how it is possible to sort an array by multiple conditions.
struct UserInformationModel: Identifiable, Hashable {
let id = UUID()
var isVip: Bool
let userIsMale: Bool
let userName: String
let age: Int
let userCountry: String
let countryIsoCode: String
let uid: String
}
And the view model contain the code:
class GetUserInformationViewModel: ObservableObject {
#Published var allUsers = [UserInformationModel]()
fun sortmyarray(){
self.allUsers = self.allUsers.sorted(by: {$0.isVip && !$1.isVip})
}
how its possible to sort first the vip users, and then sort by age and then country?
Here is a simple way to sort on multiple properties (I have assumed a sort order here for each property since it isn't mentioned in the question)
let sorted = users.sorted {
if $0.isVip == $1.isVip {
if $0.age == $1.age {
return $0.userCountry < $1.userCountry
} else {
return $0.age < $1.age
}
}
return $0.isVip && !$1.isVip
}
If the above is the natural sort order for the type then you could let the type implement Comparable and implement the < func
struct UserInformationModel: Identifiable, Hashable, Comparable {
//properties removed for brevity
static func < (lhs: UserInformationModel, rhs: UserInformationModel) -> Bool {
if lhs.isVip == rhs.isVip {
if lhs.age == rhs.age {
return lhs.userCountry < rhs.userCountry
} else {
return lhs.age < rhs.age
}
}
return lhs.isVip && !rhs.isVip
}
}
and then the sorting would be
let sorted = users.sorted()
Use tuples.
allUsers.sorted {
($1.isVip.comparable, $0.age, $0.userCountry)
<
($0.isVip.comparable, $1.age, $1.userCountry)
}
public extension Bool {
/// A way to compare `Bool`s.
///
/// Note: `false` is "less than" `true`.
enum Comparable: CaseIterable, Swift.Comparable {
case `false`, `true`
}
/// Make a `Bool` `Comparable`, with `false` being "less than" `true`.
var comparable: Comparable { .init(booleanLiteral: self) }
}
extension Bool.Comparable: ExpressibleByBooleanLiteral {
public init(booleanLiteral value: Bool) {
self = value ? .true : .false
}
}
I have to write a function in Swift where I need create a list of int if the counter is different to 0.
Here is my code in Swift :
func compteurZeros(cellule:[[Int]]) -> (compteur:Int, liste:(Int))
{
var compteur = 0;
for i in 0...2
{
for j in 0...2
{
if(cellule[i][j] == 0)
{
compteur = compteur+1
}
}
}
print(compteur)
if (compteur != 0)
{
for i in 0..<(compteur) {
var liste:(Int)
}
}
return (compteur, liste!)
}
But, in the return, I've got this error :
error: use of unresolved identifier 'liste'
return (compteur, liste!)
Thank you for your answer.
You need to change your function somewhat so that it returns an [[Int]] as the second element in the tuple
func compteurZeros(cellule:[[Int]]) -> (compteur:Int, liste:[[Int]]) {
var compteur = 0;
for i in 0...2
{
for j in 0...2
{
if(cellule[i][j] == 0)
{
compteur = compteur+1
}
}
}
var liste = [[Int]]()
if (compteur != 0)
{
for i in 0..<(compteur) {
liste.append([])
}
}
return (compteur, liste)
}
Now this code is not the best so it can be improved by skipping the loops
func compteurZeros(cellule:[[Int]]) -> (compteur:Int, liste:[[Int]]) {
let compteur = cellule.flatMap {$0}.filter {$0 == 0}.count
let liste:[[Int]] = Array(repeating: [], count: compteur)
return (compteur, liste)
}
Since the value compteur is redundant this can be further simplified to
func compteurZeros(cellule:[[Int]]) -> [[Int]] {
return Array(repeating: [], count: cellule.flatMap {$0}.filter {$0 == 0}.count)
}
im struggling in converting this function to Swift 4
func buildIndex(records: [Record]) -> [[Record]] {
var g = [NSDate: [Record]]()
for e in records {
if (g[e.time] == nil) {
g[e.time] = []
}
g[e.time]!.append(e) // grouping by `time`
}
return sorted(g.keys) { (a: NSDate, b: NSDate) in
a.compare(b) == .OrderedAscending // sorting the outer array by time
}
// sorting the inner arrays by `name`
.map { sorted(g[$0]!) { $0.name < $1.name } }
}
I hope someone can help me.
There's a lot to clean up here. The variable names are obviously a disaster, but so is the forced unwrapping. All that can be replaced with a simple call to Dictionary.init(grouping:by:)
import Foundation
struct Record {
let time: Date
let name: String
}
func buildIndex(records: [Record]) -> Any {
return Dictionary(grouping: records, by: { $0.time })
.mapValues { $0.sorted { $0.name < $1.name} }
.sorted { (pair1, pair2) in pair1.key < pair2.key }
}
Try it
func buildIndex(records: [Record]) -> [[Record]] {
var g = [NSDate: [Record]]()
for e in records {
if (g[e.time] == nil) {
g[e.time] = []
}
g[e.time]!.append(e) // grouping by `time`
}
return g.keys.sorted { (a: NSDate, b: NSDate) in
a.compare(b as Date) == .orderedAscending // sorting the outer array by time
}
// sorting the inner arrays by `name`
.map { g[$0]!.sorted { $0.name < $1.name } }
}
Hope it helped you.
