Printing a Tree With Indents. Swift - swift

Implementing a Tree data structure using swift:
class Node {
var value: String
var children: [Node] = []
weak var parent: Node?
init(_ value: String) {
self.value = value
}
func add(_ child: Node){
children.append(child)
child.parent = self
}
func printTree() {
var text = self.value
if !self.children.isEmpty {
text += "\n " + self.children.map{$0.printTree()}.joined(separator: ", ")
}
print(text)
}
}
My goal is to see something like this:
A1
B2
C3
G6
K0
H7
L8
L9
I know there should be some smart way to insert indents, but I also struggle with 'map'. Compiler gives me "ambiguous reference to member 'map'".

If you want to make it pretty, you could do this:
extension Node
{
func treeLines(_ nodeIndent:String="", _ childIndent:String="") -> [String]
{
return [ nodeIndent + value ]
+ children.enumerated().map{ ($0 < children.count-1, $1) }
.flatMap{ $0 ? $1.treeLines("┣╸","┃ ") : $1.treeLines("┗╸"," ") }
.map{ childIndent + $0 }
}
func printTree()
{ print(treeLines().joined(separator:"\n")) }
}
a1.printTree()
// A1
// ┣╸B2
// ┣╸C3
// ┃ ┗╸G6
// ┗╸K0
// ┣╸H7
// ┃ ┗╸L8
// ┗╸L9
You could also generalize it into a print function for any tree structure that will let you choose what to print for each node:
func printTree<T>(_ node:T, _ nodeInfo:#escaping (T)->(String,[T]) )
{
func lines(_ aNode:T, _ nodeIndent:String="", _ childIndent:String="") -> [String]
{
let (label,children) = nodeInfo(aNode)
return [ nodeIndent + label]
+ children.enumerated().map{ ($0 < children.count-1, $1) }
.flatMap{ $0 ? lines($1,"┣╸","┃ ") :lines($1,"┗╸"," ") }
.map{ childIndent + $0 }
}
print( lines(node).joined(separator:"\n") )
}
// print a root node providing a capture to obtain the node's label
// and its array of children
printTree(a1){ ($0.value,$0.children) }
// works for any tree structure. for example, views :
printTree(view){( "\(type(of:$0)) \($0.frame)", $0.subviews )}

I think vacawama's answer is quite smart, but if you want to practice recursive call including maps, you can write something like this:
class Node {
//...
func treeLines() -> [String] {
return [self.value] + self.children.flatMap{$0.treeLines()}.map{" "+$0}
}
func printTree() {
let text = treeLines().joined(separator: "\n")
print(text)
}
}
To test:
let a1 = Node("A1")
let b2 = Node("B2")
let c3 = Node("C3")
let g6 = Node("G6")
let k0 = Node("K0")
let h7 = Node("H7")
let l8 = Node("L8")
let l9 = Node("L9")
a1.add(b2)
a1.add(c3)
a1.add(k0)
c3.add(g6)
k0.add(h7)
k0.add(l9)
h7.add(l8)
a1.printTree()
Output:
A1
B2
C3
G6
K0
H7
L8
L9
Some points:
Your printTree() is a void function, so using it in map produces an Array of empty tuples, which cannot be joined.
(In Swift, void functions can be considered as returning empty tuple. And Swift generates ambiguous reference to member when an appropriate overload cannot be found.)
You may need an intermediate function the result of which can represent a partial tree structure.
Indentation needs to be added to each line, so the intermediate result should easily be separated into lines.

Something like this should work:
func printTree(_ indent: String = "") {
print(indent + self.value)
for child in children {
child.printTree(indent + " ")
}
}

Related

How can I build a recursive function in Swift to return a String?

