I have an array of objects like this, but with 16 properties:
class anObject: NSObject {
#objc var number: Int
#objc var name: String
#objc var price: Double
subscript(key: String) -> Any? {
return self.value(forKey: key)
}
}
I am able to sort my array on any property very easily by, for instance:
sortedArray = unsortedArray.sorted(by: { $0.name < $1.name } )
Now I am grouping my array so that I can populate a UITableView with sections and rows. I group it like this:
var groupedArray = Dictionary<String, Array<myObject>>()
for item in myArray {
// Verify each grouping is initialized only once
if groupedArray[item[byProperty] as! String] == nil {
groupedArray[item[byProperty] as! String] = Array<Items>()
}
// Add the item into the correct subarray
groupedArray[item[byProperty] as! String]?.append(item)
}
I can then sort the grouped array by doing this:
return groupedArray.sorted { $0.0 < $1.0 }
And this works great, except that two of my properties are Doubles. When I sort on those two properties, Swift sorts the groups alphabetically:
10.5, 11.5, 12, 1.5, 2.0 . . .
rather than numerically
1.5, 2.0, 10.5, 11.5, 12 . . .
I have managed to pad the Doubles by checking to see if they are too short and inserting a 0 at the front of the String. This works in that they are now sorted in correct order, but eventually I am going to have to strip that leading 0 off the front, and it seems like an ugly solution.
How do I properly sort the grouped array given that the Doubles have to be used as Strings?
When you start casting strings all over the place, you likely need to start changing the design of things. Why not make the dictionary keys some object you've designed instead of a string? This is a sample of what I mean:
struct DoubleKey {
let value: Double
}
extension DoubleKey: Hashable {
var hashValue: Int {
return value.hashValue
}
static func ==(lhs: DoubleKey, rhs: DoubleKey) -> Bool {
return lhs.value == rhs.value
}
}
extension DoubleKey: Comparable {
static func <(lhs: DoubleKey, rhs: DoubleKey) -> Bool {
return lhs.value < rhs.value
}
}
let a = DoubleKey(value: 10.0)
let b = DoubleKey(value: 20.0)
let c = DoubleKey(value: -10.0)
let dictionary: [DoubleKey: String] = [a: "10", b: "20", c: "-10"]
let sortedDictionary = dictionary.sorted { $0.0 < $1.0 }
So instead of [String: Array<myobject>] you have: [DoubleKey: Array<MyObject> or [IntegerKey: Array<MyObject>] or even [StringKey: Array<MyObject>
You could implement many variations of your own specialized key, and write some protocols if you need additional functionality. If you need to store a string in your key, then add a property, or better yet, conform it to a protocol that defines the behavior of what you need and implement it.
Additional Key
struct StringKey {
let value: String
}
extension StringKey: Hashable {
var hashValue: Int {
return value.hashValue
}
static func ==(lhs: StringKey, rhs: StringKey) -> Bool {
return lhs.value == rhs.value
}
}
extension StringKey: Comparable {
static func <(lhs: StringKey, rhs: StringKey) -> Bool {
return lhs.value < rhs.value
}
}
let a = StringKey(value: "a")
let b = StringKey(value: "c")
let c = StringKey(value: "b")
let dictionary: [StringKey: String] = [a: "10", b: "20", c: "-10"]
let sortedDictionary = dictionary.sorted { $0.0 < $1.0 }
Now what?
//EXAMPLE
protocol ViewableString {
func view() -> String
}
extension StringKey: ViewableString {
func view() -> String {
return value
}
}
extension DoubleKey: ViewableString {
func view() -> String {
return String(value)
}
}
let array: [ViewableString] = [a, b, c]
array[0].view()
Program to the protocols!
Hopefully this helps!
