Reset password call occurs 2 times, how can I remove the subscription in that block.
let digits = [firstDigit, secondDigit, thirdDigit, fourthDigit, fifthDigit, sixthDigit]
let digitsValid = digits.map { $0?.textField.rx.text.orEmpty.map({ $0.count == 1 }).share(replay: 1) }
let allDigitsFilled = Observable.combineLatest(digitsValid.map{ $0 ?? BehaviorRelay(value: false).asObservable()}).share(replay: 1)
allDigitsFilled.subscribe (onNext: { [weak self] (boolArray) in
guard let self = self else { return }
let isFilled = boolArray.allSatisfy({ $0 })
if isFilled {
self.viewModel.resetPassword()
}
}).disposed(by: disposeBag)
Your onNext block will be called every time any of the text fields changes its contents. Your first goal should be to make it so the block is only called when you want to reset the password.
So first, you want to put the .allSatisfy inside an Observable.filter call so your onNext block will only be called when all six text fields contain exactly 1 character. Once you do that, then you can simply use .take(1) which will complete the subscription once a next value is emitted.
Something like this should do it, and looks quite a bit cleaner:
let digits = [firstDigit, secondDigit, thirdDigit, fourthDigit, fifthDigit, sixthDigit]
let texts = digits.compactMap { $0?.textField.rx.text.orEmpty }
Observable.combineLatest(texts)
.filter { $0.allSatisfy { $0.count == 1 } }
.map { _ in }
.take(1)
.subscribe(onNext: { [viewModel] in
viewModel?.resetPassword()
})
.disposed(by: disposeBag)
In the first example, my terminal outputs
next(1)
next(2)
next(3)
next(4)
next(5)
next(6)
next(7)
next(8)
next(9)
next(10)
completed
In the second example it will output
next(3)
next(6)
next(8)
next(9)
completed
I know the values are different in each, but I would expect the first example to have had filtering applied so it follows the logic within my .skipWhile { $0 % 2 == 0 } block
func skipWhile() {
let bag = DisposeBag()
Observable
.from(Array(1...10))
.skipWhile { $0 % 2 == 0 }
.subscribe { print($0) }
.disposed(by: bag)
Observable
.from([2,3,6,8,9])
.skipWhile { $0 % 2 == 0 }
.subscribe { print($0) }
.disposed(by: bag)
}
skipWhile()
skipWhile is not filter. It skips elements at the start of the observable's lifetime while the predicate as true. As soon as an element comes along that no longer satisfies the predicate, it opens the flood gates and let everything else through.
Your first observable says "skip everything until the first odd number". The first element is odd, so nothing is skipped, and that's why you see all array elements being printed.
If you notice in your second observable, you didn't filter out even numbers (because there's an 8). You merely skipped over elements until the first odd number (3), causing the 2 to be skipped.
On a side note
Int.isMultiple(of: ) has been added in Swift 5, and I suggest you use it in cases like this. It just makes it clearer, and side-steps errors caused by misreading == vs !=.
Observable
.from(Array(1...10))
.skipWhile { $0.isMultiple(of: 2) }
.subscribe { print($0) }
.disposed(by: bag)
You could even name your predicate:
let isEven: (Int) -> Bool = { $0.isMultiple(of: 2) }
Observable
.from(Array(1...10))
.skipWhile(isEven)
.subscribe { print($0) }
.disposed(by: bag)
Or my favourite, add it as a computed property:
extension BinaryInteger {
var isEven: Bool { return self.isMultiple(of: 2) }
var isOdd: Bool { return !self.isEven }
}
Observable
.from(Array(1...10))
.skipWhile(\.isEven)
.subscribe { print($0) }
.disposed(by: bag)
In Swift, how can I check if an element exists in an array? Xcode does not have any suggestions for contain, include, or has, and a quick search through the book turned up nothing. Any idea how to check for this? I know that there is a method find that returns the index number, but is there a method that returns a boolean like ruby's #include??
