What is the easiest way to get a diff of two `List<Int>`'s? `minus()` won't do it, since `li...

# stdlib

[Orhan Tozan]

What is the easiest way to get a diff of two `List<Int>`'s?

minus() won't do it, since list2 - list1 only shows what list2 has what list1 hasn't.

[Orhan Tozan]

Or maybe I should just use minus() twice, list2 - list1 and list1 - list2

[Orhan Tozan]

Think I'm going to do that, thanks guys

[Klitos Kyriacou]

rubber duck

[ephemient]

fun <T> diffCounts(a: Iterable<T>, b: Iterable<T>): Map<T, Int> = buildMap {
    for (elem in a) put(elem, getOrElse(elem) { 0 } + 1)
    for (elem in b) put(elem, getOrElse(elem) { 0 } - 1)
    values.removeAll(setOf(0))
}

[nkiesel]

Would a `list2.zip(list1).map { it.first - it.second }`not suffice?

[ephemient]

@nkiesel that performs a completely different operation.

listOf(1, 2, 3) - listOf(2, 3, 4) == listOf(1)
diffCounts(listOf(1, 2, 3), listOf(2, 3, 4)) == mapOf(1 to 1, 4 to -1)
listOf(2, 3, 4).zip(listOf(1, 2, 3)) { first, second -> first - second } == listOf(1, 1, 1)

[nkiesel]

I guess it was not clear to me what "diff of two `List<Int>`" was asking for. I my view, both interpretations would make sense.

[ephemient]

since the follow-up message was that list2 - list1 and list1 - list2 together were useful for OP, I think it was clear which interpretation they meant

[nkiesel]

touché