I'd like to sum up consecutive numbers in a Kotlin list.
If the list has a 0 then it should start summing up the numbers after 0. The result would be a list of sums. Basically sum up until the first 0 then until the next 0 and so forth.
For example:
val arr = arrayOf(1, 2, 0, 2, 1, 3, 0, 4)
// list of sums = [3, 6, 4]
At the moment I got it working with fold:
val sums: List<Int> = arr.fold(listOf(0)) { sums: List<Int>, n: Int ->
if (n == 0)
sums + n
else
sums.dropLast(1) + (sums.last() + n)
}
but I wonder if there is a simpler or more efficient way of doing this.
I would personally have written it this way:
val sums = mutableListOf(0).also { acc ->
arr.forEach { if (it == 0) acc.add(0) else acc[acc.lastIndex] += it }
}
Using a mutable list, you avoid having to do any drop / concatenation. The code is also easier to follow.
You can still convert it to an immutable list using .toList() if you need to.
Related
Assuming we are given a list of integers R = [3,5,3,6,0,6,7], an threshold x (integer) and a window size (integer) p. For example, x=4 and p = 2.
I need to find the first index t that verifies the the following conditions:
R[t] >= 4, R[t+1] >= 4. Since p=2, we need to only verify for two boxes t and t+1. If p was equal to 3 we will need to verify for t, t+1 and t+2.
Here the t I am looking for is 5 (indexing is starting from 0).
How to write this in a elegant way in Kotlin (rather than looping on the elements).
A tentative that is giving an error (x=4 and p = 2. The output should be 3 since we start indexing by 0):
val numbers = listOf(1, 2, 3, 4, 6, 8, 2)
val firstIndex = numbers.find { it >= 4 for it in it..it+2-1}
val numbers = listOf(1, 2, 3, 4, 6, 8, 2)
val p = 2
val x = 4
val t = numbers.windowed(p).indexOfFirst { window -> window.all { it >= x } } // t == 3
t will be equal to -1 in case if no matches will be found
Use windowed to check groups of values for each index in the list. Use withIndex() so you are iterating with the indices, which you need in your final result. Then use firstOrNull() (which find() is a redundant alias of). And finally, take ?.index to get the index of the first entry that satisfies the condition, or null if none satisfy.
val x = 4
val p = 3
val list = listOf(2,5,3,6,0,6,7)
val t = list
.windowed(p)
.withIndex()
.firstOrNull { (_, sublist) -> sublist.all { it >= x } }
?.index
find Returns the first element matching the given predicate, or null if no such element was found.
If I've understood correctly, this should work:
fun main() {
val list = listOf(3,5,3,6,0,6,7)
val p = 2
val x = 4
val t = list.withIndex().windowed(p).firstOrNull() { window ->
window.all { it.value >= x }
}?.first()?.index
println(t)
}
Output:
5
Is there any function (like fold, map, filter), which gets 2 arrays and lambda-function (for example multiplication) as parameters and returns third array?
I've used cycle for, but is there more beautiful method?
Yes, there is zip (nice example at the bottom of the page), see this (different) example:
fun main() {
val a = arrayOf( 1, 2, 3, 4 )
val b = arrayOf( 1, 2, 3, 4 )
val c = a.zip(b) { i, j -> i * j }
println(c)
}
which outputs
[1, 4, 9, 16]
There isn't a built in specifically but you can do this:
array1.zip(array2).map { (x,y) -> x*y }
In a 3x3 matrix representation, i can find the sum of both diagonals with one liners in Swift as below,
let array = [
[1, 2, 3],
[4, 5, 6],
[-7, 8, 9]
]
let d1 = array.enumerated().map({ $1[$0] }).reduce(0, +)
let d2 = array.reversed().enumerated().map({ $1[$0] }).reduce(0, +)
print(d1) // prints 15
print(d2) // prints 1
I am able to find map and reduce equivalents in Kotlin as flatMap and fold but couldn't find for enumerated.
How can we achieve similar with higher order functions in Kotlin?
Starting with this input:
val input: Array<Array<Int>> = arrayOf(
arrayOf(1, 2, 3),
arrayOf(4, 5, 6),
arrayOf(-7, 8, 9)
)
this is how I'd phrase the diagonal sums:
val mainDiagonalSum = input.indices
.map { input[it][it] }
.reduce(Int::plus)
val counterDiagonalSum = input.indices
.map { input[input.size - 1 - it][it] }
.reduce(Int::plus)
Note that this is an improvement over your solution because it doesn't have to create the reversed array. It improves the time complexity from O(n2) to O(n).