Ok. Completely different answer. Again, you are trying to fit many objects of different types in the same container. This feels like a bad idea to me. Maybe you have to. But here is one way using enums:
enum SpecialKey {
case integer(Int)
case double(Double)
case string(String)
func asString() -> String {
switch self {
case let .integer(a):
return String(a)
case let .double(a):
return String(a)
case let .string(a):
return a
}
}
}
extension SpecialKey: Comparable {
static func <(lhs: SpecialKey, rhs: SpecialKey) -> Bool {
switch (lhs, rhs) {
case let (.double(a), .double(b)):
return a < b
case let (.integer(a), .integer(b)):
return a < b
case let (.string(a), .string(b)):
return a < b
default:
return false //Add more cases with different comparisons!
}
}
static func ==(lhs: SpecialKey, rhs: SpecialKey) -> Bool {
switch (lhs, rhs) {
case (.integer(_), .integer(_)),
(.double(_), .double(_)),
(.string(_), .string(_)):
return true
default:
return false
}
}
}
extension SpecialKey: Hashable {
var hashValue: Int {
switch self {
case let .integer(a):
return a.hashValue
case let .double(a):
return a.hashValue
case let .string(a):
return a.hashValue
}
}
}
let a = SpecialKey.integer(10)
let b = SpecialKey.string("something")
let c = SpecialKey.double(10.5)
let dictionary: [SpecialKey: String] = [a: "a", b: "b", c: "c"]
This is probably more like what you're looking for.
I've been using the sort() function but it mixes up the relative order.
This is how my code looks.
recipes.sort { $0.skill.value <= $1.skill.value }
Swift API says that:
The sorting algorithm is not stable. A nonstable sort may change the
relative order of elements that compare equal.
How can I change this so that the relative order stays the same as before?
The implementation below just work like the sorted method in the standard library, without additional limit.
extension RandomAccessCollection {
/// return a sorted collection
/// this use a stable sort algorithm
///
/// - Parameter areInIncreasingOrder: return nil when two element are equal
/// - Returns: the sorted collection
public func stableSorted(by areInIncreasingOrder: (Element, Element) throws -> Bool) rethrows -> [Element] {
let sorted = try enumerated().sorted { (one, another) -> Bool in
if try areInIncreasingOrder(one.element, another.element) {
return true
} else {
return one.offset < another.offset
}
}
return sorted.map { $0.element }
}
}
A stable sort needs to preserve the original order. So we give every element a weight of order besides its value, the index, then the original sort method will just work, as there will never be 2 equal elements.
I appreciate the elegance of leavez's answer. I adapted it to have the same signature as Sequence.sorted(by:):
extension Sequence {
func stableSorted(
by areInIncreasingOrder: (Element, Element) throws -> Bool)
rethrows -> [Element]
{
return try enumerated()
.sorted { a, b -> Bool in
try areInIncreasingOrder(a.element, b.element) ||
(a.offset < b.offset && !areInIncreasingOrder(b.element, a.element))
}
.map { $0.element }
}
}
let sortedArray = (recipes as NSArray).sortedArray(options: .stable, usingComparator: { (lhs, rhs) -> ComparisonResult in
let lhs = (lhs as! Recipe)
let rhs = (rhs as! Recipe)
if lhs.skill.value == rhs.skill.value {
return ComparisonResult.orderedSame
} else if lhs.skill.value < rhs.skill.value {
return ComparisonResult.orderedAscending
} else {
return ComparisonResult.orderedDescending
}
})
Took from here: https://medium.com/#cocotutch/a-swift-sorting-problem-e0ebfc4e46d4
In Swift 5 sort() uses stable implementation and soon it will become officially guaranted to be stable.
From Swift forums:
...
On the other hand, the actual implementation calls
/// Sorts the elements of this buffer according to `areInIncreasingOrder`,
/// using a stable, adaptive merge sort.
///
/// The adaptive algorithm used is Timsort, modified to perform a straight
/// merge of the elements using a temporary buffer.
#inlinable
public mutating func _stableSortImpl(
by areInIncreasingOrder: (Element, Element) throws -> Bool
) rethrows { ... }
And
If I recall, sort() is currently stable, but it is not yet guaranteed
to be stable (meaning, the fact that it is stable is currently an
implementation detail, and a future version of Swift could ship an
unstable algorithm instead).
