Related
The code below compiles and runs OK, and seems to indicate that the closure and String.init(describing:) functions are completely equivalent in their signature, since .map method happily takes both of them.
let someDict: [String: String] = [
"string1" : "Hello",
"string2" : "Bye",
]
//One way to call .map
var closure = { (key: String, value: String) -> String in
return "The key is \(key), the value is \(value)"
}
someDict.map(closure)
//Another way to call .map
someDict.map(String.init(describing:))
But how is it possible to place into .map a String.init(describing:) function which is a function of only 1 argument, while .map expects a function of 2 arguments? Or am i misunderstanding something here..
Btw, checking the documentation shows that it really does expect a function of 2 arguments:
transform: ((key: String, value: String)) throws -> T
Btw, checking the documentation shows that it really does expect a
function of 2 arguments:
transform: ((key: String, value: String)) throws -> T
Actually, no. Notice the extra parentheses (). It shows that it expects a function that takes one argument which is a tuple containing two elements.
Consider this example:
// function foo takes two arguments
func foo(_ a: Int, _ b: Int) -> Int {
return a + b
}
// function bar takes one tuple with two elements
func bar(_ a: (Int, Int)) -> Int {
return a.0 + a.1
}
let f1 = foo
print(type(of: f1)) // (Int, Int) -> Int
let f2 = bar
print(type(of: f2)) // ((Int, Int)) -> Int
So, the extra parentheses tell us that map is expecting one argument that is a tuple containing two elements.
The closure passed to map always operates on a single element from the sequence at a time. That element can be a tuple such as your case, and then your closure can deconstruct that tuple into multiple values.
Consider this example:
// tup is a tuple containing 3 values
let tup = (1, true, "hello")
// deconstruct the tuple through assignment
let (x, y, z) = tup
print(x) // 1
print(y) // true
print(z) // hello
So in this example:
var closure = { (key: String, value: String) -> String in
return "The key is \(key), the value is \(value)"
}
someDict.map(closure)
map's closure is given a tuple of the form (key: String, value: String) and the closure is deconstructing that into key and value just as the let did above.
In this example:
someDict.map(String.init(describing:))
which is equivalent to:
someDict.map({ String(describing: $0) })
map is taking the whole tuple and passing it to String(describing:).
As the Programming Swift book describes, tuples can be destructured either in the assignment or by value-binding in a switch
let point = (3, 2)
switch point {
case let (x, y):
print("The point is at (\(x), \(y)).")
}
let (a, b) = point
print("The point is at (\(a), \(b)).")
I can't find any mention of how to do the equivalent for structs. For example:
struct S {
let a, b: Int
}
let s = S(a: 1, b: 2)
// This doesn't work:
// let (sa, sb): s
//
// Outputs: error: expression type 'S' is ambiguous without more context
// let (sa, sb) = s
// ^
This doesn't exist as such in the language.
One option is a helper computed property:
struct S {
let i: Int
let b: Bool
}
extension S {
var destructured: (Int, Bool) {
return (self.i, self.b)
}
}
let s = S(i: 10, b: false)
let (i, b) = s.destructured
Of course, you have to manually keep that in sync. Possibly Sourcery could assist with that.
Structs cannot be destructured in Swift.
Your tuple, point = (3, 2), is of type (Int, Int), which is part of why you are able to destructure it.
The type of your struct, S, is just S. Its variables, a and b, are not included in its type in the same literal way as they are for a tuple. A struct is simply a completely different kind of object, and this behavior does not exist for it.
I'd like a function runningSum on an array of numbers a (or any ordered collection of addable things) that returns an array of the same length where each element i is the sum of all elements in A up to an including i.
Examples:
runningSum([1,1,1,1,1,1]) -> [1,2,3,4,5,6]
runningSum([2,2,2,2,2,2]) -> [2,4,6,8,10,12]
runningSum([1,0,1,0,1,0]) -> [1,1,2,2,3,3]
runningSum([0,1,0,1,0,1]) -> [0,1,1,2,2,3]
I can do this with a for loop, or whatever. Is there a more functional option? It's a little like a reduce, except that it builds a result array that has all the intermediate values.
Even more general would be to have a function that takes any sequence and provides a sequence that's the running total of the input sequence.
