N-Dimensional array swift - arrays

Is there any way to have an n dimensional array in swift? I would like to be able to make a function that creates an array with n dimensions but I cannot figure out how.
Basically something like this:
func ndarray <T> (dimensions: Int...) -> [[T]] { // What do I tell it I return?
var out
for d in dimensions {
out = Array<T>(repeating: out, count: d)
}
return out
}
The above code does not work for obvios reasons but, I think it points out the main problems I am having:
How do I define a return type
How do I actually create the array
Once created how do I traverse and populate the array

Here is the implementation of an N-Dimensional Array. It uses a normal array internally for storage and converts the multi-dimensional indices into a single index for the internal array.
struct NDimArray<T> {
let dimensions: [Int]
var data: [T]
init(dimensions: Int..., initialValue: T) {
self.dimensions = dimensions
data = Array(repeating: initialValue, count: dimensions.reduce(1, *))
}
init(dimensions: Int..., initUsing initializer: () -> T) {
self.dimensions = dimensions
data = (0 ..< dimensions.reduce(1, *)).map { _ in initializer() }
}
// Compute index into data from indices
private func computeIndex(_ indices: [Int]) -> Int {
guard indices.count == dimensions.count else { fatalError("Wrong number of indices: got \(indices.count), expected \(dimensions.count)") }
zip(dimensions, indices).forEach { dim, idx in
guard (0 ..< dim) ~= idx else { fatalError("Index out of range") }
}
var idx = indices
var dims = dimensions
var product = 1
var total = idx.removeLast()
while !idx.isEmpty {
product *= dims.removeLast()
total += (idx.removeLast() * product)
}
return total
}
subscript(_ indices: Int...) -> T {
get {
return data[computeIndex(indices)]
}
set {
data[computeIndex(indices)] = newValue
}
}
}
Example:
// Create a 3 x 4 x 5 array of String with initial value ""
var arr = NDimArray<String>(dimensions: 3, 4, 5, initialValue: "")
for x in 0 ..< 3 {
for y in 0 ..< 4 {
for z in 0 ..< 5 {
// Encode indices in the string
arr[x, y, z] = "(\(x),\(y),\(z))"
}
}
}
// Show internal storage of data
print(arr.data)
["(0,0,0)", "(0,0,1)", "(0,0,2)", "(0,0,3)", "(0,0,4)", "(0,1,0)", "(0,1,1)", "(0,1,2)", "(0,1,3)", "(0,1,4)", "(0,2,0)", "(0,2,1)", "(0,2,2)", "(0,2,3)", "(0,2,4)", "(0,3,0)", "(0,3,1)", "(0,3,2)", "(0,3,3)", "(0,3,4)", "(1,0,0)", "(1,0,1)", "(1,0,2)", "(1,0,3)", "(1,0,4)", "(1,1,0)", "(1,1,1)", "(1,1,2)", "(1,1,3)", "(1,1,4)", "(1,2,0)", "(1,2,1)", "(1,2,2)", "(1,2,3)", "(1,2,4)", "(1,3,0)", "(1,3,1)", "(1,3,2)", "(1,3,3)", "(1,3,4)", "(2,0,0)", "(2,0,1)", "(2,0,2)", "(2,0,3)", "(2,0,4)", "(2,1,0)", "(2,1,1)", "(2,1,2)", "(2,1,3)", "(2,1,4)", "(2,2,0)", "(2,2,1)", "(2,2,2)", "(2,2,3)", "(2,2,4)", "(2,3,0)", "(2,3,1)", "(2,3,2)", "(2,3,3)", "(2,3,4)"]
print(arr[2, 2, 2]) // "(2,2,2)"
print(arr[3, 0, 0]) // Fatal error: Index out of range
print(arr[0, 4, 0]) // Fatal error: Index out of range
print(arr[2]) // Fatal error: Wrong number of indices: got 1, expected 3
Initializing an Array with a Reference Type
As #DuncanC noted in the comments, you have to be careful when initializing an array with a value which is a reference type, because the array will be filled with references to the object and modifying the object at any index will modify all of them.
To solve this, I added a second initializer:
init(dimensions: Int..., initUsing initializer: () -> T)
which takes a closure () -> T which can be used to create a new object for each element of the array.
For example:
class Person {
var name = ""
}
// Pass a closure which creates a `Person` instance to fill the array
// with 25 person objects
let arr = NDimArray(dimensions: 5, 5, initUsing: { Person() })
arr[3, 3].name = "Fred"
arr[2, 2].name = "Wilma"
print(arr[3, 3].name, arr[2, 2].name)
Fred Wilma

Nope, it's not possible. Array dimensions is something that needs to be determined at compile time, while the argument you want to pass to the initializer will not be known until runtime. If you really want to achieve something like this, then you'll need to move the array indexing from compile time to runtime, e.g. by accessing the array via an array of indexes. Still you don't have compile validation, since the array length can at runtime to not match the dimensions of the array.
This problem is similar to the one that attempts to convert a tuple to an array.

