There are several idioms for declaring multidimensional arrays in Swift. Consider the following:
var ia1 = Array<Array<Int>>()
var ia2: Int[][] = []
typealias IntArray = Array<Int>
var ia3 = IntArray[]()
var ia4 = Int[][]()
ia1 += [[1, 2, 3], [2, 3, 4]]
ia2 += [[1, 2, 3], [2, 3, 4]]
ia3 += [[1, 2, 3], [2, 3, 4]]
ia4 += [[1, 2, 3], [2, 3, 4]]
let test = (ia1 == ia2) // true
let test2 = (ia3 == ia4) //true
// etc...
Is there actually any difference between the declarations that may bite the developer? And if not, is there any good reason to use one other over the others?
T[] is just syntactic sugar for Array<T> — there's no difference in implementation. Which style you prefer is a question of opinion.
Note that depending on what you're trying to model, multidimensional arrays might not be what you're looking for. It might make more sense to use a single array internally, and expose a multidimensional subscript to users of your data structure:
class GameBoard {
let width = 10
let height = 10
let board: [Int]
init() {
board = [Int](count: width * height, repeatedValue: 0)
}
subscript(i: Int, j: Int) -> Int {
return board[i + j * width]
}
}
let b = GameBoard()
b[0,0]
b[4,1]
Related
Example.
let numArray = [1,2,3,3,4,5,5]
From above array create array of non- repeating number using swift. But I don't want to use Set.
Expected output is [1,2,4]
You may try the following:
let numArray = [1, 2, 3, 3, 4, 5, 5]
// Group by value
let grouped = Dictionary(grouping: numArray, by: { $0 })
// Filter by its count, convert back to Array and sort
let unique = Array(grouped.filter { $1.count == 1 }.map(\.key)).sorted()
print(unique) // [1, 2, 4]
Here is an alternative way without using higher order functions:
let numArray = [1, 2, 3, 3, 4, 5, 5]
// Group by value
let grouped = Dictionary(grouping: numArray, by: { $0 })
var uniqueArray = [Int]()
for (key, value) in grouped {
if value.count == 1 {
uniqueArray.append(key)
}
}
print(uniqueArray.sorted()) // [1, 2, 4]
I'm trying to concatenate two observables of collection with concat() operator, however, it's not working as expected.
I've two observables:
let first = Observable<[Int]>.create { observer in
observer.onNext([1, 2])
observer.onCompleted()
return Disposables.create()
}
let second = PublishSubject<[Int]>()
Using concat():
let items = Observable.concat([first, second])
items.subscribe(onNext: {
print($0)
})
second.onNext([3, 4, 5])
Output:
[1, 2]
[3, 4, 5]
What I want:
[1, 2]
[1, 2, 3, 4, 5]
So you don't just want to concat two observables, you also want to concat the arrays that are in the events that are produced by the two observables. You aren't going far enough to get what you want.
Imagine you had two Array<[Int]> instead of two Observable<[Int]>. Concat-ing the two (as in arr1 + arr2) would not produce [[1, 2], [1, 2, 3, 4, 5]], instead it would produce [[1, 2], [3, 4, 5]]. Your Observables are behaving the same way.
To both concat and combine them, you need scan, as in:
let items = Observable.concat([first, second])
.scan([], accumulator: +)
Which will produce two events:
[1, 2]
[1, 2, 3, 4, 5]
FYI, with arrays, we don't have a scan operator, but we can approximate it with reduce. For arrays, it would be:
let arr1: Array<[Int]> = [[1, 2]]
let arr2: Array<[Int]> = [[3, 4, 5]]
let itemsArr = (arr1 + arr2).reduce([], { result, element in
return result + [(result.last ?? []) + element]
})
print(itemsArr)
How I can map a array and, in same closure, split the values in two variables?
This code work, but, I want use only one closure, not three.
let array = [1, 2, 3, 4]
let a = array.map { v -> (Int, Int) in
(v * 2, v * 10)
}
let x = a.map { $0.0 }
let y = a.map { $0.1 }
Here's a possibly less readable, but arguably more functional-style solution (immutable arrays, no for-each loop, pure function that only acts on its inputs and doesn't create side effects outside the closure):
let array = [1, 2, 3, 4]
let (x, y) = array.reduce(([Int](), [Int]())){ (result, int) in (result.0 + [int*2], result.1 + [int*10]) }
print(x) // [2, 4, 6, 8]
print(y) // [10, 20, 30, 40]
It also meets your requirement of using just one closure
Try this:
let array = [1, 2, 3, 4]
var x = [Int]()
var y = [Int]()
array.forEach() {
x.append($0 * 2)
y.append($0 * 10)
}
print(x[0]) // 2
print(y[0]) // 10
...
You can use tuple assignment to achieve this:
let array = [1, 2, 3, 4]
let (x, y) = (array.map{$0 * 2}, array.map{$0 * 10})
print(x, y)
I have two arrays of [PFObjects].
For example (simplified):
arr1: [PFObject] = [1, 2, 3, 4, 5, 6, 7, 8]
arr2: [PFObject] = [1, 2, 3, 4, 5]
What is the optimal way to compare arr1 with arr2 and only keep the duplicates (remove unique values).
So that arr1 looks like:
arr1 = [1, 2, 3, 4, 5]
let array = arr1.filter { arr2.contains($0) }
voilà !
First solution (Looping):
var arr1: [PFObject] = [1, 2, 3, 4, 5, 6, 7, 8]
var arr2: [PFObject] = [1, 2, 3, 4, 5]
var temp: [PFObject] = []
for element in arr1 {
if contains(arr2, element) {
temp.append(element)
}
}
arr1 = temp
You can loop over the first array, check if each element is contained in the array, if it is, you can add it to a temporary array. After looping over every element you can replace the value of the first array with your temporary array.