I use this wrapper
extension Array where Element: Comparable, Element: AnyObject {
public func stableSorted() -> [Element] {
let array = self as NSArray
let result = array.sortedArray(options: .stable) { (left, right) -> ComparisonResult in
let left = left as! Element
let right = right as! Element
if left < right {
return ComparisonResult.orderedAscending
}
if left > right {
return ComparisonResult.orderedDescending
}
return ComparisonResult.orderedSame
}
return result as! [Element]
}
public func stableReversed() -> [Element] {
let array = self as NSArray
let result = array.sortedArray(options: .stable) { (left, right) -> ComparisonResult in
let left = left as! Element
let right = right as! Element
if left > right {
return ComparisonResult.orderedAscending
}
if left < right {
return ComparisonResult.orderedDescending
}
return ComparisonResult.orderedSame
}
return result as! [Element]
}
}
I appreciate the elegance of Tom's answer. Harking back to my Perl days I've reworked it to use ComparisonResult and the spaceship (<=>) operator:
extension Sequence {
func sorted(with comparator: (Element, Element) throws -> ComparisonResult) rethrows -> [Element]
{
return try enumerated()
.sorted { (try comparator($0.element, $1.element) || $0.offset <=> $1.offset) == .orderedAscending }
.map { $0.element }
}
}
extension Comparable {
static func <=> (lhs: Self, rhs: Self) -> ComparisonResult {
if lhs < rhs { return .orderedAscending }
if lhs > rhs { return .orderedDescending }
return .orderedSame
}
}
extension ComparisonResult {
static func || (lhs: Self, rhs: #autoclosure () -> Self) -> ComparisonResult {
if lhs == .orderedSame {
return rhs()
}
return lhs
}
}
I have a sorted array and want to do binary search on it.
So I'm asking if something is already available in Swift library like sort etc.? Or is there a type independend version available?
Of course I could write it by my own, but I like to avoid reinventing the wheel again.
Here's my favorite implementation of binary search. It's useful not only for finding the element but also for finding the insertion index. Details about assumed sorting order (ascending or descending) and behavior with respect to equal elements are controlled by providing a corresponding predicate (e.g. { $0 < x } vs { $0 > x } vs { $0 <= x } vs { $0 >= x }). The comment unambiguously says what exactly does it do.
extension RandomAccessCollection {
/// Finds such index N that predicate is true for all elements up to
/// but not including the index N, and is false for all elements
/// starting with index N.
/// Behavior is undefined if there is no such N.
func binarySearch(predicate: (Element) -> Bool) -> Index {
var low = startIndex
var high = endIndex
while low != high {
let mid = index(low, offsetBy: distance(from: low, to: high)/2)
if predicate(self[mid]) {
low = index(after: mid)
} else {
high = mid
}
}
return low
}
}
Example usage:
(0 ..< 778).binarySearch { $0 < 145 } // 145
Here's a generic way to use binary search:
func binarySearch<T:Comparable>(_ inputArr:Array<T>, _ searchItem: T) -> Int? {
var lowerIndex = 0
var upperIndex = inputArr.count - 1
while (true) {
let currentIndex = (lowerIndex + upperIndex)/2
if(inputArr[currentIndex] == searchItem) {
return currentIndex
} else if (lowerIndex > upperIndex) {
return nil
} else {
if (inputArr[currentIndex] > searchItem) {
upperIndex = currentIndex - 1
} else {
lowerIndex = currentIndex + 1
}
}
}
}
var myArray = [1,2,3,4,5,6,7,9,10]
if let searchIndex = binarySearch(myArray, 5) {
print("Element found on index: \(searchIndex)")
}
I use an extension on RandomAccessCollection implementing bisectToFirstIndex(where:) and taking a predicate.
It takes a test predicate, and returns the index of the first element to pass the test.
If there is no such index, it returns nil.
If the Collection is empty, it returns nil.