The general combinator you're looking for is often called scan, and can be defined (like all higher-order functions on lists) in terms of reduce:
extension Array {
func scan<T>(initial: T, _ f: (T, Element) -> T) -> [T] {
return self.reduce([initial], combine: { (listSoFar: [T], next: Element) -> [T] in
// because we seeded it with a non-empty
// list, it's easy to prove inductively
// that this unwrapping can't fail
let lastElement = listSoFar.last!
return listSoFar + [f(lastElement, next)]
})
}
}
(But I would suggest that that's not a very good implementation.)
This is a very useful general function, and it's a shame that it's not included in the standard library.
You can then generate your cumulative sum by specializing the starting value and operation:
let cumSum = els.scan(0, +)
And you can omit the zero-length case rather simply:
let cumSumTail = els.scan(0, +).dropFirst()
Swift 4
The general sequence case
Citing the OP:
Even more general would be to have a function that takes any sequence
and provides a sequence that's the running total of the input
sequence.
Consider some arbitrary sequence (conforming to Sequence), say
var seq = 1... // 1, 2, 3, ... (CountablePartialRangeFrom)
To create another sequence which is the (lazy) running sum over seq, you can make use of the global sequence(state:next:) function:
var runningSumSequence =
sequence(state: (sum: 0, it: seq.makeIterator())) { state -> Int? in
if let val = state.it.next() {
defer { state.sum += val }
return val + state.sum
}
else { return nil }
}
// Consume and print accumulated values less than 100
while let accumulatedSum = runningSumSequence.next(),
accumulatedSum < 100 { print(accumulatedSum) }
// 1 3 6 10 15 21 28 36 45 55 66 78 91
// Consume and print next
print(runningSumSequence.next() ?? -1) // 120
// ...
If we'd like (for the joy of it), we could condense the closure to sequence(state:next:) above somewhat:
var runningSumSequence =
sequence(state: (sum: 0, it: seq.makeIterator())) {
(state: inout (sum: Int, it: AnyIterator<Int>)) -> Int? in
state.it.next().map { (state.sum + $0, state.sum += $0).0 }
}
However, type inference tends to break (still some open bugs, perhaps?) for these single-line returns of sequence(state:next:), forcing us to explicitly specify the type of state, hence the gritty ... in in the closure.
Alternatively: custom sequence accumulator
protocol Accumulatable {
static func +(lhs: Self, rhs: Self) -> Self
}
extension Int : Accumulatable {}
struct AccumulateSequence<T: Sequence>: Sequence, IteratorProtocol
where T.Element: Accumulatable {
var iterator: T.Iterator
var accumulatedValue: T.Element?
init(_ sequence: T) {
self.iterator = sequence.makeIterator()
}
mutating func next() -> T.Element? {
if let val = iterator.next() {
if accumulatedValue == nil {
accumulatedValue = val
}
else { defer { accumulatedValue = accumulatedValue! + val } }
return accumulatedValue
}
return nil
}
}
var accumulator = AccumulateSequence(1...)
// Consume and print accumulated values less than 100
while let accumulatedSum = accumulator.next(),
accumulatedSum < 100 { print(accumulatedSum) }
// 1 3 6 10 15 21 28 36 45 55 66 78 91
The specific array case: using reduce(into:_:)
As of Swift 4, we can use reduce(into:_:) to accumulate the running sum into an array.
let runningSum = arr
.reduce(into: []) { $0.append(($0.last ?? 0) + $1) }
// [2, 4, 6, 8, 10, 12]
By using reduce(into:_:), the [Int] accumulator will not be copied in subsequent reduce iterations; citing the Language reference:
This method is preferred over reduce(_:_:) for efficiency when the
result is a copy-on-write type, for example an Array or a
Dictionary.
See also the implementation of reduce(into:_:), noting that the accumulator is provided as an inout parameter to the supplied closure.
However, each iteration will still result in an append(_:) call on the accumulator array; amortized O(1) averaged over many invocations, but still an arguably unnecessary overhead here as we know the final size of the accumulator.
Because arrays increase their allocated capacity using an exponential
strategy, appending a single element to an array is an O(1) operation
when averaged over many calls to the append(_:) method. When an array
has additional capacity and is not sharing its storage with another
instance, appending an element is O(1). When an array needs to
reallocate storage before appending or its storage is shared with
another copy, appending is O(n), where n is the length of the array.