Example of what I need:
var elements = [1,2,3,4,5]
if elements.contains(5) {
//do something
}
Swift 2, 3, 4, 5:
let elements = [1, 2, 3, 4, 5]
if elements.contains(5) {
print("yes")
}
contains() is a protocol extension method of SequenceType (for sequences of Equatable elements) and not a global method as in
earlier releases.
Remarks:
This contains() method requires that the sequence elements
adopt the Equatable protocol, compare e.g. Andrews's answer.
If the sequence elements are instances of a NSObject subclass
then you have to override isEqual:, see NSObject subclass in Swift: hash vs hashValue, isEqual vs ==.
There is another – more general – contains() method which does not require the elements to be equatable and takes a predicate as an
argument, see e.g. Shorthand to test if an object exists in an array for Swift?.
Swift older versions:
let elements = [1,2,3,4,5]
if contains(elements, 5) {
println("yes")
}
For those who came here looking for a find and remove an object from an array:
Swift 1
if let index = find(itemList, item) {
itemList.removeAtIndex(index)
}
Swift 2
if let index = itemList.indexOf(item) {
itemList.removeAtIndex(index)
}
Swift 3, 4
if let index = itemList.index(of: item) {
itemList.remove(at: index)
}
Swift 5.2
if let index = itemList.firstIndex(of: item) {
itemList.remove(at: index)
}
Updated for Swift 2+
Note that as of Swift 3 (or even 2), the extension below is no longer necessary as the global contains function has been made into a pair of extension method on Array, which allow you to do either of:
let a = [ 1, 2, 3, 4 ]
a.contains(2) // => true, only usable if Element : Equatable
a.contains { $0 < 1 } // => false
Historical Answer for Swift 1:
Use this extension: (updated to Swift 5.2)
extension Array {
func contains<T>(obj: T) -> Bool where T: Equatable {
return !self.filter({$0 as? T == obj}).isEmpty
}
}
Use as:
array.contains(1)
If you are checking if an instance of a custom class or struct is contained in an array, you'll need to implement the Equatable protocol before you can use .contains(myObject).
For example:
struct Cup: Equatable {
let filled:Bool
}
static func ==(lhs:Cup, rhs:Cup) -> Bool { // Implement Equatable
return lhs.filled == rhs.filled
}
then you can do:
cupArray.contains(myCup)
Tip: The == override should be at the global level, not within your class/struct
I used filter.
let results = elements.filter { el in el == 5 }
if results.count > 0 {
// any matching items are in results
} else {
// not found
}
If you want, you can compress that to
if elements.filter({ el in el == 5 }).count > 0 {
}
Hope that helps.
Update for Swift 2
Hurray for default implementations!
if elements.contains(5) {
// any matching items are in results
} else {
// not found
}
(Swift 3)
Check if an element exists in an array (fulfilling some criteria), and if so, proceed working with the first such element
If the intent is:
To check whether an element exist in an array (/fulfils some boolean criteria, not necessarily equality testing),
And if so, proceed and work with the first such element,
Then an alternative to contains(_:) as blueprinted Sequence is to first(where:) of Sequence:
let elements = [1, 2, 3, 4, 5]
if let firstSuchElement = elements.first(where: { $0 == 4 }) {
print(firstSuchElement) // 4
// ...
}
In this contrived example, its usage might seem silly, but it's very useful if querying arrays of non-fundamental element types for existence of any elements fulfilling some condition. E.g.
struct Person {
let age: Int
let name: String
init(_ age: Int, _ name: String) {
self.age = age
self.name = name
}
}
let persons = [Person(17, "Fred"), Person(16, "Susan"),
Person(19, "Hannah"), Person(18, "Sarah"),
Person(23, "Sam"), Person(18, "Jane")]
if let eligableDriver = persons.first(where: { $0.age >= 18 }) {
print("\(eligableDriver.name) can possibly drive the rental car in Sweden.")
// ...
} // Hannah can possibly drive the rental car in Sweden.
let daniel = Person(18, "Daniel")
if let sameAgeAsDaniel = persons.first(where: { $0.age == daniel.age }) {
print("\(sameAgeAsDaniel.name) is the same age as \(daniel.name).")
// ...
} // Sarah is the same age as Daniel.