Related

Swift - Smart way to cycle through an Array of Array [[Int]]

I use an Array of Array [[Int]] storing coordinates in a map.
I for example want to replace all "2" with "1".
For this I created 2 for loops which define the array space and check for the content.
This is not very flexible and only works with arrays of the defined size.
private func remove2fromArray(currentArray: [[Int]]) -> [[Int]] {
var newArray : [[Int]] = currentArray
for section in 0...14 {
for row in 0...19 {
if newArray[section][row] == 2
{ newArray[section][row] = 1 }
}
}
return newArray
}
Is there a way to just scan through the content of the array to replace the numbers?
e.g.
currentArray.findAndReplace(find:2, replace: 1) or similar, that works with [[Int]]?
You can do it this way:
let result = array.map { subarray in
subarray.map { integer -> Int in
if integer == 2 {
return 1
} else {
return integer
}
}
}
Even shorter:
let result = array.map {
$0.map { integer in
return integer == 2 ? 1 : integer
}
}
And a one-liner:
let result = array.map { $0.map { $0 == 2 ? 1 : $0 } }
I'll try to explain what's happening here in simple words: What map does is that it goes through the array elements one by one and applies a function on the element.
In our example, the first map iterates over the outer array elements, so $0 here refers to the inner arrays (one after one).
The second map iterates over the inner arrays' elements. So $0 in the inner map refers to the each element of the inner arrays' elements.

Variant on Sequence-Based enumeration of permutations

This is a follow up on this thread, which main issue was to iterate over all permutations of an array, that is given ["a","b","c"], obtain ["bca","acb".. etc], using an iterator.
Thanks to Martin R's insights, and also his inputs in another thread, I came up with another possible solution for the 'Sequence-Based enumeration of permutations' problem using iterators. The issue is that I'm not sure I have all permutations although there are good indications they're all there. The algorithm is guaranteed to provide n! permutations at the most, with no duplicates.
The idea behind this approach is the following, say there's an array a=["a","b","c"...], of size n. Listing all permutations can be viewed as picking elements from bags:
■
■
■
■ ■
■ ■ ■
■ ... ■ ■ ■
0 ... n-3 n-2 n-1
So the algorithm takes the initial array, and removes a row, pass it recursively until there is no row left. At this point, if an iterator can be found, all individual permutations can be addressed independently. The iterator is hidden in FactorialSequence below, where a method next() allows to move from adjacents points.
public struct FactorialSequence : Sequence, IteratorProtocol {
private var current: [Int]
public init(size:Int) {
self.current = Array(repeating: 0, count: size)
}
public mutating func next() -> [Int]? {
return self.__next();
}
private mutating func __next() -> [Int]? {
var next = current
defer {
current = next;
}
for i in self.current.indices.reversed() {
if next[i] < current.count - i - 1 {
next[i] += 1
return next;
}
next[i] = 0
}
return nil
}
}
func permute(seq:[String],at:[Int]) -> String? {
if seq.count > 0 {
var ss = seq;
let uu = seq[at.first!]
var cc = at;
_ = ss.remove(at: cc.first!)
_ = cc.remove(at: 0);
return uu + (permute(seq:ss,at:cc) ?? "")
}
return nil ;
}
the permute() function is called passing the iterator (an array) calculated from FactorialSequence:
var fs = FactorialSequence(size: 3)
print("\(fs.current):\(permute(seq:["a","b","c"], at: fs.current)!)")
while let uu = fs.next() {
print("\(uu):\(permute(seq:["a","b","c"], at: uu)!)")
}
and gives (in flatten string format):
[-0.000][-0.000][171] [0, 0, 0]:abc
[0.0009][0.0009][174] [0, 1, 0]:acb
[0.0016][0.0007][174] [1, 0, 0]:bac
[0.0024][0.0008][174] [1, 1, 0]:bca
[0.0032][0.0008][174] [2, 0, 0]:cab
[0.0040][0.0008][174] [2, 1, 0]:cba
Note on 'no duplicates': Since permutations are accessed using an array (the iterator), if two iterators differ by one elements, they point to two different permutations. Although a little thin, I take this as an argument for not having duplicates.
The only question remaining is 'are they all there?'. One could say that there are n! distinct arrays pointing at a given permutation, but I'm not too sure about the validity of that argument, since it comes from a 'drawing'... Pointers welcome.
I didn't thoroughly scrub SO to check if this had been already formulated this way or in a similar way (although links in the original thread use other approaches). Apologies if it did.
For a given size N the FactorialSequence produces a sequence of all arrays
[ i.0, i.1, ..., i.(N-1) ]
such that
0 <= i.0 < N, 0 <= i.1 < N-1, ..., 0 <= i.(N-1) < 1
that are exactly
N * (N-1) * ... * 1 = N!
elements. The permute() function then picks the element with index i.0
from the given array with N elements, then the element with i.1 from
the remaining N-1 elements, and so on.
So yes, this indeed produces all possible permutations of the array.
However, the code can be simplified a bit. First, FactorialSequence
does not return the initial array [ 0, ..., 0 ], corresponding to
the identity permutation. Also the separate __next() method seems
unnecessary.
If we change the code to
public struct FactorialSequence : Sequence, IteratorProtocol {
private var current: [Int]
private var firstIteration = true
public init(size:Int) {
self.current = Array(repeating: 0, count: size)
}
public mutating func next() -> [Int]? {
if firstIteration {
firstIteration = false
return current
}
for i in self.current.indices.reversed() {
if current[i] < current.count - i - 1 {
current[i] += 1
return current;
}
current[i] = 0
}
return nil
}
}
then all permutations are returned (including the initial identity), and
the defer statement is no longer necessary.
The permute() function can be simplified slightly, and made generic:
func permute<E>(array: [E], at: [Int]) -> [E] {
if at.isEmpty { return [] }
var at = at
var array = array
let firstIndex = at.remove(at: 0)
let firstElement = array.remove(at: firstIndex)
return [firstElement] + permute(array: array, at: at)
}
Now
let array = ["a", "b", "c"]
let fs = FactorialSequence(size: 3)
for p in fs {
print(permute(array: array, at: p).joined())
}
produces the expected output.
Note however that permute() produces a lot of intermediate arrays,
therefore I assume it to be less efficient than the other methods
that you referenced.
An alternative would be to swap the picked element to its new
place, this avoids recursion and temporary arrays:
func permute<E>(array: [E], at: [Int]) -> [E] {
var result = array
for (i, j) in at.enumerated() {
result.swapAt(i, i + j)
}
return result
}
(It produces the permutations in a different order, though.)