Thus, knowing the final size of the accumulator, we could explicitly reserve such a capacity for it using reserveCapacity(_:) (as is done e.g. for the native implementation of map(_:))
let runningSum = arr
.reduce(into: [Int]()) { (sums, element) in
if let sum = sums.last {
sums.append(sum + element)
}
else {
sums.reserveCapacity(arr.count)
sums.append(element)
}
} // [2, 4, 6, 8, 10, 12]
For the joy of it, condensed:
let runningSum = arr
.reduce(into: []) {
$0.append(($0.last ?? ($0.reserveCapacity(arr.count), 0).1) + $1)
} // [2, 4, 6, 8, 10, 12]
Swift 3: Using enumerated() for subsequent calls to reduce
Another Swift 3 alternative (with an overhead ...) is using enumerated().map in combination with reduce within each element mapping:
func runningSum(_ arr: [Int]) -> [Int] {
return arr.enumerated().map { arr.prefix($0).reduce($1, +) }
} /* thanks #Hamish for improvement! */
let arr = [2, 2, 2, 2, 2, 2]
print(runningSum(arr)) // [2, 4, 6, 8, 10, 12]
The upside is you wont have to use an array as the collector in a single reduce (instead repeatedly calling reduce).
Just for fun: The running sum as a one-liner:
let arr = [1, 2, 3, 4]
let rs = arr.map({ () -> (Int) -> Int in var s = 0; return { (s += $0, s).1 } }())
print(rs) // [1, 3, 6, 10]
It does the same as the (updated) code in JAL's answer, in particular,
no intermediate arrays are generated.
The sum variable is captured in an immediately-evaluated closure returning the transformation.
If you just want it to work for Int, you can use this:
func runningSum(array: [Int]) -> [Int] {
return array.reduce([], combine: { (sums, element) in
return sums + [element + (sums.last ?? 0)]
})
}
If you want it to be generic over the element type, you have to do a lot of extra work declaring the various number types to conform to a custom protocol that provides a zero element, and (if you want it generic over both floating point and integer types) an addition operation, because Swift doesn't do that already. (A future version of Swift may fix this problem.)
Assuming an array of Ints, sounds like you can use map to manipulate the input:
let arr = [0,1,0,1,0,1]
var sum = 0
let val = arr.map { (sum += $0, sum).1 }
print(val) // "[0, 1, 1, 2, 2, 3]\n"
I'll keep working on a solution that doesn't use an external variable.
I thought I'd be cool to extend Sequence with a generic scan function as is suggested in the great first answer.
Given this extension, you can get the running sum of an array like this: [1,2,3].scan(0, +)
But you can also get other interesting things…
Running product: array.scan(1, *)
Running max: array.scan(Int.min, max)
Running min: array.scan(Int.max, min)
Because the implementation is a function on Sequence and returns a Sequence, you can chain it together with other sequence functions. It is efficient, having linear running time.
Here's the extension…
extension Sequence {
func scan<Result>(_ initialResult: Result, _ nextPartialResult: #escaping (Result, Self.Element) -> Result) -> ScanSequence<Self, Result> {
return ScanSequence(initialResult: initialResult, underlying: self, combine: nextPartialResult)
}
}
struct ScanSequence<Underlying: Sequence, Result>: Sequence {
let initialResult: Result
let underlying: Underlying
let combine: (Result, Underlying.Element) -> Result
typealias Iterator = ScanIterator<Underlying.Iterator, Result>
func makeIterator() -> Iterator {
return ScanIterator(previousResult: initialResult, underlying: underlying.makeIterator(), combine: combine)
}
var underestimatedCount: Int {
return underlying.underestimatedCount
}
}
struct ScanIterator<Underlying: IteratorProtocol, Result>: IteratorProtocol {
var previousResult: Result
var underlying: Underlying
let combine: (Result, Underlying.Element) -> Result
mutating func next() -> Result? {
guard let nextUnderlying = underlying.next() else {
return nil
}
previousResult = combine(previousResult, nextUnderlying)
return previousResult
}
}
One solution using reduce:
func runningSum(array: [Int]) -> [Int] {
return array.reduce([], combine: { (result: [Int], item: Int) -> [Int] in
if result.isEmpty {
return [item] //first item, just take the value
}
// otherwise take the previous value and append the new item
return result + [result.last! + item]
})
}
I'm very late to this party. The other answers have good explanations. But none of them have provided the initial result, in a generic way. This implementation is useful to me.
public extension Sequence {
/// A sequence of the partial results that `reduce` would employ.