Example
let a = [1,2,3,4]
a.map{$0>=3}
// returns [false, false, true, true]
a.bisectToFirstIndex {$0>=3}
// returns 2
Important
You need to ensure test never returns a false for any index after an index it has said true for. This is equivalent to the usual precondition that binary search requires your data to be in order.
Specifically, you must not do a.bisectToFirstIndex {$0==3}. This will not work correctly.
Why?
bisectToFirstIndex is useful because it lets you find ranges of stuff in your data. By adjusting the test, you can find the lower and upper limits of "stuff".
Here's some data:
let a = [1,1,1, 2,2,2,2, 3, 4, 5]
We can find the Range of all the 2s like this…
let firstOf2s = a.bisectToFirstIndex { $ 0>= 2 }
let endOf2s = a.bisectToFirstIndex { $0 > 2 }
let rangeOf2s = firstOf2s ..< endOf2s
Example Application
I use this in an implementation of layoutAttributesForElementsInRect. My UICollectionViewCells are stored sorted vertically in an array. It's easy to write a pair of calls that will find all cells that are within a particular rectangle and exclude any others.
Code
extension RandomAccessCollection {
public func bisectToFirstIndex(where predicate: (Element) throws -> Bool) rethrows -> Index? {
var intervalStart = startIndex
var intervalEnd = endIndex
while intervalStart != intervalEnd {
let intervalLength = distance(from: intervalStart, to: intervalEnd)
guard intervalLength > 1 else {
return try predicate(self[intervalStart]) ? intervalStart : nil
}
let testIndex = index(intervalStart, offsetBy: (intervalLength - 1) / 2)
if try predicate(self[testIndex]) {
intervalEnd = index(after: testIndex)
}
else {
intervalStart = index(after: testIndex)
}
}
return nil
}
}
Updates…
The implementation here extends RandomAccessCollection and I've updated the code to build with the current Swift version (5 or something).
A Binary Search Caution
Binary searches are notoriously hard to correctly code. You really should read that link to find out just how common mistakes in their implementation are, but here is an extract:
When Jon Bentley assigned it as a problem in a course for professional programmers, he found that an astounding ninety percent failed to code a binary search correctly after several hours of working on it, and another study shows that accurate code for it is only found in five out of twenty textbooks. Furthermore, Bentley's own implementation of binary search, published in his 1986 book Programming Pearls, contains an error that remained undetected for over twenty years.
Given that last point, here is a test for this code. It passes! The testing isn't exhaustive – so there may certainly still be errors.
Tests
final class Collection_BisectTests: XCTestCase {
func test_bisect() {
for length in 0...100 {
let collection = 0 ... length
let targets = -4 ... length + 4
for toFind in targets {
let bisectIndex = collection.bisectToFirstIndex { $0 > toFind }
let expectIndex = collection.firstIndex { $0 > toFind }
XCTAssertEqual(bisectIndex, expectIndex, "Finding \(toFind+1) in 0...\(length)")
}
}
}
}
extension ArraySlice where Element: Comparable {
func binarySearch(_ value: Element) -> Int? {
guard !isEmpty else { return nil }
let midIndex = (startIndex + endIndex) / 2
if value == self[midIndex] {
return midIndex
} else if value > self[midIndex] {
return self[(midIndex + 1)...].binarySearch(value)
} else {
return self[..<midIndex].binarySearch(value)
}
}
}
extension Array where Element: Comparable {
func binarySearch(_ value: Element) -> Int? {
return self[0...].binarySearch(value)
}
}
This is, in my opinion, very readable and leverages the fact that Swift's ArraySlice is a view on Array and retains the same indexes as the original Array with which it shares the storage so, in absence of mutations (like in this case), it is therefore very efficient.
