I am frequently needing to pass multiple objects through a series of chained Futures. I have solved the issue with code similar to the below, but it doesn't feel 'right'. Is there a better way to write this?
In particular, the lines beginning return map(to: don't smell right, as I'm taking an actual value a then turning it into a Future conn.future(a) just to be later flatMapped again in the next step.
In the contrived example below, we are creating an A, with a related object B, and optionally a C which is related to B, which is why B must be passed through the chain.
class A {
id: Int?
}
class B {
id: Int?
aId: Int?
}
class C {
id: Int?
bId: Int?
}
/// Creates an A and B, with an optionally attached C. Returns the A.
func create(withC: Bool, on conn: MySQLConnection) throws -> Future<A> {
return try A.create(on: conn)
// Returning two futures
.flatMap(to: (A, B).self) { a in
let futureB = try B(aId: a.requireID()).create(on: conn)
return map(to: (A, B).self, conn.future(a), futureB) { ($0, $1) }
}
// Returning two futures, one optional
.flatMap(to: (A, C?).self) { a, b in
guard withC else {
return map(to: (A, C?).self, conn.future(a), conn.future(nil)) { ($0, $1) }
}
let futureC = try C(bId: b.requireId()).create(on: conn)
return map(to: (A, C?).self, conn.future(a), futureC) { ($0, $1) }
}
// Convert back to the future we care about
.map(to: A.self) { a, _ in return a }
}
Related
I am trying to split a string into an array of letters, but keep some of the letters together. (I'm trying to break them into sound groups for pronunciation, for example).
So, for example, all the "sh' combinations would be one value in the array instead of two.
It is easy to find an 's' in an array that I know has an "sh" in it, using firstIndex. But how do I get more than just the first, or last, index of the array?
The Swift documentation includes this example:
let students = ["Kofi", "Abena", "Peter", "Kweku", "Akosua"]
if let i = students.firstIndex(where: { $0.hasPrefix("A") }) {
print("\(students[i]) starts with 'A'!")
}
// Prints "Abena starts with 'A'!"
How do I get both Abena and Akosua (and others, if there were more?)
Here is my code that accomplishes some of what I want (please excuse the rather lame error catching)
let message = "she sells seashells"
var letterArray = message.map { String($0)}
var error = false
while error == false {
if message.contains("sh") {
guard let locate1 = letterArray.firstIndex(of: "s") else{
error = true
break }
let locate2 = locate1 + 1
//since it keeps finding an s it doesn't know how to move on to rest of string and we get an infinite loop
if letterArray[locate2] == "h"{
letterArray.insert("sh", at: locate1)
letterArray.remove (at: locate1 + 1)
letterArray.remove (at: locate2)}}
else { error = true }}
print (message, letterArray)
Instead of first use filter you will get both Abena and Akosua (and others, if there were more?)
extension Array where Element: Equatable {
func allIndexes(of element: Element) -> [Int] {
return self.enumerated().filter({ element == $0.element }).map({ $0.offset })
}
}
You can then call
letterArray.allIndexes(of: "s") // [0, 4, 8, 10, 13, 18]
You can filter the collection indices:
let students = ["Kofi", "Abena", "Peter", "Kweku", "Akosua"]
let indices = students.indices.filter({students[$0].hasPrefix("A")})
print(indices) // "[1, 4]\n"
You can also create your own indices method that takes a predicate:
extension Collection {
func indices(where predicate: #escaping (Element) throws -> Bool) rethrows -> [Index] {
try indices.filter { try predicate(self[$0]) }
}
}
Usage:
let indices = students.indices { $0.hasPrefix("A") }
print(indices) // "[1, 4]\n"
or indices(of:) where the collection elements are Equatable:
extension Collection where Element: Equatable {
func indices(of element: Element) -> [Index] {
indices.filter { self[$0] == element }
}
}
usage:
let message = "she sells seashells"
let indices = message.indices(of: "s")
print(indices)
Note: If you need to find all ranges of a substring in a string you can check this post.
Have fun!
["Kofi", "Abena", "Peter", "Kweku", "Akosua"].forEach {
if $0.hasPrefix("A") {
print("\($0) starts with 'A'!")
