As of Swift 1.2, Apple introduces Set collection type.
Say, I have a set like:
var set = Set<Int>(arrayLiteral: 1, 2, 3, 4, 5)
Now I want to get a random element out of it. Question is how? Set does not provide subscript(Int) like Array does. Instead it has subscript(SetIndex<T>). But firstly, SetIndex<T> does not have accessible initializers (hence, I can not just create an index with the offset I need), and secondly even if I can get the index for a first element in a set (var startIndex = set.startIndex) then the only way I can get to the N-th index is through consecutive calls to successor().
Therefore, I can see only 2 options at the moment, both ugly and expensive:
Convert the set into array (var array = [Int](set)) and use its subscript (which perfectly accepts Int); or
Get index of a first element in a set, traverse the chain of successor() methods to get to the N-th index, and then read corresponding element via set's subscript.
Do I miss some other way?
Starting with Swift 4.2, you can use randomElement:
let random = set.randomElement()
Probably the best approach is advance which walks successor for you:
func randomElementIndex<T>(s: Set<T>) -> T {
let n = Int(arc4random_uniform(UInt32(s.count)))
let i = advance(s.startIndex, n)
return s[i]
}
(EDIT: Heh; noticed you actually updated the question to include this answer before I added it to my answer... well, still a good idea and I learned something too. :D)
You can also walk the set rather than the indices (this was my first thought, but then I remembered advance).
func randomElement<T>(s: Set<T>) -> T {
let n = Int(arc4random_uniform(UInt32(s.count)))
for (i, e) in enumerate(s) {
if i == n { return e }
}
fatalError("The above loop must succeed")
}
In swift 3
extension Set {
public func randomObject() -> Element? {
let n = Int(arc4random_uniform(UInt32(self.count)))
let index = self.index(self.startIndex, offsetBy: n)
return self.count > 0 ? self[index] : nil
}
}
extension Set {
func randomElement() -> Element? {
return count == 0 ? nil : self[advance(self.startIndex, Int(arc4random()) % count)]
}
}
As per comments above re Swift updates, used a minor change for an extension to Set:
func randomElement() -> Element?
{
let randomInt = Int(arc4random_uniform(UInt32(self.count)))
let index = startIndex.advancedBy(randomInt)
return count == 0 ? nil: self[index]
}
If you want a 'random' element from a Set then you use:
/// A member of the set, or `nil` if the set is empty.
var first: T? { get }
Get the 0th index or the 1,000,000th index makes no difference - they are all an arbitrary object.
But, if you want repeated calls to return a likely different element each time, then first might not fit the bill.
Related
I want to compare two Arrays with each other, that means each single item of them.
I need to run some code if two items in this arrays are the same.
I've done that so far with two For-Loops, but someone in the comments says that's not that good (because I get an Error too.
Has anybody an Idea which Code I can use to reach that?
Btw: That's the code I used before:
var index = 0
for Item1 in Array1 {
for Item2 in Array2 {
if (Item1 == Item2) {
// Here I want to put some code if the items are the same
}
}
// Index is in this case the counter which counts on which place in the Array1 I am.
index += 1
}
Okay I'll try again to describe what I mean:
I want to compare two Arrays. If there are some Items the same, I want to delete that Item from Array A / Array 1.
Okay if that is working, I want to add a few more statements that alow me to delete a Item only if an parameter of this item has a special worth, but I think I can do this step alone.
If you want to compare items from different array you need to add Equatable protocol for your Item
For example:
struct Item: Equatable {
let name: String
static func ==(l: Item, r: Item) -> Bool {
return l.name == r.name
}
}
You need to decide by which attributes you want to compare your Item. In my case I compare by name.
let array1: [Item] = [
.init(name: "John"),
.init(name: "Jack"),
.init(name: "Soer"),
.init(name: "Kate")
]
let array2: [Item] = [
.init(name: "John"),
]
for item1 in array1 {
if array2.contains(item1) {
// Array 2 contains item from the array1 and you can perform your code.
}
}
If you want to support this
Okay I'll try again to describe what I mean: I want to compare two
Arrays. If there are some Items the same, I want to delete that Item
from Array A / Array 1. Okay if that is working, I want to add a few
more statements that alow me to delete a Item only if an parameter of
this item has a special worth, but I think I can do this step alone.
I guess it can fit for you
You need to make your array1 mutable
var array1: [Item] = ...
Filter the array1 like this
let filteredArray1 = array1.filter { (item) -> Bool in
return !array2.contains(item)
}
And redefine your array1 with filtered array.
array1 = filteredArray1
Let me know it it works for you.
var itemsToRemove = array1.filter { array2.contains($0) }
for item in itemsToRemove {
if let i = array1.firstIndex(of: item) {
// run your conditional code here.
array1.remove(at: i)
}
}
Update
The question has been restated that elements are to be removed from one of the arrays.