Any chained operations using .filter { ... some condition }.first can favourably be replaced with first(where:). The latter shows intent better, and have performance advantages over possible non-lazy appliances of .filter, as these will pass the full array prior to extracting the (possible) first element passing the filter.
Check if an element exists in an array (fulfilling some criteria), and if so, remove the first such element
A comment below queries:
How can I remove the firstSuchElement from the array?
A similar use case to the one above is to remove the first element that fulfils a given predicate. To do so, the index(where:) method of Collection (which is readily available to array collection) may be used to find the index of the first element fulfilling the predicate, whereafter the index can be used with the remove(at:) method of Array to (possible; given that it exists) remove that element.
var elements = ["a", "b", "c", "d", "e", "a", "b", "c"]
if let indexOfFirstSuchElement = elements.index(where: { $0 == "c" }) {
elements.remove(at: indexOfFirstSuchElement)
print(elements) // ["a", "b", "d", "e", "a", "b", "c"]
}
Or, if you'd like to remove the element from the array and work with, apply Optional:s map(_:) method to conditionally (for .some(...) return from index(where:)) use the result from index(where:) to remove and capture the removed element from the array (within an optional binding clause).
var elements = ["a", "b", "c", "d", "e", "a", "b", "c"]
if let firstSuchElement = elements.index(where: { $0 == "c" })
.map({ elements.remove(at: $0) }) {
// if we enter here, the first such element have now been
// remove from the array
print(elements) // ["a", "b", "d", "e", "a", "b", "c"]
// and we may work with it
print(firstSuchElement) // c
}
Note that in the contrived example above the array members are simple value types (String instances), so using a predicate to find a given member is somewhat over-kill, as we might simply test for equality using the simpler index(of:) method as shown in #DogCoffee's answer. If applying the find-and-remove approach above to the Person example, however, using index(where:) with a predicate is appropriate (since we no longer test for equality but for fulfilling a supplied predicate).
An array that contains a property that equals to
yourArray.contains(where: {$0.propertyToCheck == value })
Returns boolean.
The simplest way to accomplish this is to use filter on the array.
let result = elements.filter { $0==5 }
result will have the found element if it exists and will be empty if the element does not exist. So simply checking if result is empty will tell you whether the element exists in the array. I would use the following:
if result.isEmpty {
// element does not exist in array
} else {
// element exists
}
Swift 4/5
Another way to achieve this is with the filter function
var elements = [1,2,3,4,5]
if let object = elements.filter({ $0 == 5 }).first {
print("found")
} else {
print("not found")
}
As of Swift 2.1 NSArrays have containsObjectthat can be used like so:
if myArray.containsObject(objectImCheckingFor){
//myArray has the objectImCheckingFor
}
Array
let elements = [1, 2, 3, 4, 5, 5]
Check elements presence
elements.contains(5) // true
Get elements index
elements.firstIndex(of: 5) // 4
elements.firstIndex(of: 10) // nil
Get element count
let results = elements.filter { element in element == 5 }
results.count // 2
Just in case anybody is trying to find if an indexPath is among the selected ones (like in a UICollectionView or UITableView cellForItemAtIndexPath functions):
var isSelectedItem = false
if let selectedIndexPaths = collectionView.indexPathsForSelectedItems() as? [NSIndexPath]{
if contains(selectedIndexPaths, indexPath) {
isSelectedItem = true
}
}
if user find particular array elements then use below code same as integer value.
var arrelemnts = ["sachin", "test", "test1", "test3"]
if arrelemnts.contains("test"){
print("found") }else{
print("not found") }
Here is my little extension I just wrote to check if my delegate array contains a delegate object or not (Swift 2). :) It Also works with value types like a charm.
extension Array
{
func containsObject(object: Any) -> Bool
{
if let anObject: AnyObject = object as? AnyObject
{
for obj in self
{
if let anObj: AnyObject = obj as? AnyObject
{
if anObj === anObject { return true }
}
}
}
return false
}
}
If you have an idea how to optimize this code, than just let me know.
