in belows given arrays

a = ["1", "3", "17", "12", "11", "17", "7", "7", "85", "16", "5", "30", "11", "81", "20", "49", "8", "17", "11", "8", "13", "81"]

b = ["5", "20", "78", "13", "85", "81", "12", "17", "16", "8", "17", "3", "7", "1", "11", "9", "8", "17", "49", "30", "81", "11", "11", "7"]

Below is the code I am using to find difference between two arays,

function arr_diff (a1, a2) {

var a = [], diff = [];

for (var i = 0; i < a1.length; i++) {
    a[a1[i]] = true;
}

for (var i = 0; i < a2.length; i++) {
    if (a[a2[i]]) {
        delete a[a2[i]];
    } else {
        a[a2[i]] = true;
    }
}

for (var k in a) {
    diff.push(k);
}

console.log(diff);
return diff;
}

DIFFERENCE/RESULT : I get from two arrays is : ["7", "8", "9", "78", "81", "getUnique"]

EXPECTED RESULT : from two arrays I want : ["9", "78"]

Please help, any help will be appreciated.

what should be the result for a=[1,1,1] b=[1]? should it be [1,1] or [1]? — Ziarno, Jul 27 '20 at 11:30

score 1 · Answer 1 · answered Jul 27 '20 at 10:37

In your code what is happening,

First time it checks for 7, then it removes from a. Now when next time it checks for same, since it is removed from a, it sets true in a. Same for other consecutive numbers.

You should do something like this

var a1 = ["1", "3", "17", "12", "11", "17", "7", "7", "85", "16", "5", "30", "11", "81", "20", "49", "8", "17", "11", "8", "13", "81"]

var a2  = ["5", "20", "78", "13", "85", "81", "12", "17", "16", "8", "17", "3", "7", "1", "11", "9", "8", "17", "49", "30", "81", "11", "11", "7"]

function arr_diff (a1, a2) {

var a = {}, diff = [], newArr = [];

for (var i = 0; i < a1.length; i++) {
    a[a1[i]] = true;
}

for (var i = 0; i < a2.length; i++) {
    if (a[a2[i]]) {
//         delete a[a2[i]];
    } else {
        newArr[a2[i]] = true;
    }
}

for (var k in newArr) {
    diff.push(k);
}

console.log(diff);
return diff;
}

arr_diff(a1, a2)

What I did is, I am creating a new Array (newArr) which will store only unique numbers

Will it work if a = ["1", "1"] & b = ["1"] the it will give result or difference as ["1"]...? can you please explain this. Thanks for your response. — OTP Generator, Jul 27 '20 at 11:01
@OTPGenerator No it won't. Check my answer if you want a solution for that problem aswell. — MauriceNino, Jul 27 '20 at 11:06

score 0 · Answer 2 · answered Jul 27 '20 at 10:33

0

a = ["1", "3", "17", "12", "11", "17", "7", "7", "85", "16", "5", "30", "11", "81", "20", "49", "8", "17", "11", "8", "13", "81"]
b = ["5", "20", "78", "13", "85", "81", "12", "17", "16", "8", "17", "3", "7", "1", "11", "9", "8", "17", "49", "30", "81", "11", "11", "7"]
let difference = a
                 .filter(x => !b.includes(x))
                 .concat(b.filter(x => !a.includes(x)));

console.log(difference);

This will fix your problem

const difference = a
                 .filter(x => !b.includes(x))
                 .concat(b.filter(x => !a.includes(x)));

answered Jul 27 '20 at 10:33

Qubaish Bhatti

700
5
22

can you please blend you given code in my code please. I am still learning JS. Thank you! – OTP Generator Jul 27 '20 at 10:35
This doesn't solve the problem where numbers can appear multiple times or am I wrong? Like if 1 appeared 2 times in a, but 3 times in b. – MauriceNino Jul 27 '20 at 10:37
@MauriceNino, It will also work in this scenario too. I just checked it on my console. – Qubaish Bhatti Jul 27 '20 at 10:43
I just checked it myself. No it won't work, unless I understood the question wrong. Try `const a = ['1','1']; const b = ['1']`. Won't return a difference – MauriceNino Jul 27 '20 at 10:45
yes it will not return a difference because we are not considering common values. For reference check this answer: https://stackoverflow.com/a/33034768/2824786 – Qubaish Bhatti Jul 27 '20 at 10:47
@MauriceNino, also please understand the expected result of the question. Hope it will clarify your concerns :) – Qubaish Bhatti Jul 27 '20 at 10:59
I don't know what you want to tell me with that link. It doesn't answer anything at all - completely pointless. Also, the expected result of the question doesn't include different number of occurrences, so what is your conclusion from that? I was just stating something to keep in mind when using this answer for a problem - stackoverflow is essentially a programming knowledge library and not a question/answer site. – MauriceNino Jul 27 '20 at 11:04

