I am trying to compare the values of two arrays in Swift. If a value of array2 is not found in array1 all the array2 found values need to be list and deleted.
I was trying to use the code below but its not working anymore in Swift 2:
let array1 = [["aaa","12"],["bbb","349"],["ccc","91"],["ddd","143"]]
let array2 = ["aaa","SSS","bbb","ccc","QQQ","ZZZ","ddd"]
let notNeededValues = filter(enumerate(zip(array1,array2))) { $1.0 == $1.1 }.map{ $0.0 }
print(notNeededValues)
Not sure if I understand your problem correctly, but the problem seems to be, that your code needs a simple conversion to Swift 2 syntax:
let array1 = [["aaa","12"],["bbb","349"],["ccc","91"],["ddd","143"]]
let array2 = ["aaa","SSS","bbb","ccc","QQQ","ZZZ","ddd"]
let notNeededValues = zip(array1, array2).enumerate().filter { $1.0 == $1.1 }.map { $0.0 }
print(notNeededValues)
Swift is moving away from globally defined functions, like filter and enumerate once were, and is using dot-syntax instead. This change was made possible by protocol extensions, and makes code more readable.
Update:
I assume this is what you mean(?):
let notNeededValues = array2.filter { !array1.map { $0[0] }.contains($0) }
// or like this:
let array1FirstElements = array1.map { $0[0] }
let notNeededValues = array2.filter { !array1FirstElements.contains($0) }
How about this:
let array1 = [["aaa","12"],["bbb","349"],["ccc","91"],["ddd","143"]]
let array2 = ["aaa","SSS","bbb","ccc","QQQ","ZZZ","ddd"]
extension Array where Element: Equatable {
func removeObject(object: Element) -> [Element] {
return filter {$0 != object}
}
}
var filteredArray2 = array2.reduce(array2) {
if array1.flatMap({$0}).contains($1) {
return $0.removeObject($1)
}
return $0
}
print(filteredArray2)
Related
Writing the question and answer from here, I'm curious to know if there is any simpler way to write the following:
var nums = [1,2,3]
let sum1 = nums.reduce([Int]()){
let temp = $0
temp.append($1)
return temp
}
I know I can do:
var nums = [1,2,3]
let sum1 = nums.reduce([Int]()){
return $0 + [$1]
}
But that comes off as a hack.
To explain this better, I want to get closer to the example (from docs) below, just that it should be for an array:
let numbers = [1, 2, 3, 4]
let numberSum = numbers.reduce(0, { x, y in
x + y
})
EDIT:
Since folks asked what was I trying to achieve:
I was doing the leetcode's group Anagram's challenge.
My solution was:
struct WordTraits: Equatable{
let count: Int
let charactersSet: Set<Character>
}
struct Word: Equatable{
let string: String
let wordTraits: WordTraits
}
class Solution{
func groupAnagrams(_ strs: [String]) -> [[String]]{
var words : [Word] = []
var answers : [(traits: WordTraits, words: [Word])] = []
var count = 0
strs.forEach{ str in
count += 1
let count = str.count
let string = str
let characterSet = Set(str)
let wordTraits = WordTraits(count: count, charactersSet: characterSet)
let word = Word(string: string, wordTraits: wordTraits)
words.append(word)
}
while words.count != 0{
let word = words[0]
let traits = word.wordTraits
var isWordAdded = false
for (i, answer) in answers.enumerated(){
if answer.traits == traits{
answers[i].words.append(word)
isWordAdded = true
break
}
}
if !isWordAdded{
answers.append((traits: traits, words:[word]))
}
words.removeFirst()
}
let emptyArray : [[String]] = []
let finalAnswer = answers.reduce(emptyArray, { total, answer in
let strings : [String] = answer.words.reduce([String](), {
return $0 + [$1.string]
})
return total + [strings]
})
return finalAnswer
}
}
let s = Solution()
print(s.groupAnagrams(["ate", "eta", "beta", "abet"])) // [["ate", "eta"], ["beta", "abet"]]
reduce(..) has to know which type it is working with. To infer this it can use the return type or the type of the first argument. So you can also write:
var nums = [1,2,3]
let sum1: [Int] = nums.reduce([]){
return $0 + [$1]
}
[$1] can't be replaced with $1 because +-operator between value and collection is undefined.