}
}
If you really want to use the firstIndex method, here's a recursive(!) implementation just for fun :D
extension Collection where Element: Equatable {
/// Returns the indices of an element from the specified index to the end of the collection.
func indices(of element: Element, fromIndex: Index? = nil) -> [Index] {
let subsequence = suffix(from: fromIndex ?? startIndex)
if let elementIndex = subsequence.firstIndex(of: element) {
return [elementIndex] + indices(of: element, fromIndex: index(elementIndex, offsetBy: 1))
}
return []
}
}
Recursions
Given n instances of element in the collection, the function will be called n+1 times (including the first call).
Complexity
Looking at complexity, suffix(from:) is O(1), and firstIndex(of:) is O(n). Assuming that firstIndex terminates once it encounters the first match, any recursions simply pick up where we left off. Therefore, indices(of:fromIndex:) is O(n), just as good as using filter. Sadly, this function is not tail recursive... although we can change that by keeping a running total.
Performance
[Maybe I'll do this another time.]
Disclaimer
Recursion is fun and all, but you should probably use Leo Dabus' solution.
I have the following struct defined.
struct Person {
var firstName :String
var lastName :String
var active :Bool
}
I have created a collection of Person as shown below:
var persons :[Person] = []
for var i = 1; i<=10; i++ {
var person = Person(firstName: "John \(i)", lastName: "Doe \(i)", active: true)
persons.append(person)
}
and Now I am trying to change the active property to false using the code below:
let inActionPersons = persons.map { (var p) in
p.active = false
return p
}
But I get the following error:
Cannot invoke map with an argument list of type #noescape (Person) throws
Any ideas?
SOLUTION:
Looks like Swift can't infer types sometimes which is kinda lame! Here is the solution:
let a = persons.map { (var p) -> Person in
p.active = false
return p
}
THIS DOES NOT WORK:
let a = persons.map { p in
var p1 = p
p1.active = false
return p1
}
There are exactly two cases where the Swift compiler infers the return
type of a closure automatically:
In a "single-expression closure," i.e. the closure body
consists of a single expression only (with or without explicit
closure parameters).
If the type can be inferred from the calling context.
None of this applies in
let inActionPersons = persons.map { (var p) in
p.active = false
return p
}
or
let a = persons.map { p in
var p1 = p
p1.active = false
return p1
}
and that's why
you have to specify the return type explicitly as in Kametrixom's answer.
Example of a single-expression closure:
let inActionPersons = persons.map { p in
Person(firstName: p.firstName, lastName: p.lastName, active: false)
}
and it would compile with (var p) in or (p : Person) in as well, so this has nothing to do with whether the closure arguments are given
explicitly in parentheses or not.
And here is an example where the type is inferred from the calling
context:
let a : [Person] = persons.map { p in
var p1 = p
p1.active = false
return p1
}
The result of map() must be a [Person] array, so map needs
a closure of type Person -> Person, and the compiler infers
the return type Person automatically.
For more information, see "Inferring Type From Context" and "Implicit Returns from Single-Expression Closures" in the
"Closures" chapter in the Swift book.
When using the brackets for arguments so that var works, you have to put the return type as well:
let inActionPersons = persons.map { (var p) -> Person in
p.active = false
return p
}
Swift 5
The accepted answer no longer works, as of Swift 5, anyway. Closures cannot have keyword arguments anymore which means that each element of the iteration must remain a constant. Therefore, to mutate structures using map, new elements must be initialized within each iteration:
let activePersons = persons.map { (p) -> Person in
return Person(firstName: p.firstName, lastName: p.lastName, active: true)
}
I'm currently trying to complete the Swift Course on iTunes U and we are building a calculator. I'm having trouble understanding part of the code.
I added the code below that I thought was relevant from the file.
Here is what confuses me: why does operation(operand) compute the value for the UnaryOperation (i.e. the square root)? I see that when the CalculatorBrain class is called the dictionary is initialized, but when I print the dictionary out I just get something that looks like this: [✕: ✕, -: -, +: +, ⌹: ⌹, √: √]. So where/when does the program compute the square root when I click on the square root button?
