This question already has answers here:
How to multiply two arrays element-wise
(5 answers)
Closed 3 years ago.
I am trying to compute the difference between the values in two arrays in Swift. I want to subtract values at each index between two arrays.
I tried doing the following:
var array1 : [Double] = [1, 2, 3, 4, 5]
var array2 : [Double] = [2, 3, 4, 5, 6]
var result = array2 - array1
Expected answer:
result = [1, 1, 1, 1, 1]
I get the following error message:
Binary operator '-' cannot be applied to two '[Double]' operands
The following worked:
var array1 : [Double] = [1, 2, 3, 4, 5]
var array2 : [Double] = [2, 3, 4, 5, 6]
let velocity = (0..<5).map { array2[$0] - array1[$0] }
print(velocity)
I wanted to understand if there is an efficient way to accomplish this.
Your attempt works quite well. In general, you would need to check which array has the fewer elements (just in case):
(0..<(min(array1.count, array2.count))).map { array2[$0] - array1[$0] }
Or, as Connor mentioned in their answer, use zip, which handles this comparison of array lengths for you.
zip(lhs, rhs).map { $0.0 - $0.1 }
You can go one step further and overload the - operator to achieve the syntax you wanted (array1 - array2):
func -<T: Numeric>(lhs: [T], rhs: [T]) -> [T] {
return zip(lhs, rhs).map(-)
}
// usage:
print([1,2,3] - [0, 1, 2])
But do note that, to other people, it might be quite unclear what an array “minus” another array means.
You can zip the 2 arrays together to get pairs of elements, and then map over them:
let result = zip(array1, array2).map { $0 - $1 }
Note that, with this method, if one array has more elements than the other, those extra elements will be ignored.
Related
So, something is bugging me with the syntax in Swift for performing operations on Arrays of Ints.
What I wanna do is this : I have an array of Ints which is outputted from a function, its size (count) varies between say 2 and 6 for now, depending on buttons I press in my app.
For each array that is outputted and that contain n ints, I want to create n arrays on which to perform an other action later on.
These "sub" arrays are supposed to be calculated this way :
newArray1's values should be array's values - the value of the first index of newArray1
newArray2's values should be array's values - the value of the second index of newArray2
etc... (I'll automate the number of newArrays according to the array.count)
An other condition applying for those new arrays is that if at a given index the value is negative, I add 12 (so it'll occur for newArray2 at index 1, for newArray3 at indexes 1 & 2, etc... as long as those newArrays are created).
Here's how I wanted to perform that (I created this with dummy arbitrary array in the playground for the sake of testing before inserting the correct stuff in my app code) :
var array : [Int] = [2,4,6,8,9]
var newArray2 = [Int]()
var increment2 = Int()
increment2 = array[1]
newArray2 = array.map {$0 - increment2}
for i in 0..<newArray2.count {
if array[i] < 0 {
newArray2[i] = array[i] + 12
} else {
newArray2[i] = array[i]
}
}
print(array)
print(newArray2)
So of course it doesn't work because I can't seem to figure how to correctly perform operations on Arrays...
Intuitively it seems in my first if statement I'm comparing not the element at index i but i itself, not sure how to reformat that though...
Any help is most welcome, thanks in advance ! :)
[EDIT: I just edited the names of newArray1 to newArray2, same for increments, so that I have negative values and it matches the index value of 1 which is the second element of my main array]
You seem to mean this:
let arr = [2,4,6,8,9]
var results = [[Int]]()
for i in arr.indices {
results.append(arr.map {
var diff = $0-arr[i]
if diff < 0 { diff += 12 }
return diff
})
}
// results is now:
// [[0, 2, 4, 6, 7],
// [10, 0, 2, 4, 5],
// [8, 10, 0, 2, 3],
// [6, 8, 10, 0, 1],
// [5, 7, 9, 11, 0]]
This question already has answers here:
Removing duplicate elements from an array in Swift
(49 answers)
Closed 5 years ago.