You don't provide the actual code where you get the index out of bounds error, but it's almost certainly because you don't account for having removed elements when using the index, so you're probably indexing into a region that was valid at the start, but isn't anymore.
My first advice is don't do the actual removal inside the search loop. There is a solution to achieve the same result, which I'll give, but apart from having to be very careful about indexing into the shortened array, there is also a performance issue: Every deletion requires Array to shift all the later elements down one. That means that every deletion is an O(n) operation, which makes an the overall algorithim O(n^3).
So what do you do instead? The simplest method is to create a new array containing only the elements you wish to keep. For example, let's say you want to remove from array1 all elements that are also in array2:
array1 = array1.filter { !array2.contains($0) }
I should note that one of the reasons I kept my original answer below is because you can use those methods to replace array2.contains($0) to achieve better performance in some cases, and the original answer describes those cases and how to achieve it.
Also even though the closure is used to determine whether or not to keep the element, so it has to return a Bool, there is nothing that prevents you from putting additional code in it to do any other work you might want to:
array1 = array1.filter
{
if array2.contains($0)
{
doSomething(with: $0)
return false // don't keep this element
}
return true
}
The same applies to all of the closures below.
You could just use the removeAll(where:) method of Array.
array1.removeAll { array2.contains($0) }
In this case, if the closure returns true, it means "remove this element" which is the opposite sense of the closure used in filter, so you have to take that into account if you do additional work in it.
I haven't looked up how the Swift library implements removeAll(where:) but I'm assuming it does its job the efficient way rather than the naive way. If you find the performance isn't all that good you could roll your own version.
extension Array where Element: Equatable
{
mutating func efficientRemoveAll(where condition: (Element) -> Bool)
{
guard count > 0 else { return }
var i = startIndex
var iEnd = endIndex
while i < iEnd
{
if condition(self[i])
{
iEnd -= 1
swapAt(i, iEnd)
continue // note: skips incrementing i,iEnd has been decremented
}
i += 1
}
removeLast(endIndex - iEnd)
}
}
array1.efficientRemoveAll { array2.contains($0) }
Instead of actually removing the elements inside the loop, this works by swapping them with the end element, and handling when to increment appropriately. This collects the elements to be removed at the end, where they can be removed in one go after the loop finishes. So the deletion is just one O(n) pass at the end, which avoids increasing the algorithmic complexity that removing inside the loop would entail.
Because of the swapping with the current "end" element, this method doesn't preserve the relative order of the elements in the mutating array.
If you want to preserve the order you can do it like this:
extension Array where Element: Equatable
{
mutating func orderPreservingRemoveAll(where condition: (Element) -> Bool)
{
guard count > 0 else { return }
var i = startIndex
var j = i
repeat
{
swapAt(i, j)
if !condition(self[i]) { i += 1 }
j += 1
} while j != endIndex
removeLast(endIndex - i)
}
}
array1.orderPreservingRemoveAll { array2.contains($0) }
If I had to make a bet, this would be very close to how standard Swift's removeAll(where:) for Array is implemented. It keeps two indices, one for the current last element to be kept, i, and one for the next element to be examined, j. This has the effect of accumulating elements to be removed at the end (those past i).
Original answer
The previous solutions are fine for small (yet still surprisingly large) arrays, but they are O(n^2) solutions.
If you're not mutating the arrays, they can be expressed more succinctly
array1.filter { array2.contains($0) }.forEach { doSomething($0) }
where doSomething doesn't actually have to be a function call - just do whatever you want to do with $0 (the common element).
You'll normally get better performance by putting the smaller array inside the filter.
Special cases for sorted arrays
If one of your arrays is sorted, then you might get better performance in a binary search instead of contains, though that will require that your elements conform to Comparable. There isn't a binary search in the Swift Standard Library, and I also couldn't find one in the new swift-algorithms package, so you'd need to implement it yourself:
extension RandomAccessCollection where Element: Comparable, Index == Int
{
func sortedContains(_ element: Element) -> Bool
{
var range = self[...]
while range.count > 0
{
let midPoint = (range.startIndex + range.endIndex) / 2
if range[midPoint] == element { return true }
range = range[midPoint] > element
? self[..<midPoint]
: self[index(after: midPoint)...]
}
return false
}
}
unsortedArray.filter { sortedArray.sortedContains($0) }.forEach { doSomething($0) }
This will give O(n log n) performance. However, you should test it for your actual use case if you really need performance, because binary search is not especially friendly for the CPU's branch predictor, and it doesn't take that many mispredictions to result in slower performance than just doing a linear search.