here is binary search using while syntax
func binarySearch<T: Comparable>(_ a: [T], key: T) -> Int? {
var lowerBound = 0
var upperBound = a.count
while lowerBound < upperBound {
let midIndex = lowerBound + (upperBound - lowerBound) / 2
if a[midIndex] == key {
return midIndex
} else if a[midIndex] < key {
lowerBound = midIndex + 1
} else {
upperBound = midIndex
}
}
return nil
}
Here is an implementation for a sorted array of strings.
var arr = ["a", "abc", "aabc", "aabbc", "aaabbbcc", "bacc", "bbcc", "bbbccc", "cb", "cbb", "cbbc", "d" , "defff", "deffz"]
func binarySearch(_ array: [String], value: String) -> String {
var firstIndex = 0
var lastIndex = array.count - 1
var wordToFind = "Not founded"
var count = 0
while firstIndex <= lastIndex {
count += 1
let middleIndex = (firstIndex + lastIndex) / 2
let middleValue = array[middleIndex]
if middleValue == value {
wordToFind = middleValue
return wordToFind
}
if value.localizedCompare(middleValue) == ComparisonResult.orderedDescending {
firstIndex = middleIndex + 1
}
if value.localizedCompare(middleValue) == ComparisonResult.orderedAscending {
print(middleValue)
lastIndex = middleIndex - 1
}
}
return wordToFind
}
//print d
print(binarySearch(arr, value: "d"))
Another implementation: if you want to have your structs or classes searchable without making them Comparable, make them BinarySearchable instead:
public protocol BinarySearchable {
associatedtype C: Comparable
var searchable: C { get }
}
public extension Array where Element: BinarySearchable {
func binarySearch(_ prefix: Element.C) -> Index {
var low = 0
var high = count
while low != high {
let mid = (low + high) / 2
if self[mid].searchable < prefix {
low = mid + 1
} else {
high = mid
}
}
return low
}
}
Example usage for a struct that should be sorted and searched by name:
struct Country: BinraySearchable {
var code: String
var name: String
var searchable: String { name }
}
// Suppose you have a list of countries sorted by `name`, you want to find
// the index of the first country whose name starts with "United", others
// will follow:
let index = listOfCountries.binarySearch("United")
And for completeness, here's a entirely pattern matching based implementation:
extension Collection where Element: Comparable {
func binarySearch(for element: Element) -> Index? {
switch index(startIndex, offsetBy: distance(from: startIndex, to: endIndex) / 2) {
case let i where i >= endIndex: return nil
case let i where self[i] == element: return i
case let i where self[i] > element: return self[..<i].binarySearch(for: element)
case let i: return self[index(after: i)..<endIndex].binarySearch(for: element)
}
}
}
The above code should work with any kind of collections, sliced or not sliced, zero offset-ed or non-zero offset-ed.
Here is how you create a binary search function in swift 5, in this example I assume that the item you are looking for is guaranteed to be in the list, however if your item is not guaranteed to be in the list then you can run this code to check first:
yourList.contains(yourItem) //will return true or false
Here is the binary search function:
override func viewDidLoad() {
super.viewDidLoad()
print(binarySearch(list: [1, 2, 4, 5, 6], num: 6)) //returns 4
}
func binarySearch(list: [Int], num: Int) -> Int //returns index of num
{
var firstIndex = 0
var lastIndex = list.count - 1
var middleIndex = (firstIndex + lastIndex) / 2
var middleValue = list[middleIndex]
while true //loop until we find the item we are looking for
{
middleIndex = (firstIndex + lastIndex) / 2 //getting the list's middle index
middleValue = list[middleIndex]
if middleValue > num
{
lastIndex = middleIndex - 1 //get the left side of the remaining list
}
else if middleValue < num
{
firstIndex = middleIndex + 1 //get the right side of the remaining list
}
else if middleValue == num
{
break //found the correct value so we can break out of the loop
}
}
return middleIndex
}
I have made a youtube video explaining this here
Here's a better implementation that returns more than one index, if there are more than 1 in the array.