I have an [Int] array and I need to check for all elements, that don't have identical value with some another element. Those non- identical elements I would like to insert to a new array.
var spojeni1 = [1, 1, 2, 3, 4, 4] // Here it is values 2,3
var NewArray = [Int]()
for i in spojeni1 {
if { // the value hasn't another identical value in the array
NewArray.append(i)
}
}
Hope it is clear, thank you
This can be done in a single line:
let numbers = [1, 1, 2, 3, 4, 4] // Here it is values 2,3
let uniques = Set(numbers).filter{ (n) in numbers.filter{$0==n}.count == 1 }
UPDATE
With Swift 4, you could also use a dictionary constructor to do it:
let uniques = Dictionary(grouping:numbers){$0}.filter{$1.count==1}.map{$0.0}
I would make use of 2 sets:
var seenOnce: Set<Int> = []
var seenMorethanOnce: Set<Int> = []
let input = [1, 1, 2, 3, 4, 4]
for number in input
{
if seenMorethanOnce.contains(number)
{
// Skip it since it's already been detected as a dupe.
}
else if seenOnce.contains(number)
{
// We saw it once, and now we have a dupe.
seenOnce.remove(number)
seenMorethanOnce.insert(number)
}
else
{
// First time seeing this number.
seenOnce.insert(number)
}
}
When that for loop finishes, the seenOnce set will have your values, and you can easily convert that to an array if you wish.
This question already has answers here:
In Swift, an efficient function that separates an array into 2 arrays based on a predicate
(7 answers)
Closed 6 months ago.
Problem
Given an array of values how can I split it into sub-arrays made of elements that are equal?
Example
Given this array
let numbers = [1, 1, 1, 3, 3, 4]
I want this output
[[1,1,1], [3, 3], [4]]
What I am NOT looking for
A possible way of solving this would be creating some sort of index to indicate the occurrences of each element like this.
let indexes = [1:3, 3:2, 4:1]
And finally use the index to rebuild the output array.
let subsequences = indexes.sort { $0.0.0 < $0.1.0 }.reduce([Int]()) { (res, elm) -> [Int] in
return res + [Int](count: elm.1, repeatedValue: elm.0)
}
However with this solution I am losing the original values. Of course in this case it's not a big problem (an Int value is still and Inteven if recreated) but I would like to apply this solution to more complex data structures like this
struct Starship: Equatable {
let name: String
let warpSpeed: Int
}
func ==(left:Starship, right:Starship) -> Bool {
return left.warpSpeed == right.warpSpeed
}
Final considerations
The function I am looking for would be some kind of reverse of flatten(), infact
let subsequences: [[Int]] = [[1,1,1], [3, 3], [4]]
print(Array(subsequences.flatten())) // [1, 1, 1, 3, 3, 4]
I hope I made myself clear, let me know should you need further details.
// extract unique numbers using a set, then
// map sub-arrays of the original arrays with a filter on each distinct number
let numbers = [1, 1, 1, 3, 3, 4]
let numberGroups = Set(numbers).map{ value in return numbers.filter{$0==value} }
print(numberGroups)
[EDIT] changed to use Set Initializer as suggested by Hamish
[EDIT2] Swift 4 added an initializer to Dictionary that will do this more efficiently:
let numberGroups = Array(Dictionary(grouping:numbers){$0}.values)
For a list of objects to be grouped by one of their properties:
let objectGroups = Array(Dictionary(grouping:objects){$0.property}.values)
If you could use CocoaPods/Carthage/Swift Package Manager/etc. you could use packages like oisdk/SwiftSequence which provides the group() method:
numbers.lazy.group()
// should return a sequence that generates [1, 1, 1], [3, 3], [4].
or UsrNameu1/TraverSwift which provides groupBy:
groupBy(SequenceOf(numbers), ==)
If you don't want to add external dependencies, you could always write an algorithm like:
func group<S: SequenceType where S.Generator.Element: Equatable>(seq: S) -> [[S.Generator.Element]] {
var result: [[S.Generator.Element]] = []
var current: [S.Generator.Element] = []
for element in seq {
if current.isEmpty || element == current[0] {
current.append(element)
} else {
result.append(current)
current = [element]
}
}
result.append(current)
return result
}
group(numbers)
// returns [[1, 1, 1], [3, 3], [4]].