score 0 · Answer 3 · answered Jul 27 '20 at 10:58

Try this, let me know if it works for you

let a = ["1", "3", "17", "12", "11", "17", "7", "7", "85", "16", "5", "30", "11", "81", "20", "49", "8", "17", "11", "8", "13", "81"]

let b = ["5", "20", "78", "13", "85", "81", "12", "17", "16", "8", "17", "3", "7", "1", "11", "9", "8", "17", "49", "30", "81", "11", "11", "7"];

function arr_diff(a1,a2){
    let a = [...a1];
    let b = [...a2];

    let c = [];
    for(let i=0; i < b.length; i++){
        if(a.includes(b[i])) {b.splice(i, 1);}
        else c.push(b[i]);
    }

console.log(c);
return c;
}

arr_diff(a,b)

score 0 · Accepted Answer · answered Jul 27 '20 at 11:20

If you want to keep the extra values in the difference array too, ex. for

a = [1,1,1,1]
b = [1]

the result: [1,1,1], I'd do this: count occurrences in first array and subtract the count from the second array.

function getDifference(array1, array2) {
  const counter = {}
  const result = []

  function addNumberToCounter(number) {
    const currentCount = counter[number] || 0
    counter[number] = currentCount + 1
  }
  function subtractNumberFromCounter(number) {
    const currentCount = counter[number] || 0
    counter[number] = currentCount - 1
  }
  function addNumberToResult(number, times) {
    for(let i = 0; i < times; i++) {
      result.push(number)
    }
  }
  
  array1.forEach(addNumberToCounter)
  array2.forEach(subtractNumberFromCounter)
  
  for (let number in counter) {
    const numberCount = Math.abs(counter[number])
    addNumberToResult(number, numberCount)
  }
  
  return result
}

const a = ["1", "3", "17", "12", "11", "17", "7", "7", "85", "16", "5", "30", "11", "81", "20", "49", "8", "17", "11", "8", "13", "81"]
const b = ["5", "20", "78", "13", "85", "81", "12", "17", "16", "8", "17", "3", "7", "1", "11", "9", "8", "17", "49", "30", "81", "11", "11", "7"]
const result = getDifference(a, b)
console.log(result)

Sven.hig · Answer 5 · 2020-07-27T10:53:52.410

-1

For your code to work you could use Set to deduplicate values first

a = ["1", "3", "17", "12", "11", "17", "7", "7", "85", "16", "5", "30", "11", "81", "20", "49", "8", "17", "11", "8", "13", "81"]

b = ["5", "20", "78", "13", "85", "81", "12", "17", "16", "8", "17", "3", "7", "1", "11", "9", "8", "17", "49", "30", "81", "11", "11", "7"]

 var set1=new Set([...a]) // remove duplicates from each array
 var set2=new Set([...b])
 a1=[...set1]
 a2=[...set2]

function arr_diff (a1, a2) {

  var a = [], diff = [];
  
  for (var i = 0; i < a1.length; i++) {
      a[a1[i]] = true;
  }
  
  for (var i = 0; i < a2.length; i++) {
      if (a[a2[i]]) {
          delete a[a2[i]];
      } else {
          a[a2[i]] = true;
      }
  }
  
  for (var k in a) {
      diff.push(k);
  }
  
  console.log(diff);
  return diff;
  }

  arr_diff (a1, a2)

Alternatively you could use filter and some

a = ["1", "3", "17", "12", "11", "17", "7", "7", "85", "16", "5", "30", "11", "81", "20", "49", "8", "17", "11", "8", "13", "81"]

b = ["5", "20", "78", "13", "85", "81", "12", "17", "16", "8", "17", "3", "7", "1", "11", "9", "8", "17", "49", "30", "81", "11", "11", "7"]

res=b.filter(n=>!a.some(u=>n==u))
console.log(res)

edited Jul 27 '20 at 10:53

answered Jul 27 '20 at 10:47

Sven.hig

4,449
2
8
18

I believe the duplicates in arrays are the whole point of this question, so you can't use Sets – Ziarno Jul 27 '20 at 11:23
I did again now, well tbh it doesn't say what should the result be for: a=[1,1,1]; b=[1] :P – Ziarno Jul 27 '20 at 11:29
so what do you think the result should be then ? – Sven.hig Jul 27 '20 at 11:31
I think it should be [1,1]. Asked the OP in a comment – Ziarno Jul 27 '20 at 11:33
there is another reason you can't user Sets though - for a=[1,1] and b=[1] we want to get result [1], and when you remove duplicates the result is [] – Ziarno Jul 27 '20 at 11:42
the question was about getting the **difference** between 2 arrays does a=[1,1] and b=[1] has any different elements??? from the question he said his solution doesn't work because arrays with even duplicates make his code produce wrong results so he want to **cancel** them introducing set to his code gives the expected results he mentioned, now you may have understood what he needs but his question was misleading in many levels hence why pretty much all the answers here point to the same direction as my answer, and so far he has jumped from one answer to the other each time accepting one – Sven.hig Jul 27 '20 at 11:49

How to find difference between two same array in JS?

5 Answers5