I have a Node class defined as follows. value: T is a String
class Node<T> {
var value: T
weak var parent: Node?
var children = [Node<T>]()
init(_ value: T) {
self.value = value
}
func add(_ node: Node<T>) {
children.append(node)
node.parent = self
}
}
I'd like to build a function to return a String of the current Node's value and all Parent values. Ideally, the function would be defined in the class. For example,
currentnode.listAllValues()
would return -> "/parent2value/parent1value/currentnodevalue"
So far the following function works with a simple print(), and I've also considered using an inout parameter.
func listAllValues(node: Node<String>) {
print(node.value)
if node.parent?.value != nil {
listAllValues(node: node.parent!)
}
}
You can achieve this with an instance method which calls itself on the parent node first, if there is one:
func listAllValues() -> String {
if let p = parent {
return "\(p.listAllValues())/\(value)"
} else {
return "/\(value)"
}
}
Or as an obfuscated one-liner:
func listAllValues() -> String {
return "\(parent?.listAllValues() ?? "")/\(value)"
}
Example:
let p2 = Node("parent2value")
let p1 = Node("parent1value") ; p2.add(p1)
let n = Node("currentNodeValue") ; p1.add(n)
print(n.listAllValues())
// --> /parent2value/parent1value/currentNodeValue
Here you go:
func desc(_ s:String? = nil) -> String {
var me = String(describing:self.value)
if let prev = s {
me += "/" + prev
}
return self.parent?.desc(me) ?? me
}
Example:
let n = Node("hey")
n.add(Node("ho"))
n.children.first?.add(Node("nonny nonny no"))
let start = n.children.first!.children.first!
print(start.desc())

How to fix this Binary Tree Deserialize problem?

Where is the bug in this serialize() function
I wrote a function deserialize() in Swift to construct a binary tree from an array of optional Integer. I tested this function by comparing the output with the output of a tree constructed using manual method with same array. They look the same, which is good !
However, when I run another function isSameTree(), that is used to compare 2 trees (I am sure this function is working correctly), on deserialize()'s output and manual method's output, I have different results !
I assume deserialize() is not correct, but I could not find the bug!
// helper code
public class BinaryNode {
public var value: Int
public var left: BinaryNode?
public var right: BinaryNode?
public init(_ value: Int) {
self.value = value
self.left = nil
self.right = nil
}
}
extension BinaryNode {
public var description: String {
return diagram(for: self)
}
private func diagram(for node: BinaryNode?,
_ top: String = "",
_ root: String = "",
_ bottom: String = "") -> String {
guard let node = node else {
return root + "nil\n"
}
if node.left == nil && node.right == nil {
return root + "\(node.value)\n"
}
return diagram(for: node.right,
top + " ", top + "┌──", top + "│ ")
+ root + "\(node.value)\n"
+ diagram(for: node.left,
bottom + "│ ", bottom + "└──", bottom + " ")
}
}
public func deserialize(_ array: inout [Int?]) -> BinaryNode? {
guard !array.isEmpty, let value = array.removeFirst() else {
return nil
}
let node = BinaryNode(value)
node.left = deserialize(&array)
node.right = deserialize(&array)
return node
}
func isSameTree(_ p: BinaryNode?, _ q: BinaryNode?) -> Bool {
guard let p = p else {
return q == nil
}
guard let q = q else {
return p == nil
}
if p.value != q.value {
return false
}
return isSameTree(p.left, q.left) && isSameTree(p.right, q.right)
}
// Using deserialize to construct trees
var a1: [Int?] = [1,nil,2,3]
var a2: [Int?] = [1,nil,2,nil,3]
if let tree = deserialize(&a1) {
print(tree.description)
}
if let tree = deserialize(&a2) {
print(tree.description)
}
// Using manual to construct trees
let a3: BinaryNode = {
let one = BinaryNode(1)
let two = BinaryNode(2)
let three = BinaryNode(3)
one.right = two
two.left = three
return one
}()
print(a3.description)
let a4: BinaryNode = {
let one = BinaryNode(1)
let two = BinaryNode(2)
let three = BinaryNode(3)
one.right = two
two.right = three
return one
}()
print(a4.description)
// The print statements above show similar trees are constructed
// However, below results are not same
isSameTree(deserialize(&a1), deserialize(&a2)) // true <- this is wrong
isSameTree(a3, a4) // false <--- this is correct
Seems you have forgotten that your deserialize(_:) is destructive for its parameter... Please remember why you need &.
//Re-load a1 & a2...
a1 = [1,nil,2,3]
a2 = [1,nil,2,nil,3]
print(isSameTree(deserialize(&a1), deserialize(&a2))) //-> false
print(isSameTree(a3, a4)) //-> false