Let's assume that you have an unsorted array of items. You will need to sort the initial array then you will have something like this:
[1, 1, 1, 3, 3, 4]
After that you will initialize two arrays: one for storing arrays and another one to use it as a current array.
Loop through the initial array and:
if the current value isn't different from the last one, push it to the current array
otherwise push the current array to the first one then empty the current array.
Hope it helps!
Worth mentioning, using Swift Algorithms this is now a one-liner:
import Algorithms
let numbers = [1, 1, 1, 3, 3, 4]
let chunks: [[Int]] = numbers.chunked(by: ==).map { .init($0) }
print(chunks) // [[1, 1, 1], [3, 3], [4]]
I have a multi-dimensional array and I need a count of all of the items within all of the arrays, excluding container arrays themselves from the count.
What would be the most generic and idiomatic solution in Swift? I'm guessing it's going to be something functional (a reduce() operation?), but not sure on the best overall approach.
The obvious non-functional approach would be to simply iterate over the array and tally up the number of items.
With the latest Swift 2.0 beta 6 you can use flatten()
let array = [[1, 2, 3], [4, 5], [6]]
array.flatten().count
EDIT: Just tested it: Lazy is not needed, the values are never evaluated, it just calculates endIndex - startIndex of every subcollection.
You can do the following :
let array = [[1, 2, 3], [4, 5], [6]]
let countOfAll = array.map { (nested) -> Int in
return nested.count
}.reduce(0, combine: +) // 6
For Swift 2 you can use flatMap.
var anArray = [[1,0,0], ["asdf","df","lef"], [0,0,1]]
var flatArray = anArray.flatMap { $0 }
print(flatArray.count) // 9
I have a big array of objects and would like to split it into two arrays containing the objects in alternate order.
Example:
[0, 1, 2, 3, 4, 5, 6]
Becomes these two arrays (they should alternate)
[0, 2, 4, 6] and [1, 3, 5]
There are a ton of ways to split an array. But, what is the most efficient (least costly) if the array is huge.
There are various fancy ways to do it with filter but most would probably require two passes rather than one, so you may as well just use a for-loop.
Reserving space up-front could make a big difference in this case since if the source is large it’ll avoid unnecessary re-allocation as the new arrays grow, and the calculation of space needed is in constant time on arrays.
// could make this take a more generic random-access collection source
// if needed, or just make it an array extension instead
func splitAlternating<T>(source: [T]) -> ([T],[T]) {
var evens: [T] = [], odds: [T] = []
evens.reserveCapacity(source.count / 2 + 1)
odds.reserveCapacity(source.count / 2)
for idx in indices(source) {
if idx % 2 == 0 {
evens.append(source[idx])
}
else {
odds.append(source[idx])
}
}
return (evens,odds)
}
let a = [0,1,2,3,4,5,6]
splitAlternating(a) // ([0, 2, 4, 6], [1, 3, 5])
If performance is truly critical, you could use source.withUnsafeBufferPointer to access the source elements, to avoid the index bounds checking.
If the arrays are really huge, and you aren’t going to use the resulting data except to sample a small number of elements, you could consider using a lazy view instead (though the std lib lazy filter isn’t much use here as it returns sequence not a collection – you’d possibly need to write your own).