How to split a sequential integer array, if any sequence is missing

Supposre I have an input array of integers. I want to split this array in multiple array based on the missing integer and append it in a new Array. I think split can be used here but not sure how to do it. I want arrayFinal only.
myArray = [0,1,2,4,7,8]
Desired Output
arrayOne = [0,1,2]
arrayTwo = [4]
arrayThree = [7,8]
arrayFinal = [[0,1,2], [4], [7,8]]
That's an algorithm you're asking for so there are a dozen different ways to do it. Since you are going to have to walk through the array's contents to find the missing integers, I would just create an array and append the numbers to it as you go, then create a new array whenever you hit a gap.
You'll probably have to adjust this for any special cases you might have. "Will this always start at 0 and move in a positive direction?" etc.
Try this out:
func splitByMissingInteger(array: [Int]) -> [[Int]]? {
var arrayFinal :[[Int]] = [ [Int]() ]
var i = 0
for num in array{
if arrayFinal[i].isEmpty || (arrayFinal[i].last == nil){
arrayFinal[i].append(num)
} else if num == (arrayFinal[i].last! + 1){
arrayFinal[i].append(num)
} else {
i += 1
arrayFinal.append([Int]())
arrayFinal[i].append(num)
}
}
return arrayFinal
}
You can just sort your array and iterate them in order. Check if the element minus one is equal to the last element of your 2D array, if true append otherwise append a new array with the element and increase the index of the subarrays:
extension Collection where Element == Int {
func grouped() -> [[Element]] {
let elements = Set(self).sorted()
guard let first = elements.first else { return [] }
var result = [[first]]
var i = 0
for element in elements.dropFirst() {
if element-1 == result[i].last! {
result[i].append(element)
} else {
result.append([element])
i += 1
}
}
return result
}
}
let myArray = [0,1,2,4,7,8]
let grouped = myArray.grouped() // [[0, 1, 2], [4], [7, 8]]