Can anyone explain me the backtracking algorithm? I'm trying to make a permutation program for example if n=2, program will print
1,2
2,1
Although I did this in high-school the teacher didn't really care about us hard enough for us to understand it and I didn't really understand it, neither the algorithms on the internet.
This is what I did so far in the playground but I find myself stuck. I get
1
0
1
0
import Foundation
let n = 2
var tempMatrix : [Int] = []
var resultString : String = ""
//1,2 || 2,1
//0,0
func initTempMatrix(){
for _ in 0..<n {
tempMatrix.append(0)
}
}
func resetMatrixValue(val: Int) {
tempMatrix[val] = 0
}
func isSolution() -> Bool {
for i in 0..<n-1 {
if tempMatrix[i] == 0 {
return false
}
}
return true
}
func isValid() -> Bool {
for i in 0..<n-1 {
if tempMatrix[n-1] == tempMatrix[i] {
return false
}
}
return true
}
func increaseMatrix(val: Int) -> Bool {
if tempMatrix[val] < n {
tempMatrix[val] += 1
return true
}
return false
}
func backtrack() {
initTempMatrix()
var k = 1
while k != 0 {
while !isSolution() {
if increaseMatrix(val: k-1) {
k+=1
}
}
if isValid() {
for i in 0..<tempMatrix.count {
resultString += "\(tempMatrix[i])"
print(resultString)
resultString = ""
}
}
resetMatrixValue(val: k-1)
k -= 1
}
}
backtrack()
I like many of the features in Swift, but using manipulating strings are still a big pain in the ass.
func checkPalindrome(word: String) -> Bool {
print(word)
if word == "" {
return true
} else {
if word.characters.first == word.characters.last {
return checkPalindrome(word.substringWithRange(word.startIndex.successor() ..< word.endIndex.predecessor()))
} else {
return false
}
}
}
This code fails miserably whenever the string's length is an odd number. Of course I could make it so the first line of the block would be if word.characters.count < 2, but is there a way in Swift to get substrings and check easily?
Update
I like many of the suggestions, but I guess the original question could be misleading a little, since it's a question about String more than getting the right results for the function.
For instance, in Python, checkPalindrome(word[1:-1]) would work fine for the recursive definition, whereas Swift code is much less graceful since it needs other bells and whistles.
return word == String(word.reversed())
func isPalindrome(myString:String) -> Bool {
let reverseString = String(myString.characters.reversed())
if(myString != "" && myString == reverseString) {
return true
} else {
return false
}
}
print(isPalindrome("madam"))
I have used the below extension to find whether the number is Palindrome or Not.
extension String {
var isPalindrome: Bool {
return self == String(self.reversed())
}
}
Sometimes having a front end for a recursion can simplify life. I sometimes do this when the arguments which are most convenient to use are not what I want in the user interface.
Would the following meet your needs?
func checkPalindrome(str: String) -> Bool {
func recursiveTest(var charSet: String.CharacterView) -> Bool {
if charSet.count < 2 {
return true
} else {
if charSet.popFirst() != charSet.popLast() {
return false
} else {
return recursiveTest(charSet)
}
}
}
return recursiveTest(str.characters)
}
just add on more condition in if
func checkPalindrome(word: String) -> Bool {
print(word)
if (word == "" || word.characters.count == 1){
return true
}
else {
if word.characters.first == word.characters.last {
return checkPalindrome(word.substringWithRange(word.startIndex.successor() ..< word.endIndex.predecessor()))
} else {
return false
}
}
}
extension StringProtocol where Self: RangeReplaceableCollection {
var letters: Self { filter(\.isLetter) }
var isPalindrome: Bool {
let letters = self.letters
return String(letters.reversed()).caseInsensitiveCompare(letters) == .orderedSame
}
}
"Dammit I'm Mad".isPalindrome // true
"Socorram-me subi no onibus em marrocos".isPalindrome // true
You can also break your string into an array of characters and iterate through them until its half comparing one by one with its counterpart:
func checkPalindrome(_ word: String) -> Bool {
let chars = Array(word.letters.lowercased())
for index in 0..<chars.count/2 {
if chars[index] != chars[chars.count - 1 - index] {
return false
}
}
return true
}
And the recursive version fixing the range issue where can't form a range with endIndex < startIndex:
func checkPalindrome<T: StringProtocol>(_ word: T) -> Bool {
let word = word.lowercased()
.components(separatedBy: .punctuationCharacters).joined()
.components(separatedBy: .whitespacesAndNewlines).joined()
if word == "" || word.count == 1 {
return true
} else {
if word.first == word.last {
let start = word.index(word.startIndex,offsetBy: 1, limitedBy: word.endIndex) ?? word.startIndex
let end = word.index(word.endIndex,offsetBy: -1, limitedBy: word.startIndex) ?? word.endIndex
return checkPalindrome(word[start..<end])
} else {
return false
}
}
}
checkPalindrome("Dammit I'm Mad")
I think if you make an extension to String like this one then it will make your life easier:
extension String {
var length: Int { return characters.count }
subscript(index: Int) -> Character {
return self[startIndex.advancedBy(index)]
}
subscript(range: Range<Int>) -> String {
return self[Range<Index>(start: startIndex.advancedBy(range.startIndex), end: startIndex.advancedBy(range.endIndex))]
}
}
With it in place, you can change your function to this:
func checkPalindrome(word: String) -> Bool {
if word.length < 2 {
return true
}
if word.characters.first != word.characters.last {
return false
}
return checkPalindrome(word[1..<word.length - 1])
}
Quick test:
print(checkPalindrome("aba")) // Prints "true"
print(checkPalindrome("abc")) // Prints "false"
extension String {
func trimmingFirstAndLastCharacters() -> String {
guard let startIndex = index(self.startIndex, offsetBy: 1, limitedBy: self.endIndex) else {
return self
}
guard let endIndex = index(self.endIndex, offsetBy: -1, limitedBy: self.startIndex) else {
return self
}
guard endIndex >= startIndex else {
return self
}
return String(self[startIndex..<endIndex])
}
var isPalindrome: Bool {
guard count > 1 else {
return true
}
return first == last && trimmingFirstAndLastCharacters().isPalindrome
}
}
We first declare a function that removes first and last characters from a string.