If both arrays are sorted, you can get even better performance by exploiting that, though again you have to implement the algorithm because it's not supplied by the Swift Standard Library.
extension Array where Element: Comparable
{
// Assumes no duplicates
func sortedIntersection(_ sortedOther: Self) -> Self
{
guard self.count > 0, sortedOther.count > 0 else { return [] }
var common = Self()
common.reserveCapacity(Swift.min(count, sortedOther.count))
var i = self.startIndex
var j = sortedOther.startIndex
var selfValue = self[i]
var otherValue = sortedOther[j]
while true
{
if selfValue == otherValue
{
common.append(selfValue)
i += 1
j += 1
if i == self.endIndex || j == sortedOther.endIndex { break }
selfValue = self[i]
otherValue = sortedOther[j]
continue
}
if selfValue < otherValue
{
i += 1
if i == self.endIndex { break }
selfValue = self[i]
continue
}
j += 1
if j == sortedOther.endIndex { break }
otherValue = sortedOther[j]
}
return common
}
}
array1.sortedIntersection(array2).forEach { doSomething($0) }
This is O(n) and is the most efficient of any of the solutions I'm including, because it makes exactly one pass through both arrays. That's only possible because both of the arrays are sorted.
Special case for large array of Hashable elements
However, if your arrays meet some criteria, you can get O(n) performance by going through Set. Specifically if
The arrays are large
The array elements conform to Hashable
You're not mutating the arrays
Make the larger of the two arrays a Set:
let largeSet = Set(largeArray)
smallArray.filter { largeSet.contains($0) }.forEach { doSomething($0) }
For the sake of analysis if we assume both arrays have n elements, the initializer for the Set will be O(n). The intersection will involve testing each element of the smallArray's elements for membership in largeSet. Each one of those tests is O(1) because Swift implements Set as a hash table. There will be n of those tests, so that's O(n). Then the worst case for forEach will be O(n). So O(n) overall.
There is, of course, overhead in creating the Set, the worst of which is the memory allocation involved, so it's not worth it for small arrays.
I have a simple dictionary:
struct AnnotationDict: Encodable, Decodable {
let id: Int
let index: Int
}
var sortedAnnotation: [AnnotationDict] = []
let ADict = AnnotationDict(id: 1, index: 2)
sortedAnnotation.append(ADict)
How get index from AnnotationDict searching by id=1 ?
Like the others have said, you don't have a dictionary, you have an array.
let item = sortedAnnotation.first { $0.id == 1 } returns the first item whose id == 1. Note that the type of the returned item is optional i.e. AnnotationDict?.
To get its index, you do item!.index, though force-unwrapping is not recommended.
You don't have a dictionary. You have an array of Structs.
Larme gave you the answer:
let itemWithID1 = sortedAnnotation.first(where: { $0.id == 1})
or more generally
func itemWithIndex(_ index: Int, inAnnotationDictArray annotationDictArray: [AnnotationDict]) -> AnnotationDict? {
return annotationDictArray.first(where: { $0.id == index})
}
Note that first(where:) operates in O(n) time, so it will be slow, and expressions that run in O(n^2) time are easy to create without realizing it. (For those without a formal CS background, that means "The time to complete goes up with the square of the number of elements, so it's easy to write code that gets REALLY slow for more than a very small number of items.") Finding items in a real Dictionary takes ≈constant time, which is much better.
I am a beginner of Swift for iOS development, and want to find an elegant way to loop through an array from the second element.
If it is in Objective-C, I think I can write the code like below.
if(int i = 1; i < array.count; i++){
array[i].blablabla....
}
And I have tried several methods to do this in swift
for i in 1..<array.count {
array[i].blablabla....
}
But in this way, if the array's count is zero, the app will crash, because it cannot create an array from 1 to 0.
I also tried code like this, but it seems that it cannot enumerate from the second element.
for (index, object) in array.enumerate() {
object.blablabla...
}
So what's the best way to do this, do you guys have any idea?
You can use your array indices combined with dropfirst:
let array = [1,2,3,4,5]
for index in array.indices.dropFirst() {
print(array[index])
}
or simply:
for element in array.dropFirst() {
print(element)
}
you can also use custom subscript to make sure you avoid crashing
extension Array {
subscript (gurd index: Index) -> Array.Element? {
return (index < self.count && index >= 0) ? self[index] : nil
}
}
use example:
let myarray = ["a","b","c","d","e","f","g","h"]
for i in -15...myarray.count+20{
print(myarray[gurd: i])
}
In general, there is a nice pattern for getting a sub-array like this:
let array = [1, 2, 3, 4, 5]
array[2..<array.count].forEach {
print($0) // Prints 3, 4, 5 (array from index 2 to index count-1)
}
Leo's answer is more succinct and a better approach for the case you asked about (starting with the second element). This example starts with the third element and just shows a more general pattern for getting arbitrary ranges of elements from an array and performing an operation on them.