Check if array contains element in Swift 4

I am trying to check if array categories contain number 1 as Int since categories = [Int]() for example categories = {1, 2, 3, 4, 5}
I have tried the below code which gives me error Binary operator '==' cannot be applied to operands of type 'Any' and 'Int'
if categories.contains (where: {$0 == 1}) {
// 1 is found
}
also tried it without the where and brackets as below which gives me the same error
if categories.contains { $0 == 1 } {
// 1 is found
}
I tried using just the element as below which gives me error Missing argument label 'where:' in call
if categories.contains(1) {
// 1 is found
}
How can I do that?
It seems like your category array is of type Any
Ways to fix it
You can declare your array as an Int array
var categories: [Int]
OR
You can change the following piece of code
if categories.contains { $0 == 1 } {
// 1 is found
}
to
if categories.contains { ($0 as! Int) == 1 } {
// 1 is found
}
Note: This method might cause your app to crash if your category array has an element other than of type Int
it is working See my output in PlayGround
Code used:
var categories : [Int] = [0,1,2,3,4,5,6,7,8,9]
if categories.contains(5)
{
print("Yes it contains")
}
else
{
print("it do not")
}
and also This condition is working
if categories.contains (where: {$0 == 1}) {
print("yes")
}
see your Array Declaration I think there is main Issue
Declaration 1 :
var categories = [Int]()
categories = [0,1,2,3,4,5,6,7,8,9]
Declaration 2 :
var categories : [Int] = [0,1,2,3,4,5,6,7,8,9]
Regarding error message
Binary operator '==' cannot be applied to operands of type 'Any' and 'Int'
Your array is not an Int array instead it contains Any so it needs typecasting before comparision. Declaration of array is also wrong use [] instead of {}. And typecast object as an int ($0 as! Int) == 1 (I'm using force casting here because I know its an Int array).
There are many ways to check if array contains any element.
1> Just try to get the index of element with guard if index is nil means array doesn't contain the element. Although you didn't declare array in right way still I'm considering it a valid array.
let categories: [Int] = [1, 2, 3, 4, 5]
guard categories.index(of: 1) != nil else {
print("Doesn't Contain")
return
}
print("Contains")
2> Use contains method
if (categories.contains(1)) {
print("Contains")
}
else {
print("Doesn't Contain")
}
3> Not Recommended for this case But still you can get this
let result = categories.filter({$0 == 1})
if result.count == 0 {
print("Doesn't Contain")
}
else {
print("Contains")
}
filter returns an array of element which matches with condition. So that if there are multiple 1 in array so it will give you an array of all elements. And $0 describes the object while enumerating the array.
4> Not Recommended for this case
let contains = categories.contains(where: {$0 == 1})
if contains {
print("Contains")
}
else {
print("Doesn't Contain")
}
Thanks to your comments made me check the declaration of the array and the problem was that was declared as [Any] after I get it's value from the UserDefaults. I have checked and found the solution on How do I save an Int array in Swift using NSUserDefaults?
// old declaration
let categories = userDefaults.array(forKey:"categories") ?? [Int]()
// new correct declaration
var categories = [Int]()
if let temp = userDefaults.array(forKey:"categories") as? [Int] {
categories = temp
}

How to Remove Every Other Element in an Array in Swift?

So say I have an array:
var stringArray = ["a","b","c","d","e","f","g","h","i","j"]
Now, how do I delete "a", "c", "e", "g", and "i" (all the even number indexes from the array)?
Thanks!
Instead of using C-style for-loops (which are set to be deprecated in an upcoming version of Swift), you could accomplish this using strides:
var result = [String]()
for i in stride(from: 1, through: stringArray.count - 1, by: 2) {
result.append(stringArray[i])
}
Or for an even more functional solution,
let result = stride(from: 1, to: stringArray.count - 1, by: 2).map { stringArray[$0] }
Traditional
var filteredArray = []
for var i = 1; i < stringArray.count; i = i + 2 {
filteredArray.append(stringArray[i])
}
Functional alternative
var result = stringArray.enumerate().filter({ index, _ in
index % 2 != 0
}).map { $0.1 }
enumerate takes a array of elements and returns an array of tuples where each tuple is an index-array pair (e.g. (.0 3, .1 "d")). We then remove the elements that are odd using the modulus operator. Finally, we convert the tuple array back to a normal array using map. HTH
There are a bunch of different ways to accomplish this, but here are a couple that I found interesting:
Using flatMap() on indices:
let result: [String] = stringArray.indices.flatMap {
if $0 % 2 != 0 { return stringArray[$0] }
else { return nil }
}
Note: result needs to be defined as a [String] otherwise the compiler doesn't know which version of flatMap() to use.
Or, if you want to modify the original array in place:
stringArray.indices.reverse().forEach {
if $0 % 2 == 0 { stringArray.removeAtIndex($0) }
}
In this case you have to call reverse() on indices first so that they're enumerated in reverse order. Otherwise by the time you get to the end of the array you'll be attempting to remove an index that doesn't exist anymore.
Swift 4.2
A function accepting generics and producing reduced result
func stripElements<T>(in array:[T]) -> [T] {
return array.enumerated().filter { (arg0) -> Bool in
let (offset, _) = arg0
return offset % 2 != 0
}.map { $0.element }
}

Resources