In Swift How do I iterate over an array getting 2 variables when each pair of elements is a String? and a String

I want to iterate over an array String?, String repeated pair but I cannot form the "for case let (a,b) in array" correctly.
The best I have come up with is to create a temp struct of {String?, String} and create an array of the temp structs and then iterate it but I would like to skip this step.
Below is the basic example with the last for loop showing the error Xcode reports.
class Foo {
var s1: String?
var s2: String?
var s3: String?
}
let foo = Foo()
foo.s1="Test1"
foo.s2=nil
foo.s3="Test3"
let fooArray = [foo.s1, ", ", foo.s2, "; ", foo.s3,"."]
let fooArray1 = [foo.s1,foo.s2, foo.s3]
var text:String = ""
for case let prop? in fooArray1 {
text = text + prop + " / "
}
print(text)
// The above works but now I want to use a different separator
//base on the property name
text=""
for case let (prop, sep) in fooArray { // Error <= Expression Type
// [String?] is ambiguous without more context
text = text + prop + sep
}
print(text)
Here is what I have come up with
struct temp {
var prop:String?
var sep:String
init(_ prop:String?, _ sep:String) {
self.prop=prop
self.sep=sep
}
let ary:[temp] = [ temp(foo.s1,", "), temp(foo.s2,"; "), temp(foo.s3,".") ]
text = ""
for a in ary {
if let p = a.prop {
text = text + p + a.sep
}
}
print (text)
is there another way just using the for loop
for (a,b) in fooArray {
...
}
As noted by #RJE, the inferred type of fooArray, as defined in your code, is [String?].
Here is one way to make it work:
class Foo {
var s1: String?
var s2: String?
var s3: String?
}
let foo = Foo()
foo.s1 = "Test1"
foo.s2 = nil
foo.s3 = "Test3"
let fooArray1 = [foo.s1, foo.s2, foo.s3]
let separators = [", ", "; ", "."]
var text = ""
for i in fooArray1.indices {
if let p = fooArray1[i] {
text = text + p + separators[i]
}
}
print (text) //Test1, Test3.
Or
let zipped = zip(fooArray1, separators)
let text = zipped.map { tuple -> String in
if case let (x?, y) = tuple {
return x + y
} else {
return ""
}
}.joined()
print (text) //Test1,Test3.
Or
let fooArray = [foo.s1, ", ", foo.s2, "; ", foo.s3, "."]
var text = ""
var step = 1
var index = 0
while index < fooArray.count {
if let str = fooArray[index] {
step = 1
text += str
} else {
step = 2
}
index += step
}
print(text) //Test1, Test3.
It would be better to define the initializer this way :
class Foo {
var s1: String?
var s2: String?
var s3: String?
init(s1: String?, s2: String?, s3: String?) {
self.s1 = s1
self.s2 = s2
self.s3 = s3
}
}
let foo = Foo(s1: "Test1", s2: nil, s3: "Test3")
P.S: The desired output seems to be more appropriate for a description property of the Foo class.
Thanks for the answer I was hoping through this question to get a better understanding of how to use [for] parameters. But the while solution is the solution I would probably use with the following modifications
text = ""
var index = 0
while index < fooArray.count {
if let prop = fooArray[index] {
index += 1
let sep = fooArray[index]!
index += 1
text = text + prop + sep
} else {
index += 2
}
}

