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How can I split a List into two at specific index ...
# getting-started
l
How can I split a List into two at specific index in the most concise way?
I suppose
Copy code
list.withIndex().partition { it.index < selectedIndex }
Although then I end up with a 
Pair<List<IndexedValue<T>>>
b
That will create new List objects I think, if that's acceptable, and will need to iterate through the whole list. There's the dumb and simple "sublist" you could call twice.
l
Yes, I suppose I won't get a nice one-liner 😉
😄 1
b
I love Kotlin, very very much. But the call of "Go" is strong. Dumbest, simplest, most obvious code, 0 cleverness (or elegance)
h
🙏 1
a
val (before, after) = list.take(n) to list.drop(n)
👍 1
l
I went with 
subList()
, as it doesn't create additional lists
e
Yes, I suppose I won't get a nice one-liner
Extension function containing two calls to 
subList
if it's a readonly 
List<T>
l
👆 That's @hho’s example above
✔️ 1
e
If you need to split in 
n
parts, you can even overload the 
div
operator.
Copy code
val (one, two, three, four) = list / 4
l
Although I just realized my list can be smaller than the split index, so I have to go with @Alejandro Serrano.Mena’s approach 
take(n)
and 
drop(n)
e
k
Although I just realized my list can be smaller than the split index, so I have to go with @Alejandro Serrano.Mena’s approach 
take(n)
and 
drop(n)
Instead of creating two new lists, I would stick to Henning's 
subList
and just use 
splitIndex.coerceAtMost(size)
.
👍 1
e
Frankly, I also use this functionality so much that I've made my own private libraries for what Haskell has as splitAt and span(). I should really give some live examples someday because those are among the few things I would like to have extra in Kotlin (though of course it would burden the language with even more things, so there's a tradeoff...)
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