func scan<Result>(
_ initialResult: Result,
_ nextPartialResult: #escaping (Result, Element) -> Result
) -> AnySequence<Result> {
var iterator = makeIterator()
return .init(
sequence(first: initialResult) { partialResult in
iterator.next().map {
nextPartialResult(partialResult, $0)
}
}
)
}
}
extension Sequence where Element: AdditiveArithmetic & ExpressibleByIntegerLiteral {
var runningSum: AnySequence<Element> { scan(0, +).dropFirst() }
}
I'd like to write an extension for tuples of (e.g.) two value in Swift. For instance, I'd like to write this swap method:
let t = (1, "one")
let s = t.swap
such that s would be of type (String, Int) with value ("one", 1). (I know I can very easily implement a swap(t) function instead, but that's not what I'm interested in.)
Can I do this? I cannot seem to write the proper type name in the extension declaration.
Additionally, and I suppose the answer is the same, can I make a 2-tuple adopt a given protocol?
You cannot extend tuple types in Swift.
According to
Types, there are named types (which
can be extended) and compound types. Tuples and functions are compound
types.
See also (emphasis added):
Extensions
Extensions add new functionality to an existing
class, structure, or enumeration type.
As the answer above states, you cannot extend tuples in Swift. However, rather than just give you a no, what you can do is box the tuple inside a class, struct or enum and extend that.
struct TupleStruct {
var value: (Int, Int)
}
extension TupleStruct : Hashable {
var hashValue: Int {
return hash()
}
func hash() -> Int {
var hash = 23
hash = hash &* 31 &+ value.0
return hash &* 31 &+ value.1
}
}
func ==(lhs: TupleStruct, rhs: TupleStruct) -> Bool {
return lhs.value == rhs.value
}
As a side note, in Swift 2.2, tuples with up to 6 members are now Equatable.
Details
Xcode 11.2.1 (11B500), Swift 5.1
Solution
struct Tuple<T> {
let original: T
private let array: [Mirror.Child]
init(_ value: T) {
self.original = value
array = Array(Mirror(reflecting: original).children)
}
func getAllValues() -> [Any] { array.compactMap { $0.value } }
func swap() -> (Any?, Any?)? {
if array.count == 2 { return (array[1].value, array[0].value) }
return nil
}
}
Usage
let x = (1, "one")
let tuple = Tuple(x)
print(x) // (1, "one")
print(tuple.swap()) // Optional((Optional("one"), Optional(1)))
if let value = tuple.swap() as? (String, Int) {
print("\(value) | \(type(of: value))") // ("one", 1) | (String, Int)
}
If you wanted to be a Bad Person™ you can define custom operators on tuples, like this:
postfix operator <->
postfix func <-> <A, B>(lhs: (A, B)) -> (B, A) {
return (lhs.1, lhs.0)
}
let initial = (1, "one")
let reversed = initial<->
FWIW I can't think of a place where my 'clever' code trumps the readability of just writing your swap function.
Is it possible to pass in a tuple into a function as long as their types match up?
When I try it, I get a missing argument in parameter error:
var myTuple = ("Text",10,"More Text")
func myFunction(a:String, b:Int, c:String) {
// etc...
}
myFunction(myTuple)
It was possible, although was deprecated in Swift 2.2:
In Swift 2.1 and earlier it was possible to use a carefully crafted tuple to fill the parameters of a function. So, if you had a function that took two parameters, you could call it with a two-element tuple as long as the tuple had the correct types and element names.
...
This syntax — affectionately called “tuple splat syntax” — is the antithesis of idiomatic Swift’s self-documenting, readable style, and so it’s deprecated in Swift 2.2.
https://swift.org/blog/swift-2-2-new-features/
I came here wanting to know how to pass a tuple as a function parameter. The answers here focus on a different case. I'm not entirely clear what the OP was after.