extension Array where Element: Comparable {
/* Array Must be sorted */
func binarySearch(key: Element) -> [Index]? {
return self.binarySearch(key, initialIndex: 0)
}
private func binarySearch(key: Element, initialIndex: Index) -> [Index]? {
guard count > 0 else { return nil }
let midIndex = count / 2
let midElement = self[midIndex]
if key == midElement {
// Found!
let foundIndex = initialIndex + midIndex
var indexes = [foundIndex]
// Check neighbors for same values
// Check Left Side
var leftIndex = midIndex - 1
while leftIndex >= 0 {
//While there is still more items on the left to check
print(leftIndex)
if self[leftIndex] == key {
//If the items on the left is still matching key
indexes.append(leftIndex + initialIndex)
leftIndex--
} else {
// The item on the left is not identical to key
break
}
}
// Check Right side
var rightIndex = midIndex + 1
while rightIndex < count {
//While there is still more items on the left to check
if self[rightIndex] == key {
//If the items on the left is still matching key
indexes.append(rightIndex + initialIndex)
rightIndex++
} else {
// The item on the left is not identical to key
break
}
}
return indexes.sort{ return $0 < $1 }
}
if count == 1 {
guard let first = first else { return nil }
if first == key {
return [initialIndex]
}
return nil
}
if key < midElement {
return Array(self[0..<midIndex]).binarySearch(key, initialIndex: initialIndex + 0)
}
if key > midElement {
return Array(self[midIndex..<count]).binarySearch(key, initialIndex: initialIndex + midIndex)
}
return nil
}
}
By recursive binary search,
func binarySearch(data : [Int],search: Int,high : Int,low:Int) -> Int? {
if (low > high)
{
return nil
}
let mid = low + (low + high)/2
if (data[mid] == search) {
return mid
}
else if (search < data[mid]){
return binarySearch(data: data, search: search, high: high-1, low: low)
}else {
return binarySearch(data: data, search: search, high: high, low: low+1)
}
}
Input : let arry = Array(0...5) // [0,1,2,3,4,5]
print(binarySearch(data: arry, search: 0, high: arry.count-1, low: 0))
Here's a full example with several test cases for Swift 3.1. There is no chance that this is faster than the default implementation, but that's not the point. Array extension is at the bottom:
// BinarySearchTests.swift
// Created by Dan Rosenstark on 3/27/17
import XCTest
#testable import SwiftAlgos
class BinarySearchTests: XCTestCase {
let sortedArray : [Int] = [-25, 1, 2, 4, 6, 8, 10, 14, 15, 1000]
func test5() {
let traditional = sortedArray.index(of: 5)
let newImplementation = sortedArray.indexUsingBinarySearch(of: 5)
XCTAssertEqual(traditional, newImplementation)
}
func testMembers() {
for item in sortedArray {
let traditional = sortedArray.index(of: item)
let newImplementation = sortedArray.indexUsingBinarySearch(of: item)
XCTAssertEqual(traditional, newImplementation)
}
}
func testMembersAndNonMembers() {
for item in (-100...100) {
let traditional = sortedArray.index(of: item)
let newImplementation = sortedArray.indexUsingBinarySearch(of: item)
XCTAssertEqual(traditional, newImplementation)
}
}
func testSingleMember() {
let sortedArray = [50]
for item in (0...100) {
let traditional = sortedArray.index(of: item)
let newImplementation = sortedArray.indexUsingBinarySearch(of: item)
XCTAssertEqual(traditional, newImplementation)
}
}
func testEmptyArray() {
let sortedArray : [Int] = []
for item in (0...100) {
let traditional = sortedArray.index(of: item)
let newImplementation = sortedArray.indexUsingBinarySearch(of: item)
XCTAssertEqual(traditional, newImplementation)
}
}
}
extension Array where Element : Comparable {
// self must be a sorted Array
func indexUsingBinarySearch(of element: Element) -> Int? {
guard self.count > 0 else { return nil }
return binarySearch(for: element, minIndex: 0, maxIndex: self.count - 1)
}
private func binarySearch(for element: Element, minIndex: Int, maxIndex: Int) -> Int? {
let count = maxIndex - minIndex + 1
// if there are one or two elements, there is no futher recursion:
// stop and check one or both values (and return nil if neither)
if count == 1 {
return element == self[minIndex] ? minIndex : nil
} else if count == 2 {
switch element {
case self[minIndex]: return minIndex
case self[maxIndex]: return maxIndex
default: return nil
}
}
let breakPointIndex = Int(round(Double(maxIndex - minIndex) / 2.0)) + minIndex
let breakPoint = self[breakPointIndex]
let splitUp = (breakPoint < element)
let newMaxIndex : Int = splitUp ? maxIndex : breakPointIndex
let newMinIndex : Int = splitUp ? breakPointIndex : minIndex
return binarySearch(for: element, minIndex: newMinIndex, maxIndex: newMaxIndex)
}
}
This is quite homemade, so... caveat emptor. It does work and does do binary search.