If you're dealing with large matrices, it would pay to reduce the space complexity from O(n) to O(1) as well, by using fold instead of reduce:
val mainDiagonalSum = input.indices
.fold(0) { sum, i -> sum + input[i][i] }
val counterDiagonalSum = input.indices
.fold(0) { sum, i -> sum + input[input.size - 1 - i][i] }
It looks like you're looking for withIndex
I'm trying to find a way to get the product of a List or Array without using "repeat" or any loop on Kotlin but after some research I couldn't find anything similar.
Something like this in Python would be:
>>> reduce(lambda x, y: x*y, [1,2,3,4,5,6])
output: 720
You can use reduce in Kotlin.
From the doc:
Accumulates value starting with the first element and applying
operation from left to right to current accumulator value and each
element.
val list = listOf<Int>(1, 2, 3, 4, 5, 6)
val array = intArrayOf(1, 2, 3, 4, 5, 6)
list.reduce { acc, i -> acc * i } // returns 720
array.reduce { acc, i -> acc * i } // returns 720
An even simpler solution might be: (1..6).reduce(Int::times)
Use the fold function
val total = listOf(1, 2, 3, 4, 5).fold(1, { total, next -> total * next })
Hope this helps:
fun main(args: Array<String>){
val array = intArrayOf(1, 2, 3, 4, 5, 6)
val product = array.fold(1){acc, i -> acc * i}
println("The result is: $product")
}
This will output the product of the array.
Use the fold or reduce function. Both will work.
val array = arrayOf(1, 2, 3, 4, 5, 6)
println("Product of list: ${listOfMultiplication(array)}")
fun listOfMultiplication(array: Array<Int>): Int {
return array.reduce { total, next -> total * next }
}
val array = arrayOf(1, 2, 3, 4, 5, 6)
println("Product of list: ${listOfMultiplication(array)}")
fun listOfMultiplication(array: Array<Int>): Int {
return array.fold(1) { total, next -> total * next }
}
I have a sequence of interleaved data (with fixed stride) and I'd like to reduce it to a single value for each "structure" (n*stride values to n values).
I could just use loop writing into the mutable list with selected step for reader index, but I'm looking for more functional and readable approach. Any thoughts?
For example:
Input sequence consists of RGB triplets (stride 3) and output is grayscale.
Imperative way is like:
fun greyscale(stream:List<Byte>):List<Byte>{
val out = ArrayList(stream.size / 3)
var i = 0; var o = 0
while(i < stream.size)
out[o++]=(stream[i++] + stream[i++] + stream[i++])/3
return out
}
How can I make something like that without explicitly implementing a function and mutable container, but purely on functional extensions like .map and so on?
Kotlin 1.2 (Milestone 1 was released yesterday) brings the chunked method on collections. It chunks up the collection into blocks of a given size. You can use this to implement your function:
fun greyscale(stream: List<Byte>): List<Byte> =
stream.chunked(3)
.map { (it.sum() / 3).toByte() }
A possible way would be grouping by the index of the elements (in this case /3) and mapping these groups to their sum.
stream.withIndex()
.groupBy { it.index / 3 }
.toSortedMap()
.values
.map { (it.sumBy { it.value } / 3).toByte() }
Also strictly functional, but using Rx, would be possible by using window(long)
Observable.from(stream)
.window(3)
.concatMap { it.reduce(Int::plus).toObservable() }
.map { (it / 3).toByte() }
Similar to #marstran's answer, in Kotlin 1.2 you can use chunked function, but providing the transform lambda to it:
fun greyscale(stream: List<Byte>): List<Byte> =
stream.chunked(3) { it.average().toByte() }
This variant has an advantage that it doesn't instantiate a new List for every triple, but rather creates a single List and reuses it during the entire operation.
Excludes remaining elements:
const val N = 3
fun greyscale(stream: List<Byte>) = (0 until stream.size / N)
.map { it * N }
.map { stream.subList(it, it + N).sum() / N }
.map(Int::toByte)
Output
[1, 2, 3, 4, 5, 6] => [2, 5]
[1, 2, 3, 4, 5] => [2]
Includes remaining elements:
const val N = 3
fun greyscale(stream: List<Byte>) = (0 until (stream.size + N - 1) / N)
.map { it * N }
.map { stream.subList(it, minOf(stream.size, it + N)).sum() / N }
.map(Int::toByte)
Output
[1, 2, 3, 4, 5, 6] => [2, 5]
[1, 2, 3, 4, 5] => [2, 3]
Best what I'm capable of is this:
fun grayscale(rgb:List<Byte>):List<Byte>
= rgb.foldIndexed(
IntArray(rgb.size / 3),
{ idx, acc, i ->
acc[idx / 3] = acc[idx / 3] + i; acc
}).map{ (it / 3).toByte() }
Output
in: [1, 2, 3, 4, 5, 6]
out: [2, 5]
And variations with ArrayList with add and last