In any case, here is how to pass a tuple as a parameter. And, for good measure, how to do it variadically.
func acceptTuple(tuple : (Int, String)) {
print("The Int is: \(tuple.0)")
print("The String is '\(tuple.1)'")
}
acceptTuple((45, "zebras"))
// Outputs:
// The Int is: 45
// The String is 'zebras'
func acceptTuples(tuples : (Int, String) ...) {
var index = 0
// note: you can't use the (index, tuple) pattern in the for loop,
// the compiler thinks you're trying to unpack the tuple, hence
/// use of a manual index
for tuple in tuples {
print("[\(index)] - Int is: \(tuple.0)")
print("[\(index)] - String is '\(tuple.1)'")
index++
}
}
acceptTuples((45, "zebras"), (17, "armadillos"), (12, "caterpillars"))
//Outputs
//[0] - Int is: 45
//[0] - String is 'zebras'
//[1] - Int is: 17
//[1] - String is 'armadillos'
//[2] - Int is: 12
//[2] - String is 'caterpillars'
Passing tuples in can be a quick and convenient approach, saving you from having to create wrappers etc. For example, I have a use case where I am passing a set of tokens and parameters to create a game level. Tuples makes this nice and compact:
// function signature
class func makeLevel(target: String, tokens: (TokenType, String)...) -> GameLevel
// The function is in the class Level. TokenType here is an Enum.
// example use:
let level = Level("Zoo Station", tokens:
(.Label, "Zebra"),
(.Bat, "LeftShape"),
(.RayTube, "HighPowered"),
(.Bat, "RightShape"),
(.GravityWell, "4"),
(.Accelerator, "Alpha"))
Yes, it's possible under these conditions:
the tuple must be immutable
the number of values in the tuple, their type, and their order must match the parameters expected by the function
named parameters must match external names in the function signature
non-named parameters must match parameters without external name in the function signature
So, your code is ok, the only thing you have to do is turning the tuple into an immutable one (i.e. using let and not var):
let myTuple = ("Text", 10, "More Text")
func myFunction(a:String, b:Int, c:String) {
// etc...
}
myFunction(myTuple)
One more example with external names:
let myTuple = ("Text", paramB: 10, paramC: "More Text")
func myFunction(a:String, paramB b:Int, paramC c:String) {
// etc...
}
myFunction(myTuple)
In your tuple, it appears as though you must name them and then refer to them as such:
so your code should be
var myTuple = (val1: "Text", val2: 10, val3: "More Text")
func myFunction(a:String, b:Int, c:String) {
// etc...
}
myFunction(myTuple.val1, myTuple.val2, myTuple.val3)
The tuple has named values (val1, val2, val3) which you set and then reference, when you pass in myTuple, to the function myFunction(), it appears as though you are just filling 1 of the 3 available arguements - and with the wrong type to boot! This is the equivalent of storing the types in a tuple, then taking them out for a function call. However, if you want a function to actually take a tuple as a parameter, see below:
var myTuple = (val1: "Text", val2: 10, val3: "More Text")
func tupleFunc(a:(String, Int, String)) {
}
tupleFunc(myTuple)
Yes, but that's the wrong structure: you're passing three variables called a, b, and c rather than a tuple with those components.
You need parentheses around the whole thing:
var myTuple = ("Text", 10, "More Text")
func myFunction(a:(x: String, y: Int, z: String)) {
println(a)
}
myFunction(myTuple)
You can use the following feature: Swift allows you to pass a function (f1) with any number of parameters (but without inout parameters) as a parameter of type (TIn) -> TOut to another function. In this case, TIn will represent a tuple from the parameters of the function f1:
precedencegroup ApplyArgumentPrecedence {
higherThan: BitwiseShiftPrecedence
}
infix operator <- :ApplyArgumentPrecedence
func <-<TIn, TOut>(f: ((TIn) -> TOut), arg: TIn) -> TOut {
return f(arg)
}
func sum(_ a: Int, _ b: Int) -> Int {
return a + b
}
print(sum <- (40, 2))
In swift 3.0, we should not able to pass the tuple directly to the function.If we did so, it shows the error message as "This type has been removed in swift 3.0"
func sum(x: Int, y: Int) -> Int
return x+y }
let params = (x: 1, y: 1)
let x = params.0
let y = params.1
sum(x: x, y: y)
Hope it helps you!!
The best option for now seems to be to just save it to a compound variable or use the build in dot syntax
let (val1, val2) = (1, 2)
func f(first: Int, second: Int) { }
f(first: val1, second: val2)
let vals = (1, 2)
f(first: vals.0, second: vals.1)
That feature called implicit tuple splat was removed in swift 3.
You can find more detailed explanation on the removal proposal here
Some suggestions to keep using tuple as an argument is by doing so:
func f1(_ a : (Int, Int)) { ... }
let x = (1, 2)
f1(x)
func f2<T>(_ a : T) -> T { ... }
let x = (1, 2)
f2(x)