Find the Range of the Nth word in a String

What I want is something like
"word1 word2 word3".rangeOfWord(2) => 6 to 10
The result could come as a Range or a tuple or whatever.
I'd rather not do the brute force of iterating over the characters and using a state machine. Why reinvent the lexer? Is there a better way?
In your example, your words are unique, and you can use the following method:
let myString = "word1 word2 word3"
let wordNum = 2
let myRange = myString.rangeOfString(myString.componentsSeparatedByString(" ")[wordNum-1])
// 6..<11
As pointed out by Andrew Duncan in the comments below, the above is only valid if your words are unique. If you have non-unique words, you can use this somewhat less neater method:
let myString = "word1 word2 word3 word2 word1 word3 word1"
let wordNum = 7 // 2nd instance (out of 3) of "word1"
let arr = myString.componentsSeparatedByString(" ")
var fromIndex = arr[0..<wordNum-1].map { $0.characters.count }.reduce(0, combine: +) + wordNum - 1
let myRange = Range<String.Index>(start: myString.startIndex.advancedBy(fromIndex), end: myString.startIndex.advancedBy(fromIndex+arr[wordNum-1].characters.count))
let myWord = myString.substringWithRange(myRange)
// string "word1" (from range 36..<41)
Finally, lets use the latter to construct an extension of String as you have wished for in your question example:
extension String {
private func rangeOfNthWord(wordNum: Int, wordSeparator: String) -> Range<String.Index>? {
let arr = myString.componentsSeparatedByString(wordSeparator)
if arr.count < wordNum {
return nil
}
else {
let fromIndex = arr[0..<wordNum-1].map { $0.characters.count }.reduce(0, combine: +) + (wordNum - 1)*wordSeparator.characters.count
return Range<String.Index>(start: myString.startIndex.advancedBy(fromIndex), end: myString.startIndex.advancedBy(fromIndex+arr[wordNum-1].characters.count))
}
}
}
let myString = "word1 word2 word3 word2 word1 word3 word1"
let wordNum = 7 // 2nd instance (out of 3) of "word1"
if let myRange = myString.rangeOfNthWord(wordNum, wordSeparator: " ") {
// myRange: 36..<41
print(myString.substringWithRange(myRange)) // prints "word1"
}
You can tweak the .rangeOfNthWord(...) method if word separation is not unique (say some words are separated by two blankspaces " ").
Also pointed out in the comments below, the use of .rangeOfString(...) is not, per se, pure Swift. It is, however, by no means bad practice. From Swift Language Guide - Strings and Characters:
Swift’s String type is bridged with Foundation’s NSString class. If
you are working with the Foundation framework in Cocoa, the entire
NSString API is available to call on any String value you create when
type cast to NSString, as described in AnyObject. You can also use a
String value with any API that requires an NSString instance.
See also the NSString class reference for rangeOfString method:
// Swift Declaration:
func rangeOfString(_ searchString: String) -> NSRange
I went ahead and wrote the state machine. (Grumble..) FWIW, here it is:
extension String {
private func halfOpenIntervalOfBlock(n:Int, separator sep:Character? = nil) -> (Int, Int)? {
enum State {
case InSeparator
case InPrecedingSeparator
case InWord
case InTarget
case Done
}
guard n > 0 else {
return nil
}
var state:State
if n == 1 {
state = .InPrecedingSeparator
} else {
state = .InSeparator
}
var separatorNum = 0
var startIndex:Int = 0
var endIndex:Int = 0
for (i, c) in self.characters.enumerate() {
let inSeparator:Bool
// A bit inefficient to keep doing this test.
if let s = sep {
inSeparator = c == s
} else {
inSeparator = c == " " || c == "\n"
}
endIndex = i
switch state {
case .InPrecedingSeparator:
if !inSeparator {
state = .InTarget
startIndex = i
}
case .InTarget:
if inSeparator {
state = .Done
}
case .InWord:
if inSeparator {
separatorNum += 1
if separatorNum == n - 1 {
state = .InPrecedingSeparator
} else {
state = .InSeparator
}
}
case .InSeparator:
if !inSeparator {
state = .InWord
}
case .Done:
break
}
if state == .Done {
break
}
}
if state == .Done {
return (startIndex, endIndex)
} else if state == .InTarget {