Second solution (Sets):
var arr1: [PFObject] = [1, 2, 3, 4, 5, 6, 7, 8]
var arr2: [PFObject] = [1, 2, 3, 4, 5]
let set1 = Set(arr1)
let set2 = Set(arr2)
var arr1= Array(set1.intersect(set2)) // [1, 2, 3, 4, 5]
What you do here is:
First you create sets from your arrays
Then you use the intersect method from sets to determine common elements
Finally you transform your set to an array before passing it back to arr1
Of course since you will be using sets, duplicate elements will be lost but I'm guessing that shouldn't be a problem in your case
Third solution (filter):
From the answer of Pham Hoan you can use filters to obtain a subset of arr1, the closure gives you the conditions, here it is that arr2 contains the value you are looking at.
let array = arr1.filter { arr2.contains($0) }
This is obviously the shorter solution in terms of code length.
I do not know which technique would be more efficient if you have very large arrays however.
I have an array and I want to remove a bunch of indices
var arr = [0,1,2,3,4,5,6]
var rmIndices = [1,4,5]
What is the best way to remove indices 1,4,5 from arr?
Note that PermutationGenerator is going away in Swift 3 and also doesn't keep the ordering the same, though perhaps it did at one time. Using the accepted answer results in [2, 6, 0, 3] which may be unexpected. A couple of alternative approaches that give the expected result of [0, 2, 3, 6] are:
let flatArr = arr.enumerate().flatMap { rmIndices.contains($0.0) ? nil : $0.1 }
or
let filterArr = arr.enumerate().filter({ !rmIndices.contains($0.0) }).map { $0.1 }
Rather than a list of indices to remove, it may be easier to have a list of indices to keep, which you can do using the Set type:
let rmIndices = [1,4,5]
let keepIndices = Set(arr.indices).subtract([1,4,5])
Then you can use PermutationGenerator to create a fresh array of just those indices:
arr = Array(PermutationGenerator(elements: arr, indices: keepIndices))
rmIndices.sort({ $1 < $0 })
for index in rmIndices
{
arr.removeAtIndex(index)
}
Note that I've sorted the indices in descending order. This is because everytime you remove an element E, the indices of the elements beyond E reduce by one.
For Swift 3
var arr = [0,1,2,3,4,5,6]
let rmIndices = [1,4,5]
arr = arr.filter{ !rmIndices.contains($0) }
print(arr)
if you want to produce output very fastly then you can use
var arr = [0,1,2,3,4,5,6]
let rmIndices = [1,4,5]
arr = Array(Set(arr).subtracting(rmIndices))
print(array)
But it will change order of your array
Remove elements using indexes array:
Array of Strings and indexes
let animals = ["cats", "dogs", "chimps", "moose", "squarrel", "cow"]
let indexAnimals = [0, 3, 4]
let arrayRemainingAnimals = animals
.enumerated()
.filter { !indexAnimals.contains($0.offset) }
.map { $0.element }
print(arrayRemainingAnimals)
//result - ["dogs", "chimps", "cow"]
Array of Integers and indexes
var numbers = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
let indexesToRemove = [3, 5, 8, 12]
numbers = numbers
.enumerated()
.filter { !indexesToRemove.contains($0.offset) }
.map { $0.element }
print(numbers)
//result - [0, 1, 2, 4, 6, 7, 9, 10, 11]
Remove elements using element value of another array
Arrays of integers
let arrayResult = numbers.filter { element in
return !indexesToRemove.contains(element)
}
print(arrayResult)
//result - [0, 1, 2, 4, 6, 7, 9, 10, 11]
Arrays of strings
let arrayLetters = ["a", "b", "c", "d", "e", "f", "g", "h", "i"]
let arrayRemoveLetters = ["a", "e", "g", "h"]
let arrayRemainingLetters = arrayLetters.filter {
!arrayRemoveLetters.contains($0)
}
print(arrayRemainingLetters)
//result - ["b", "c", "d", "f", "i"]
In Swift 4:
let newArr = arr.enumerated().compactMap {
rmIndices.contains($0.0) ? nil : $0.1
}
enumerated() generates (index, value) pairs
compactMap concatenates non-nil values
In the closure, $0.0 is the index (first element of enumerated pair) as $0.1$ is the value
compactMap gathers values whose indices are not found in rmIndices
The problem with flatmap is that it gives incorrect results if your array contains optionals.
The following is much faster than the functional style solutions provided and works with optionals. You just have to make sure rmIndices is sorted and unique. It's also fairly language agnostic.
var numRemoved: Int = 0
for index in rmIndices {
let indexToRemove = index - numRemoved
arr.remove(at: indexToRemove)
numRemoved += 1
}
If you need to make sure rmIndices is sorted and unique:
rmIndices = Set(rmIndices).sorted()
Using XCTest to remove 500 elements (including the operation to ensure uniqueness and sorted):
0.006 sec
vs.
arr.enumerated().filter({ !rmIndices.contains($0.0) }).map { $0.1 }:
0.206 sec
I use this as an extension on Array
extension Array {
mutating func remove(at indices: [Int]) {
let rmIndices = Set(indices).sorted()
var numRemoved: Int = 0
for index in rmIndices {
let indexToRemove = index - numRemoved
self.remove(at: indexToRemove)
numRemoved += 1
}
}
}
Using lodash https://lodash.com/
var arr = [0,1,2,3,4,5,6]
var rmIndices = [1,4,5]
_.pullAt(arr, rmIndices);