Class CalculatorBrain
{
private enum Op: Printable
{
case Operand(Double)
case UnaryOperation(String, Double -> Double)
case BinaryOperation(String, (Double, Double) -> Double)
var description: String {
get {
switch self {
case .Operand(let operand):
return "\(operand)"
case .UnaryOperation(let symbol, _):
return symbol
case .BinaryOperation(let symbol, _):
return symbol
}
}
}
}
private var opStack = [Op]()
private var knownOps = [String: Op]()
init() {
func learnOp(op: Op) {
knownOps[op.description] = op
}
learnOp(Op.BinaryOperation("✕", *))
learnOp(Op.BinaryOperation("⌹") { $1 / $0 })
learnOp(Op.BinaryOperation("+", +))
learnOp(Op.BinaryOperation("-") { $0 - $1 })
learnOp(Op.UnaryOperation ("√", sqrt))
}
private func evaluate(ops: [Op]) -> (result: Double?, remainingOps: [Op])
{
if !ops.isEmpty {
var remainingOps = ops
let op = remainingOps.removeLast()
switch op {
case .Operand(let operand):
return (operand, remainingOps)
case .UnaryOperation(_, let operation):
let operandEvaluation = evaluate(remainingOps)
if let operand = operandEvaluation.result {
**return (operation(operand), operandEvaluation.remainingOps)**
}
// case.BinaryOperation(.....)
}
}
return (nil, ops)
}
func evaluate() -> Double? {
let (result, remainder) = evaluate(opStack)
return result
}
func pushOperand(operand: Double) -> Double? {
opStack.append(Op.Operand(operand))
return evaluate()
}
func performOperation(symbol: String) -> Double? {
if let operation = knownOps[symbol] {
opStack.append(operation)
}
return evaluate()
}
}
The Op enum implements the Printable protocol, which means it has a description: String property. When you print the Dictionary, you are sending [String : Op] to the println function which then tries to print the Op using its description.
The reason the description of the operators is the same as its key in the Dictionary is because the learnOp(op: Op) function sets the key to be op.description (knownOps[op.description] = op)
To see the effects of this, you could add a new operator learnOp(Op.UnaryOperation ("#", sqrt)) which will be printed as #:# inside of the knownOps Dictionary. (And if you add a new button for the # operator, it will also perform the square root operation)
Since the calculator is stack based, the operands get pushed on, then the operations. When evaluate() gets called, it calls evaluate(opStack) passing the entire stack through.
evaluate(ops: [Op]) then takes the to item off of the stack and evaluates the function after having calculated the operands.
As an example, lets say you want to calucalte sqrt(4 + 5).
You would push the items onto the stack, and it would look like: [ 4, 5, +, sqrt ]
Then evaluate(ops: [Op]) sees the sqrt and evaluates the operand with a recursive call. That call then evaluates + with two more recursive calls which return 5 and 4.
The tree of calls would look like this:
ops: [4, 5, +, sqrt] // Returns sqrt(9) = 3
|
ops: [4, 5, +] // Returns 4 + 5 = 9
____|_____
| |
ops: [4, 5] ops: [4]
return 5 return 4
I strongly recommend you put a breakpoint on the evaluate() -> Double? function and step through the program to see where it goes with different operands and operations.
learnOp(Op.UnaryOperation ("√", sqrt))
sqrt is a built in function, so you're teaching the calculator that "√" means it should perform the sqrt operation.
Given a Dictionary<String, Arrary<Int>> find the how many entries have the same two specified values in the first 5 entries in the Array<Int>.
For example:
Given:
let numberSeries = [
"20022016": [07,14,36,47,50,02,05],
"13022016": [16,07,32,36,41,07,09],
"27022016": [14,18,19,31,36,04,05],
]
And the values: 7 and 36, the result should be 2 since the first and second entry have both the values 7 and 36 in the first 5 entries of the entry's array.
I've tried to accomplish this many ways, but I haven't been able to get it to work.
This is my current attempt:
//created a dictionary with (key, values)
let numberSeries = [
"20022016": [07,14,36,47,50,02,05],
"13022016": [16,07,32,36,41,07,09],
"27022016": [14,18,19,31,36,04,05],
]
var a = 07 //number to look for
var b = 36 // number to look for
// SearchForPairAB // search for pair // Doesn't Work.
var ab = [a,b] // pair to look for
var abPairApearedCount = 0
for (kind, numbers) in numberSeries {
for number in numbers[0...4] {
if number == ab { //err: Cannot invoke '==' with argument listof type Int, #Value [Int]
abPairApearedCount++
}
}
}
This gives the error: Cannot invoke '==' with argument listof type Int, #Value [Int] on the line: if number == ab
You can't use == to compare an Int and Array<Int>, that just doesn't make any sense from a comparison perspective. There are lots of different ways you can achieve what you're trying to do though. In this case I'd probably use map/reduce to count your pairs.