You can use the for in stride loop to fill two resulting arrays as follow:
extension Array {
var groupOfTwo:(firstArray:[T],secondArray:[T]) {
var firstArray:[T] = []
var secondArray:[T] = []
for index in stride(from: 0, to: count, by: 2) {
firstArray.append(self[index])
if index + 1 < count {
secondArray.append(self[index+1])
}
}
return (firstArray,secondArray)
}
}
[0, 1, 2, 3, 4, 5, 6].groupOfTwo.firstArray // [0, 2, 4, 6]
[0, 1, 2, 3, 4, 5, 6].groupOfTwo.secondArray // [1, 3, 5]
update: Xcode 7.1.1 • Swift 2.1
extension Array {
var groupOfTwo:(firstArray:[Element],secondArray:[Element]) {
var firstArray:[Element] = []
var secondArray:[Element] = []
for index in 0.stride(to: count, by: 2) {
firstArray.append(self[index])
if index + 1 < count {
secondArray.append(self[index+1])
}
}
return (firstArray,secondArray)
}
}
A more concise, functional approach would be to use reduce
let a = [0,1,2,3,4,5,6]
let (evens, odds) = a.enumerate().reduce(([Int](),[Int]())) { (cur, next) in
let even = next.index % 2 == 0
return (cur.0 + (even ? [next.element] : []),
cur.1 + (even ? [] : [next.element]))
}
evens // [0,2,4,6]
odds // [1,3,5]
Big/huge array always pose problems when being partially processed, like in this case, as creating two extra (even if half-sized) arrays can be both time and memory consuming. What if, for example, you just want to compute the mean and standard deviation of oddly and evenly positioned numbers, but this will require calling a dedicated function which requires a sequence as input?
Thus why not creating two sub-collections that instead of duplicating the array contents, they point to the original array, in a transparent manner to allow querying them for elements:
extension Collection where Index: Strideable{
func stride(from: Index, to: Index, by: Index.Stride) -> StridedToCollection<Self> {
return StridedToCollection(self, from: from, to: to, by: by)
}
}
struct StridedToCollection<C>: Collection where C: Collection, C.Index: Strideable {
private let _subscript : (C.Index) -> C.Element
private let step: C.Index.Stride
fileprivate init(_ collection: C, from: C.Index, to: C.Index, by: C.Index.Stride) {
startIndex = from
endIndex = Swift.max(to, startIndex)
step = by
_subscript = { collection[$0] }
}
let startIndex: C.Index
let endIndex: C.Index
func index(after i: C.Index) -> C.Index {
let next = i.advanced(by: step)
return next >= endIndex ? endIndex : next
}
subscript(_ index: C.Index) -> C.Element {
return _subscript(index)
}
}
The Collection extension and the associated struct would create a pseudo-array that you can use to access only the elements you are interested into.
Usage is simple:
let numbers: [Int] = [1, 2, 3, 4]
let stride1 = numbers.stride(from: 0, to: numbers.count, by: 2)
let stride2 = numbers.stride(from: 1, to: numbers.count, by: 2)
print(Array(stride1), Array(stride2))
With the above you can iterate the two strides without worrying you'll double the amount of memory. And if you actually need two sub-arrays, you just Array(stride)-ify them.
Use for loops. If the index value is even then send that to one array and if the index value is odd, then send that to odd array.
Here's, in my opinion, the easiest way
old_list = [0, 1, 2, 3, 4, 5, 6]
new_list1 =[]
new_list2 = []
while len(old_list)>0:
new_list1.append(old_list.pop(-1))
if len(old_list) != 0:
new_list2.append(old_list.pop(-1))
new_list1.reverse()
new_list2.reverse()
I just had to do this where I split an array into two in one place, and three into another. So I built this:
extension Array {
/// Splits the receiving array into multiple arrays
///
/// - Parameter subCollectionCount: The number of output arrays the receiver should be divided into
/// - Returns: An array containing `subCollectionCount` arrays. These arrays will be filled round robin style from the receiving array.
/// So if the receiver was `[0, 1, 2, 3, 4, 5, 6]` the output would be `[[0, 3, 6], [1, 4], [2, 5]]`. If the reviever is empty the output
/// Will still be `subCollectionCount` arrays, they just all will be empty. This way it's always safe to subscript into the output.
func split(subCollectionCount: Int) -> [[Element]] {
precondition(subCollectionCount > 1, "Can't split the array unless you ask for > 1")
var output: [[Element]] = []
(0..<subCollectionCount).forEach { (outputIndex) in
let indexesToKeep = stride(from: outputIndex, to: count, by: subCollectionCount)
let subCollection = enumerated().filter({ indexesToKeep.contains($0.offset)}).map({ $0.element })
output.append(subCollection)
}
precondition(output.count == subCollectionCount)
return output
}
}
It works on Swift 4.2 and 5.0 (as of 5.0 with Xcode 10.2 beta 2)