Simple solution in Swift 5:
func binarySerach(list: [Int], item: Int) -> Int? {
var low = 0
var high = list.count - 1
while low <= high {
let mid = (low + high) / 2
let guess = list[mid]
if guess == item {
return mid
} else if guess > item {
high = mid - 1
} else {
low = mid + 1
}
}
return nil
}
let myList = [1,3,4,7,9]
print(binarySerach(list: myList, item: 9))
//Optional(4)
Details
Swift 5.2, Xcode 11.4 (11E146)
Solution
import Foundation
extension RandomAccessCollection where Element: Comparable {
private func binarySearchIteration(forIndexOf value: Element, in range: Range<Index>? = nil,
valueDetected: ((Index, _ in: Range<Index>) -> Index?)) -> Index? {
let range = range ?? startIndex..<endIndex
guard range.lowerBound < range.upperBound else { return nil }
let size = distance(from: range.lowerBound, to: range.upperBound)
let middle = index(range.lowerBound, offsetBy: size / 2)
switch self[middle] {
case value: return valueDetected(middle, range) ?? middle
case ..<value: return binarySearch(forIndexOf: value, in: index(after: middle)..<range.upperBound)
default: return binarySearch(forIndexOf: value, in: range.lowerBound..<middle)
}
}
func binarySearch(forIndexOf value: Element, in range: Range<Index>? = nil) -> Index? {
binarySearchIteration(forIndexOf: value, in: range) { currentIndex, _ in currentIndex }
}
func binarySearch(forFirstIndexOf value: Element, in range: Range<Index>? = nil) -> Index? {
binarySearchIteration(forIndexOf: value, in: range) { currentIndex, range in
binarySearch(forFirstIndexOf: value, in: range.lowerBound..<currentIndex)
}
}
func binarySearch(forLastIndexOf value: Element, in range: Range<Index>? = nil) -> Index? {
binarySearchIteration(forIndexOf: value, in: range) { currentIndex, range in
binarySearch(forFirstIndexOf: value, in: index(after: currentIndex)..<range.upperBound)
}
}
func binarySearch(forIndicesRangeOf value: Element, in range: Range<Index>? = nil) -> Range<Index>? {
let range = range ?? startIndex..<endIndex
guard range.lowerBound < range.upperBound else { return nil }
guard let currentIndex = binarySearchIteration(forIndexOf: value, in: range, valueDetected: { index, _ in index
}) else { return nil }
let firstIndex = binarySearch(forFirstIndexOf: value, in: range.lowerBound ..< index(after: currentIndex)) ?? currentIndex
let lastIndex = binarySearch(forFirstIndexOf: value, in: index(after: currentIndex) ..< range.upperBound) ?? currentIndex
return firstIndex..<index(after: lastIndex)
}
}
Usage
//let array = ["one", "two", "three", "three", "three", "three", "three", "four", "five", "five"]
//let value = "three"
let array = [1, 2, 3, 3, 3, 3, 3, 4, 5, 5]
let value = 3
print(array.binarySearch(forFirstIndexOf: value))
print(array.binarySearch(forLastIndexOf: value))
print(array.binarySearch(forIndicesRangeOf: value))
Tests
protocol _BinarySearchTestable: class where Collection: RandomAccessCollection, Collection.Element: Comparable {
associatedtype Collection
var array: Collection! { get set }
var elementToSearch: Collection.Element! { get set }
func testFindFirstIndexOfValueInCollection()
func testFindLastIndexOfValueInCollection()
func testFindIndicesRangeOfValueInCollection()
}
extension _BinarySearchTestable where Self: XCTest {
typealias Element = Collection.Element
typealias Index = Collection.Index
func _testFindFirstIndexOfValueInCollection() {
_testfindFirstIndex(comparableArray: array, testableArray: array)
}
func _testFindLastIndexOfValueInCollection() {
let index1 = array.lastIndex(of: elementToSearch)
let index2 = array.binarySearch(forLastIndexOf: elementToSearch)
_testElementsAreEqual(indexInComparableArray: index1, comparableArray: array,
indexInTestableArray: index2, testableArray: array)
}
func _testFindIndicesRangeOfValueInCollection() {
var range1: Range<Index>?