The idea is to map the values in your ab array to Bool values determined by whether or not the value is in your numbers array. Then, reduce those mapped Bools to a single value: true if they're all true, or false. If that reduced value is true, then we found the pair so we increment the count.
var ab = [a,b] // pair to look for
var abPairApearedCount = 0
for (kind, numbers) in numberSeries {
let found = ab.map({ number in
// find is a built-in function that returns the index of the value
// in the array, or nil if it's not found
return find(numbers[0...4], number) != nil
}).reduce(true) { (result, value: Bool) in
return result && value
}
if found {
abPairApearedCount++
}
}
That can actually be compacted quite a bit by using some of Swift's more concise syntax:
var ab = [a,b] // pair to look for
var abPairApearedCount = 0
for (kind, numbers) in numberSeries {
let found = ab.map({ find(numbers[0...4], $0) != nil }).reduce(true) { $0 && $1 }
if found {
abPairApearedCount++
}
}
And, just for fun, can be compacted even further by using reduce instead of a for-in loop:
var ab = [a,b] // pair to look for
var abPairApearedCount = reduce(numberSeries, 0) { result, series in
result + (ab.map({ find(series.1[0...4], $0) != nil }).reduce(true) { $0 && $1 } ? 1 : 0)
}
That's getting fairly unreadable though, so I'd probably expand some of that back out.
So here's my FP solution, aimed at decomposing the problem into easily digestible and reusable bite-sized chunks:
First, we define a functor that trims an array to a given length:
func trimLength<T>(length: Int) -> ([T]) -> [T] {
return { return Array($0[0...length]) }
}
Using this we can trim all the elements using map(array, trimLength(5))
Now, we need an predicate to determine if all the elements of one array are in the target array:
func containsAll<T:Equatable>(check:[T]) -> ([T]) -> Bool {
return { target in
return reduce(check, true, { acc, elem in return acc && contains(target, elem) })
}
}
This is the ugliest bit of code here, but essentially it's just iterating over check and insuring that each element is in the target array. Once we've got this we can use filter(array, containsAll([7, 26])) to eliminate all elements of the array that don't contain all of our target values.
At this point, we can glue the whole thing together as:
filter(map(numberSeries.values, trimLength(5)), containsAll([7, 36])).count
But long lines of nested functions are hard to read, let's define a couple of helper functions and a custom operator:
func rmap<S:SequenceType, T>(transform:(S.Generator.Element)->T) -> (S) -> [T] {
return { return map($0, transform) }
}
func rfilter<S:SequenceType>(predicate:(S.Generator.Element)->Bool) -> (S) -> [S.Generator.Element] {
return { sequence in return filter(sequence, predicate) }
}
infix operator <^> { associativity left }
func <^> <S, T>(left:S, right:(S)->T) -> T {
return right(left)
}
And a convenience function to count it's inputs:
func count<T>(array:[T]) -> Int {
return array.count
}
Now we can condense the whole thing as:
numberSeries.values <^> rmap(trimLength(5)) <^> rfilter(containsAll([7, 36])) <^> count
I have these two objects in my model:
Message:
class Message: Object {
//Precise UNIX time the message was sent
dynamic var sentTime: NSTimeInterval = NSDate().timeIntervalSince1970
let images = List<Image>()
}
Image:
class Image: Object {
dynamic var mediaURL: String = ""
var messageContainingImage: Message {
return linkingObjects(Message.self, forProperty: "images")[0]
}
}
I want to form a query which returns messages and images, messages sorted by sentTime and images sorted by their messageContainingImage's sent time. They'd be sorted together.
The recommended code for a query is this:
let messages = Realm().objects(Message).sorted("sentTime", ascending: true)
This returns a Result<Message> object. A Result doesn't have a way to be joined to another Result. There are other issues in my way too, such as, if I could combine them, how would I then perform a sort.
Additional thoughts:
I could also add a property to Image called sentTime, then once they're combined I'd be able to call that property on both of them.