However, you still need to explicitly check to make sure you are using a valid range to avoid crashing as you noted, e.g.:
if array.count > 1 {
array[2..<array.count].forEach {
print($0)
}
}
if array.count>=2 {
for n in 1...array.count-1{
print(array[n]) //will print from second element forward
}
}
You could use array.dropFirts(), but if you need to iterate starting from 2nd element or 3rd I would do something like this:
for (index, item) in array.enumerated() where index > 1 {
item.blablabla()
}
var index: Int=0
for index in 1...3{
print(index)
}
print(index)//prints 0
If I run this code, the last print gives 0, which means the index inside the for-in is not the same as outside. Seems like this force declares a new constant.
I am looking for similar way to retain the last value in the sequence
I know I can do
var index_out: Int=0
for index in 1...3{
print(index)
index_out = index
}
print(index_out)
If you're gonna loop through something and want to know the end index just use the amount of times you looped through it:
let n = 3
for index in 1...n{
print(index)
}
print(n)
Or with an array:
let array = [Int](count: 10, repeatedValue: 2)
for index in 0..<array.count {
print(index)
}
print(array.count)
The way you know you can do it is how to do it, for loops create their own scope. Anything declared inside of a set of { } means that it's for use within that scope only.
You can also use the way for in is implemented with an outer scope generator and while:
var generator = sequence.generate()
var element = generator.next()
while let x = generator.next() {
element = x
}
// element is nil if the sequence is empty
print(element)
this is only another way to do this but personally I think you should avoid that.
A much nicer solution would be with reduce (in Swift 1.x: global function, in Swift 2: as method)
// with default value
let lastElement = sequence.reduce(default) { $1 }
// if default should be nil you have to provide the explicit type because it cannot be inferred; probably use an extension:
// as extension
extension SequenceType {
var last: Self.Generator.Element? {
return self.reduce(nil) { $1 }
}
}
There is no way to do it, since it is in different scope than your method. But you also SHOULD NOT do what you do, to assign to your index on each iteration - that is wasteful. Instead, just increase index after you are done:
var data = []
var lastIndex = 0
for (index, object) in enumerate(data) {
...
}
lastIndex += data.count
Or, if you need to break for cycle for some reason, just assign it from inside of the cycle just before you break.
As of Swift 1.2, Apple introduces Set collection type.
Say, I have a set like:
var set = Set<Int>(arrayLiteral: 1, 2, 3, 4, 5)
Now I want to get a random element out of it. Question is how? Set does not provide subscript(Int) like Array does. Instead it has subscript(SetIndex<T>). But firstly, SetIndex<T> does not have accessible initializers (hence, I can not just create an index with the offset I need), and secondly even if I can get the index for a first element in a set (var startIndex = set.startIndex) then the only way I can get to the N-th index is through consecutive calls to successor().
Therefore, I can see only 2 options at the moment, both ugly and expensive:
Convert the set into array (var array = [Int](set)) and use its subscript (which perfectly accepts Int); or
Get index of a first element in a set, traverse the chain of successor() methods to get to the N-th index, and then read corresponding element via set's subscript.
Do I miss some other way?
Starting with Swift 4.2, you can use randomElement:
let random = set.randomElement()
Probably the best approach is advance which walks successor for you:
func randomElementIndex<T>(s: Set<T>) -> T {
let n = Int(arc4random_uniform(UInt32(s.count)))
let i = advance(s.startIndex, n)
return s[i]
}
(EDIT: Heh; noticed you actually updated the question to include this answer before I added it to my answer... well, still a good idea and I learned something too. :D)
You can also walk the set rather than the indices (this was my first thought, but then I remembered advance).
func randomElement<T>(s: Set<T>) -> T {
let n = Int(arc4random_uniform(UInt32(s.count)))
for (i, e) in enumerate(s) {
if i == n { return e }
}
fatalError("The above loop must succeed")
}
In swift 3
extension Set {
public func randomObject() -> Element? {
let n = Int(arc4random_uniform(UInt32(self.count)))
let index = self.index(self.startIndex, offsetBy: n)
return self.count > 0 ? self[index] : nil
}
}
extension Set {
func randomElement() -> Element? {
return count == 0 ? nil : self[advance(self.startIndex, Int(arc4random()) % count)]
}
}
As per comments above re Swift updates, used a minor change for an extension to Set:
func randomElement() -> Element?
{
let randomInt = Int(arc4random_uniform(UInt32(self.count)))
let index = startIndex.advancedBy(randomInt)
return count == 0 ? nil: self[index]
}
If you want a 'random' element from a Set then you use:
/// A member of the set, or `nil` if the set is empty.
var first: T? { get }
Get the 0th index or the 1,000,000th index makes no difference - they are all an arbitrary object.
But, if you want repeated calls to return a likely different element each time, then first might not fit the bill.