return (startIndex, endIndex + 1) // We ran off end.
} else {
return nil
}
}
func rangeOfWord(n:Int) -> Range<Index>? {
guard let (s, e) = self.halfOpenIntervalOfBlock(n) else {
return nil
}
let ss = self.startIndex.advancedBy(s)
let ee = self.startIndex.advancedBy(e)
return Range(start:ss, end:ee)
}
}
It's not really clear whether the string has to be considered divided in words by separators it may contains, or if you're just looking for a specific substring occurrence.
Anyway both cases could be addressed in this way in my opinion:
extension String {
func enumerateOccurencies(of pattern: String, _ body: (Range<String.Index>, inout Bool) throws -> Void) rethrows {
guard
!pattern.isEmpty,
count >= pattern.count
else { return }
var stop = false
var lo = startIndex
while !stop && lo < endIndex {
guard
let r = self[lo..<endIndex].range(of: pattern)
else { break }
try body(r, &stop)
lo = r.upperBound
}
}
}
You'll then set stop to true in the body closure once reached the desired occurrence number and capture the range passed to it:
let words = "word1, word1, word2, word3, word1, word3"
var matches = 0
var rangeOfThirdOccurencyOfWord1: Range<String.Index>? = nil
words.enumerateOccurencies(of: "word1") { range, stop in
matches +=1
stop = matches == 3
if stop {
rangeOfThirdOccurencyOfWord1 = range
}
}
Regarding the DFA: recently I've wrote one leveraging on Hashable and using a an Array of Dictionaries as its state nodes, but I've found that the method above is faster, cause maybe range(of:) uses finger-printing.
UPDATE
Otherwise you could also achieve that API you've mentioned in this way:
import Foundation
extension String {
func rangeOfWord(order: Int, separator: String) -> Range<String.Index>? {
precondition(order > 0)
guard
!isEmpty,
!separator.isEmpty,
separator.count < count
else { return nil }
var wordsSoFar = 0
var lo = startIndex
while let r = self[lo..<endIndex].range(of: separator) {
guard
r.lowerBound != lo
else {
lo = r.upperBound
continue
}
wordsSoFar += 1
guard
wordsSoFar < order
else { return lo..<r.lowerBound }
lo = r.upperBound
}
if
lo < endIndex,
wordsSoFar + 1 == order
{
return lo..<endIndex
}
return nil
}
}
let words = "word anotherWord oneMore lastOne"
if let r = words.rangeOfWord(order: 4, separator: " ") {
print(words[r])
} else {
print("not found")
}
Here order parameter refers to the nth order of the word in the string, starting from 1. I've also added the separator parameter to specify a string token to use for finding words in the string (it can also be defaulted to " " to be able to call the function without having to specify it).
Here's my attempt at an updated answer in Swift 5.5:
import Foundation
extension String {
func rangeOfWord(atPosition wordAt: Int) -> Range<String.Index>? {
let fullrange = self.startIndex..<self.endIndex
var count = 0
var foundAt: Range<String.Index>? = nil
self.enumerateSubstrings(in: fullrange, options: .byWords) { _, substringRange, _, stop in
count += 1
if count == wordAt {
foundAt = substringRange
stop = true // Stop the enumeration after the word range is found.
}
}
return foundAt
}
}
let lorem = "Morbi leo risus, porta ac consectetur ac, vestibulum at eros."
if let found = lorem.rangeOfWord(atPosition: 8) {
print("found: \(lorem[found])")
} else {
print("not found.")
}
This solution doesn't make a new array to contain the words so uses less memory (I have not tested but in theory it should use less memory). As much as possible, the build in method is used therefore less chance of bugs.
Swift 5 solution, which allows you to specify the word separator
extension String {
func rangeOfWord(atIndex wordIndex: Int) -> Range<String.Index>? {
let wordComponents = self.components(separatedBy: " ")
guard wordIndex < wordComponents.count else {
return nil
}
let characterEndCount = wordComponents[0...wordIndex].map { $0.count }.reduce(0, +)
let start = String.Index(utf16Offset: wordIndex + characterEndCount - wordComponents[wordIndex].count, in: self)
let end = String.Index(utf16Offset: wordIndex + characterEndCount, in: self)
return start..<end
}
}