if let firstIndex = array.firstIndex(of: elementToSearch),
let lastIndex = array.lastIndex(of: elementToSearch) {
range1 = firstIndex ..< array.index(after: lastIndex)
}
let range2 = array.binarySearch(forIndicesRangeOf: elementToSearch)
XCTAssertEqual(range1, range2)
}
private func _testElementsAreEqual(indexInComparableArray: Index?, comparableArray: Collection,
indexInTestableArray: Index?, testableArray: Collection) {
XCTAssertEqual(indexInComparableArray, indexInTestableArray)
var valueInComparableArray: Element?
if let index = indexInComparableArray { valueInComparableArray = comparableArray[index] }
var valueInTestableArray: Element?
if let index = indexInComparableArray { valueInTestableArray = testableArray[index] }
XCTAssertEqual(valueInComparableArray, valueInTestableArray)
}
private func _testfindFirstIndex(comparableArray: Collection, testableArray: Collection) {
let index1 = comparableArray.firstIndex(of: elementToSearch)
let index2 = testableArray.binarySearch(forFirstIndexOf: elementToSearch)
_testElementsAreEqual(indexInComparableArray: index1, comparableArray: comparableArray,
indexInTestableArray: index2, testableArray: testableArray)
}
}
class TestsInEmptyArray: XCTestCase, _BinarySearchTestable {
var array: [String]!
var elementToSearch: String!
override func setUp() {
array = []
elementToSearch = "value"
}
func testFindFirstIndexOfValueInCollection() { _testFindFirstIndexOfValueInCollection() }
func testFindLastIndexOfValueInCollection() { _testFindLastIndexOfValueInCollection() }
func testFindIndicesRangeOfValueInCollection() { _testFindIndicesRangeOfValueInCollection() }
}
class TestsInArray: XCTestCase, _BinarySearchTestable {
var array: [Int]!
var elementToSearch: Int!
override func setUp() {
array = [1, 2, 3, 3, 3, 3, 3, 4, 5, 5]
elementToSearch = 3
}
func testFindFirstIndexOfValueInCollection() { _testFindFirstIndexOfValueInCollection() }
func testFindLastIndexOfValueInCollection() { _testFindLastIndexOfValueInCollection() }
func testFindIndicesRangeOfValueInCollection() { _testFindIndicesRangeOfValueInCollection() }
}
class TestsInArrayWithOneElement: XCTestCase, _BinarySearchTestable {
var array: [Date]!
var elementToSearch: Date!
override func setUp() {
let date = Date()
array = [date]
elementToSearch = date
}
func testFindFirstIndexOfValueInCollection() { _testFindFirstIndexOfValueInCollection() }
func testFindLastIndexOfValueInCollection() { _testFindLastIndexOfValueInCollection() }
func testFindIndicesRangeOfValueInCollection() { _testFindIndicesRangeOfValueInCollection() }
}