Get Difference Between Two Lists

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);
    }

    return diff;
}

console.log(arr_diff(['a', 'b'], ['a', 'b', 'c', 'd']));
console.log(arr_diff("abcd", "abcde"));
console.log(arr_diff("zxc", "zxc"));

Difference between two lists of lists in Python

Just use list comprehensions like so:

a = [[1,2,3],[4,5,6],[7,8,9]]
b = [[1,2,3],[9,9,9]]
c = [d for d in a if d not in b]
print(c)

Output:

[[4, 5, 6], [7, 8, 9]]

How to get difference between two lists in dart?

You can loop through one of the loop and then check if the item is present in the other list or not.

void main() {
  List x = ['one' , 'two' , 'three' , 'four'];
  List y = ['one' , 'two',];
  List output = [];

  for(final e in x){
    bool found = false;
    for(final f in y) {
      if(e == f) {
        found = true;
        break;
      }
    }
    if(!found){
      output.add(e);
    }
  }
  print(output);
}

How can I get the difference between two lists?

The other solution runs in O(N^2), where as this solution runs in O(N + M) time complexity.

x1 = [["x1", "y1"], ["x1", "x2"]]
x2 = [["x1", "y1"], ["x1", "x2"], ["x2", "y2"]]
set1 = {tuple(item) for item in x1}
print [item for item in x2 if tuple(item) not in set1]
# [['x2', 'y2']]

Just convert the first set of items to a list of tuples and then create a set with the tuples. Then, for every item in the next list of lists, convert that to a tuple and check if it is not there in the set.

Difference between two lists

Using Except is exactly the right way to go. If your type overrides Equals and GetHashCode, or you're only interested in reference type equality (i.e. two references are only "equal" if they refer to the exact same object), you can just use:

var list3 = list1.Except(list2).ToList();

If you need to express a custom idea of equality, e.g. by ID, you'll need to implement IEqualityComparer<T>. For example:

public class IdComparer : IEqualityComparer<CustomObject>
{
    public int GetHashCode(CustomObject co)
    {
        if (co == null)
        {
            return 0;
        }
        return co.Id.GetHashCode();
    }

    public bool Equals(CustomObject x1, CustomObject x2)
    {
        if (object.ReferenceEquals(x1, x2))
        {
            return true;
        }
        if (object.ReferenceEquals(x1, null) ||
            object.ReferenceEquals(x2, null))
        {
            return false;
        }
        return x1.Id == x2.Id;
    }
}

Then use:

var list3 = list1.Except(list2, new IdComparer()).ToList();

Note that this will remove any duplicate elements. If you need duplicates to be preserved, it would probably be easiest to create a set from list2 and use something like:

var list3 = list1.Where(x => !set2.Contains(x)).ToList();

How can I return the difference between two lists?

You can convert them to Set collections, and perform a set difference operation on them.

Like this:

Set<Date> ad = new HashSet<Date>(a);
Set<Date> bd = new HashSet<Date>(b);
ad.removeAll(bd);

Python - Differences between two lists

Trying out your code, it seemed to work fine with me regardless of the length of the lists - when I added 100 to list1, it showed up for both difference functions.

However, there appear to be a few issues with your code that could be causing the problems. Firstly, you accept arguments list1 and list2 for both functions, but these variables are the same name as your list variables. This seems not to cause an issue, but it means that the global variables are no longer accessible, and it is generally a better practice to avoid confusion by using different names for global variables and variables within functions.

Additionally, your function does not take the symmetric difference - it only loops over the variables in the first list, so unique variables in the second list will not be counted. To fix this easily, you could add a line combining your lists into a sum list, then looping over that entire list and checking if each value is in only one of the lists - this would use ^ to do an xor comparison of whether or not the variable is in the two lists, so that it returns true if it is in only one of the lists. This could be done like so:

def difference_extra_credit(l1,l2):
    list  = l1 + l2
    return [value for value in list if (value in l1) ^ (value in l2)]

Testing this function on my own has resulted in the list [4, 11, 7, 1], and [4, 11, 100, 7, 1] if 100 is added to list1 or list2.

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