Swift
If you are not using object then you can user this code for contains.
let elements = [ 10, 20, 30, 40, 50]
if elements.contains(50) {
print("true")
}
If you are using NSObject Class in swift. This variables is according to my requirement. you can modify for your requirement.
var cliectScreenList = [ATModelLeadInfo]()
var cliectScreenSelectedObject: ATModelLeadInfo!
This is for a same data type.
{ $0.user_id == cliectScreenSelectedObject.user_id }
If you want to AnyObject type.
{ "\($0.user_id)" == "\(cliectScreenSelectedObject.user_id)" }
Full condition
if cliectScreenSelected.contains( { $0.user_id == cliectScreenSelectedObject.user_id } ) == false {
cliectScreenSelected.append(cliectScreenSelectedObject)
print("Object Added")
} else {
print("Object already exists")
}
what about using a hash table for the job, like this?
first, creating a "hash map" generic function, extending the Sequence protocol.
extension Sequence where Element: Hashable {
func hashMap() -> [Element: Int] {
var dict: [Element: Int] = [:]
for (i, value) in self.enumerated() {
dict[value] = i
}
return dict
}
}
This extension will work as long as the items in the array conform to Hashable, like integers or strings, here is the usage...
let numbers = Array(0...50)
let hashMappedNumbers = numbers.hashMap()
let numToDetect = 35
let indexOfnumToDetect = hashMappedNumbers[numToDetect] // returns the index of the item and if all the elements in the array are different, it will work to get the index of the object!
print(indexOfnumToDetect) // prints 35
But for now, let's just focus in check if the element is in the array.
let numExists = indexOfnumToDetect != nil // if the key does not exist
means the number is not contained in the collection.
print(numExists) // prints true
Swift 4.2 +
You can easily verify your instance is an array or not by the following function.
func verifyIsObjectOfAnArray<T>(_ object: T) -> Bool {
if let _ = object as? [T] {
return true
}
return false
}
Even you can access it as follows. You will receive nil if the object wouldn't be an array.
func verifyIsObjectOfAnArray<T>(_ object: T) -> [T]? {
if let array = object as? [T] {
return array
}
return nil
}
You can add an extension for Array as such:
extension Array {
func contains<T>(_ object: T) -> Bool where T: Equatable {
!self.filter {$0 as? T == object }.isEmpty
}
}
This can be used as:
if myArray.contains(myItem) {
// code here
}
let numbers = [1,3,4,5,5,9,0,1]
To find the first 5, use:
numbers.indexOf(5)
How do I find the second occurence?
List item
You can perform another search for the index of element at the remaining array slice as follow:
edit/update: Swift 5.2 or later
extension Collection where Element: Equatable {
/// Returns the second index where the specified value appears in the collection.
func secondIndex(of element: Element) -> Index? {
guard let index = firstIndex(of: element) else { return nil }
return self[self.index(after: index)...].firstIndex(of: element)
}
}
extension Collection {
/// Returns the second index in which an element of the collection satisfies the given predicate.
func secondIndex(where predicate: (Element) throws -> Bool) rethrows -> Index? {
guard let index = try firstIndex(where: predicate) else { return nil }
return try self[self.index(after: index)...].firstIndex(where: predicate)
}
}
Testing:
let numbers = [1,3,4,5,5,9,0,1]
if let index = numbers.secondIndex(of: 5) {
print(index) // "4\n"
} else {
print("not found")
}
if let index = numbers.secondIndex(where: { $0.isMultiple(of: 3) }) {
print(index) // "5\n"
} else {
print("not found")
}
Once you've found the first occurrence, you can use indexOf on the remaining slice of the array to locate the second occurrence:
let numbers = [1,3,4,5,5,9,0,1]
if let firstFive = numbers.indexOf(5) { // 3
let secondFive = numbers[firstFive+1..<numbers.count].indexOf(5) // 4
}
I don't think you can do it with indexOf. Instead you'll have to use a for-loop. A shorthand version:
let numbers = [1,3,4,5,5,9,0,1]
var indexes = [Int]()
numbers.enumerate().forEach { if $0.element == 5 { indexes += [$0.index] } }
print(indexes) // [3, 4]
Here's a general use extension of Array that will work for finding the nth element of a kind in any array:
extension Array where Element: Equatable {
// returns nil if there is no nth occurence
// or the index of the nth occurence if there is
func findNthIndexOf(n: Int, thing: Element) -> Int? {
guard n > 0 else { return nil }
var count = 0
for (index, item) in enumerate() where item == thing {
count += 1
if count == n {
return index
}
}
return nil
}
}
let numbers = [1,3,4,5,5,9,0]
numbers.findNthIndexOf(2, thing: 5) // returns 4
EDIT: as per #davecom's comment, I've included a similar but slightly more complex solution at the bottom of the answer.