I could make them both subclass from a type which has sentTime. The problem is, doing Realm().objects(Message) would only returns things which are messages, and not subclasses of Message.
How would I be able to do this?
My end goal is to display these message and image results in a tableview, messages separately from their attached image.
I think, inheritance is not the right solution here, this introduces more drawbacks by complicating your object schema, than it's worth for your use case.
Let's go back to what you wrote is your end goal: I guess you want to display messages and images together in one table view as separated rows, where the images follow their message. Do I understand that correctly?
You don't need to sort both, sorting the messages and accessing them and their images in a suitable way will ensure that everything is sorted correctly. The main challenge is more how to enumerate / random-access this two-dimensional data structure as an one-dimensional sequence.
Depending on the amount of data, you query, you have to decide, whether you can go a simple approach by keeping them all in memory at once, or introducing a view object on top of Results, which takes care of accessing all objects in order.
The first solution could just look like this:
let messages = Realm().objects(Message).sorted("sentTime", ascending: true)
array = reduce(messages, [Object]()) { (var result, message) in
result.append(message)
result += map(message.images) { $0 }
return result
}
While the latter solution is more complex, but could look like this:
// Let you iterate a list of nodes with their related objects as:
// [a<list: [a1, a2]>, b<list: [b1, b2, b3]>]
// in pre-order like:
// [a, a1, a2, b, b1, b2, b3]
// where listAccessor returns the related objects of a node, e.g.
// listAccessor(a) = [a1, a2]
//
// Usage:
// class Message: Object {
// dynamic var sentTime = NSDate()
// let images = List<Image>()
// }
//
// class Image: Object {
// …
// }
//
// FlattenedResultsView(Realm().objects(Message).sorted("sentTime"), listAccessor: { $0.images })
//
class FlattenedResultsView<T: Object, E: Object> : CollectionType {
typealias Index = Int
typealias Element = Object
let array: Results<T>
let listAccessor: (T) -> (List<E>)
var indexTransformVectors: [(Int, Int?)]
var notificationToken: NotificationToken? = nil
init(_ array: Results<T>, listAccessor: T -> List<E>) {
self.array = array
self.listAccessor = listAccessor
self.indexTransformVectors = FlattenedResultsView.computeTransformVectors(array, listAccessor)
self.notificationToken = Realm().addNotificationBlock { note, realm in
self.recomputeTransformVectors()
}
}
func recomputeTransformVectors() {
self.indexTransformVectors = FlattenedResultsView.computeTransformVectors(array, listAccessor)
}
static func computeTransformVectors(array: Results<T>, _ listAccessor: T -> List<E>) -> [(Int, Int?)] {
let initial = (endIndex: 0, array: [(Int, Int?)]())
return reduce(array, initial) { (result, element) in
var array = result.array
let list = listAccessor(element)
let vector: (Int, Int?) = (result.endIndex, nil)
array.append(vector)
for i in 0..<list.count {
let vector = (result.endIndex, Optional(i))
array.append(vector)
}
return (endIndex: result.endIndex + 1, array: array)
}.array
}
var startIndex: Index {
return indexTransformVectors.startIndex
}
var endIndex: Index {
return indexTransformVectors.endIndex
}
var count: Int {
return indexTransformVectors.count
}
subscript (position: Index) -> Object {
let vector = indexTransformVectors[position]
switch vector {
case (let i, .None):
return array[i]
case (let i, .Some(let j)):
return listAccessor(array[i])[j]
}
}
func generate() -> GeneratorOf<Object> {
var arrayGenerator = self.array.generate()
var lastObject: T? = arrayGenerator.next()
var listGenerator: GeneratorOf<E>? = nil
return GeneratorOf<Object> {
if listGenerator != nil {
let current = listGenerator!.next()
if current != nil {
return current
} else {
// Clear the listGenerator to jump back on next() to the first branch
listGenerator = nil
}
}
if let currentObject = lastObject {
// Get the list of the currentObject and advance the lastObject already, next
// time we're here the listGenerator went out of next elements and we check
// first whether there is anything on first level and start over again.
listGenerator = self.listAccessor(currentObject).generate()
lastObject = arrayGenerator.next()
return currentObject
} else {
return nil
}
}
}
}