Split a String without removing the delimiter in Swift

This might be a duplicate. I couldn't find the answer in Swift, so I am not sure.
componentsSeparatedByCharactersInSet removes the delimiter. If you separate by only one possible character it is easy to add it back. But what when you have a set?
Is there another method to split?
Swift 3 and 4 Versions
extension Collection {
func splitAt(isSplit: (Iterator.Element) throws -> Bool) rethrows -> [SubSequence] {
var p = self.startIndex
var result:[SubSequence] = try self.indices.flatMap {
i in
guard try isSplit(self[i]) else {
return nil
}
defer {
p = self.index(after: i)
}
return self[p...i]
}
if p != self.endIndex {
result.append(suffix(from: p))
}
return result
}
}
Thanks to Oisdk for getting me thinking.
This method works on CollectionTypes, rather than Strings, but it should be easy enough to adapt:
extension CollectionType {
func splitAt(#noescape isSplit: Generator.Element throws -> Bool) rethrows -> [SubSequence] {
var p = startIndex
return try indices
.filter { i in try isSplit(self[i]) }
.map { i in
defer { p = i }
return self[p..<i]
} + [suffixFrom(p)]
}
}
extension CollectionType where Generator.Element : Equatable {
func splitAt(splitter: Generator.Element) -> [SubSequence] {
return splitAt { el in el == splitter }
}
}
You could use it like this:
let sentence = "Hello, my name is oisdk. This should split: but only at punctuation!"
let puncSet = Set("!.,:".characters)
sentence
.characters
.splitAt(puncSet.contains)
.map(String.init)
// ["Hello", ", my name is oisdk", ". This should split", ": but only at punctuation", "!"]
Or, this version, which uses a for-loop, and splits after the delimiter:
extension CollectionType {
func splitAt(#noescape isSplit: Generator.Element throws -> Bool) rethrows -> [SubSequence] {
var p = startIndex
var result: [SubSequence] = []
for i in indices where try isSplit(self[i]) {
result.append(self[p...i])
p = i.successor()
}
if p != endIndex { result.append(suffixFrom(p)) }
return result
}
}
extension CollectionType where Generator.Element : Equatable {
func splitAt(splitter: Generator.Element) -> [SubSequence] {
return splitAt { el in el == splitter }
}
}
let sentence = "Hello, my name is oisdk. This should split: but only at punctuation!"
let puncSet = Set("!.,:".characters)
sentence
.characters
.splitAt(puncSet.contains)
.map(String.init)
// ["Hello,", " my name is oisdk.", " This should split:", " but only at punctuation!"]
Or, if you wanted to get the most Swift features into one function (defer, throws, a Protocol extension, an evil flatMap, guard, and Optionals):
extension CollectionType {
func splitAt(#noescape isSplit: Generator.Element throws -> Bool) rethrows -> [SubSequence] {
var p = startIndex
var result: [SubSequence] = try indices.flatMap { i in
guard try isSplit(self[i]) else { return nil }
defer { p = i.successor() }
return self[p...i]
}
if p != endIndex { result.append(suffixFrom(p)) }
return result
}
}
I came here looking for an answer to this question. Didn't find what I was looking for and ended up building this by repeated calls to .split(...) It isn't elegant but you can choose which delimiters are preserved and which aren't. There's probably a way to avoid the String <--> Substring conversions, anyone know?
var input = """
{All those moments will be (lost in time)},
like tears [in rain](. ([(Time to)] die))
"""
var separator: Character = "!"
var output: [String] = []
repeat {
let tokens = input.split(
maxSplits: 1,
omittingEmptySubsequences: false,
whereSeparator: {
switch $0 {
case "{", "}", "(", ")", "[", "]": // preserve
separator = $0; return true
case " ", "\n", ",", ".": // omit
separator = " "; return true
default:
return false
}
}
)
if tokens[0] != "" {
output.append(String(tokens[0]))
}
guard tokens.count == 2 else { break }
if separator != " " {
output.append(String(separator))
}
input = String(tokens[1])
} while true
for token in output { print("\(token)") }
In the case above, the selectors are not in actual sets. I didn't need that, but if you do, simply make these declarations,
let preservedDelimiters: Set<Character> = [ "{", "}", "(", ")", "[", "]" ]
let omittedDelimiters: Set<Character> = [ " ", "\n", ",", "." ]
and replace the whereSeparator function with:
whereSeparator: {
if preservedDelimiters.contains($0) {
separator = $0
return true
} else if omittedDelimiters.contains($0) {
separator = " "
return true
} else {
return false
}
}