Next we declare a computer property which will contain the actual recursive code that checks if a string is palindrome.
If string's size is less than or equal 1 we immediately return true (strings composed by one character like "a" or the empty string "" are considered palindrome), otherwise we check if first and last characters of the string are the same and we recursively call isPalindrome on the current string deprived of the first and last characters.
Convert the string into an Array. When the loop is executed get the first index and compare with the last index.
func palindrome(string: String)-> Bool{
let char = Array(string)
for i in 0..<char.count / 2 {
if char[i] != char[char.count - 1 - i] {
return false
}
}
return true
}
This solution is not recursive, but it is a O(n) pure index based solution without filtering anything and without creating new objects. Non-letter characters are ignored as well.
It uses two indexes and walks outside in from both sides.
I admit that the extension type and property name is stolen from Leo, I apologize. 😉
extension StringProtocol where Self: RangeReplaceableCollection {
var isPalindrome : Bool {
if isEmpty { return false }
if index(after: startIndex) == endIndex { return true }
var forward = startIndex
var backward = endIndex
while forward < backward {
repeat { formIndex(before: &backward) } while !self[backward].isLetter
if self[forward].lowercased() != self[backward].lowercased() { return false }
repeat { formIndex(after: &forward) } while !self[forward].isLetter
}
return true
}
}
Wasn't really thinking of this, but I think I came up with a pretty cool extension, and thought I'd share.
extension String {
var subString: (Int?) -> (Int?) -> String {
return { (start) in
{ (end) in
let startIndex = start ?? 0 < 0 ? self.endIndex.advancedBy(start!) : self.startIndex.advancedBy(start ?? 0)
let endIndex = end ?? self.characters.count < 0 ? self.endIndex.advancedBy(end!) : self.startIndex.advancedBy(end ?? self.characters.count)
return startIndex > endIndex ? "" : self.substringWithRange(startIndex ..< endIndex)
}
}
}
}
let test = ["Eye", "Pop", "Noon", "Level", "Radar", "Kayak", "Rotator", "Redivider", "Detartrated", "Tattarrattat", "Aibohphobia", "Eve", "Bob", "Otto", "Anna", "Hannah", "Evil olive", "Mirror rim", "Stack cats", "Doom mood", "Rise to vote sir", "Step on no pets", "Never odd or even", "A nut for a jar of tuna", "No lemon, no melon", "Some men interpret nine memos", "Gateman sees name, garageman sees nametag"]
func checkPalindrome(word: String) -> Bool {
if word.isEmpty { return true }
else {
if word.subString(nil)(1) == word.subString(-1)(nil) {
return checkPalindrome(word.subString(1)(-1))
} else {
return false
}
}
}
for item in test.map({ $0.lowercaseString.stringByReplacingOccurrencesOfString(",", withString: "").stringByReplacingOccurrencesOfString(" ", withString: "") }) {
if !checkPalindrome(item) {
print(item)
}
}
A simple solution in Swift:
func isPalindrome(word: String) -> Bool {
// If no string found, return false
if word.count == 0 { return false }
var index = 0
var characters = Array(word) // make array of characters
while index < characters.count / 2 { // repeat loop only for half length of given string
if characters[index] != characters[(characters.count - 1) - index] {
return false
}
index += 1
}
return true
}
func checkPalindrome(_ inputString: String) -> Bool {
if inputString.count % 2 == 0 {
return false
} else if inputString.count == 1 {
return true
} else {
var stringCount = inputString.count
while stringCount != 1 {
if inputString.first == inputString.last {
stringCount -= 2
} else {
continue
}
}
if stringCount == 1 {
return true
} else {
return false
}
}
}