I see a couple of good solutions here, especially considering the limitations the relatively new language of Swift. There is a really concise way to do it too, but beware...it is rather quick-and-dirty. May not be the perfect solution, but it is pretty quick. Also very versatile (not to brag).
extension Array where Element: Equatable {
func indexes(search: Element) -> [Int] {
return enumerate().reduce([Int]()) { $1.1 == search ? $0 + [$1.0] : $0 }
}
}
Using this extension, you could access the second index as follows:
let numbers = [1, 3, 4, 5, 5, 9, 0, 1]
let indexesOf5 = numbers.indexes(5) // [3, 4]
indexesOf5[1] // 4
And you're done!
Basically, the method works like this: enumerate() maps the array to tuples including the index of each element with the element itself. In this case, [1, 3, 4, 5, 5, 9, 0, 1].enumerate() returns a collection of the type EnumerateSequence<Array<Int>> which, translated to an Integer array, returns [(0,1), (1,3), (2,4), (3,5), (4,5), (5,9), (6,0), (7,1)].
The rest of the work is done using reduce (called 'inject' in some languages), which is an extremely powerful tool that many coders are not familiar with. If the reader is among those coders, I'd recommend checking out this article regarding use of the function in JS (keep in mind the placement of the non-block argument passed in is inputted after the block in JS, rather than before as seen here).
Thanks for reading.
P.S. not to be too long-winded on this relatively simple solution, but if the syntax for the indexes method shown above is a bit too quick-and-dirty, you could try something like this in the method body, where the closure's parameters are expanded for a bit more clarity:
return enumerate().reduce([Int]()) { memo, element in
element.1 == search ? memo + [element.0] : memo
}
EDIT: Here's another option that allows the implementer to scan for a specific "index at index" (e.g. the second occurrence of 5) for a more efficient solution.
extension Array where Element: Equatable {
func nIndex(search: Element, n: Int) -> Int? {
let info = enumerate().reduce((count: 0, index: 0), combine: { memo, element in
memo.count < n && element.1 == search ? (count: memo.count + 1, index: element.0) : memo
})
return info.count == n ? info.index : nil
}
}
[1, 3, 4, 5, 5, 9, 0, 1].nIndex(5, n: 2) // 4
[1, 3, 4, 5, 5, 9, 0, 1].nIndex(5, n: 3) // nil
The new method still iterates over the entire array, but is much more efficient due to the lack of "array-building" in the previous method. That performance hit would be negligible with the 8-object array used for the majority. But consider a list of 10,000 random numbers from 0 to 99:
let randomNumbers = (1...10000).map{_ in Int(rand() % 100)}
let indexes = randomNumbers.indexes(93) // count -> 100 (in my first run)
let index1 = indexes[1] // 238
// executed in 29.6603130102158 sec
let index2 = randomNumbers.nIndex(93, n: 2) // 238
// executed in 3.82625496387482 sec
As can be seen, this new method is considerably faster with the (very) large dataset; it is a bit more cumbersome and confusing though, so depending on your application, you may prefer the simpler solution, or a different one entirely.
(Again) thanks for reading.
extension Collection where Element: Equatable {
func nth(occurance: Int, of element: Element) -> Index? {
var level : Int = occurance
var position = self.startIndex
while let index = self[position...].index(of: element) {
level -= 1
guard level >= 0 else { return nil }
guard level != 0 else { return index }
position = self.index(after: index)
}
return nil
}
}