Nope. But you can add it:
extension Array {
func appending(_ newElement: Element) -> Array<Element> {
return self + [newElement]
}
func appending(contentsOf sequence: Sequence) -> Array<Element> {
return self + sequence
}
}
Um, how about the + operator?
let nums = [1, 3, 5]
let more = nums + [7]
Your code is trying to convert a complex structure to an array of arrays. You can use map for this.
This should work:
let finalAnswer = answers.map { answer in
answer.words.map {
$0.string
}
}
Edit:
I was able to solve it using minimal code:
class Solution {
func groupAnagrams(_ words: [String]) -> [[String]] {
let processedWords = words.map {
(key: String($0.sorted()), value: $0)
}
return Dictionary(grouping: processedWords, by: { $0.key }).map { groupedValue in
groupedValue.value.map {
$0.value
}
}
}
}
You've greatly overcomplicated your computation of "final answers". It could just be:
return answers.map { $0.words.map { $0.string } }
So I have this method to get an array of random ints between 1-9, a random number of times between 1 and 7.
let n = arc4random_uniform(7) + 1
var arr: [UInt32] = []
for _ in 0 ... n {
var temp = arc4random_uniform(9) + 1
while arr.contains(temp) {
temp = arc4random_uniform(9) + 1
}
print(temp)
arr.append(temp)
}
print(arr)
So that gets me an array like [1,4,2] or [5,7,3,4,6]. And I have a method to turn another array of strings into a enumerated dictionary.
var someArray: [String] = ["War", "Peanuts", "Cats", "Dogs", "Nova", "Bears", "Pigs", "Packers", "Mango", "Turkey"]
extension Collection {
var indexedDictionary: [Int: Element] {
return enumerated().reduce(into: [:]) { $0[$1.offset] = $1.element }
}
}
let dict1 = someArray.indexedDictionary
print(dict1)
giving me the indexed dictionary
[1:"War", 2:"Peanuts",..etc]
MY question is using the Ints of the random array how do I create a new dictionary that only includes those keys and their values?
So for example if arr = [3,1,5]
how do I get a new dictionary of
[3:"dogs", 1:"Peanuts",5:"Bears"].
This should do it:
let finalDict = dict1.filter { arr.contains($0.key) }
Update:
You can even go a step further and skip the whole strings to array mapping. So remove
extension Collection {
var indexedDictionary: [Int: Element] {
return enumerated().reduce(into: [:]) { $0[$1.offset] = $1.element }
}
}
let dict1 = someArray.indexedDictionary
print(dict1)
and just use this:
Swift 4:
let finalArray = someArray.enumerated().flatMap { arr.contains($0.offset) ? $0.element : nil }
Swift 4.1:
let finalArray = someArray.enumerated().compactMap { arr.contains($0.offset) ? $0.element : nil }
Update 2:
If you need a dictionary and not an array in the end use this:
Swift 4:
let finalDict = someArray.enumerated().flatMap { randomInts.contains($0.offset) ? ($0.offset, $0.element) : nil }.reduce(into: [:]) { $0[$1.0] = $1.1 }
Swift 4.1:
let finalDict = someArray.enumerated().compactMap { randomInts.contains($0.offset) ? ($0.offset, $0.element) : nil }.reduce(into: [:]) { $0[$1.0] = $1.1 }
For example say I have an array like so:
var someArray = ["1", "1", "2"]
I need to put this into two arrays that look like:
["1","1"]
["2"]
How can I go about this?
Any help would be great!
Use Dictionary initializer init(grouping:by:)
Then just get arrays by accessing values property.
Example:
let dic = Dictionary(grouping: someArray) { $0 }
let values = Array(dic.values)
print(values)
Result:
[["2"], ["1", "1"]]
Here are some facts (the upvote and answer should go to #kirander)
With #kirander method's is using the Dictionary to map the objects in a O(N) runtime and O(N) memory.
The other solutions are mostly running in O(N*N) runtime and O(N) memory. Because of this, grouping a random array of 1000 items will take: 0.07s with #kirander solution and 34s. with other solutions.
func benchmark(_ title:String, code: ()->()) {
let startTime = CFAbsoluteTimeGetCurrent()
code()
let timeElapsed = CFAbsoluteTimeGetCurrent() - startTime
print("Time elapsed for \(title): \(timeElapsed) s.")
}
var array:[String] = []
for _ in 0...1000 {
array.append("\(Int(arc4random_uniform(10)))")
}
// #kirander solution 0.07s
benchmark("Dictionary", code: {
let dic = Dictionary(grouping: array, by: { $0 })
let values = Array(dic.values)
})
// #Bruno solution ~34s
benchmark("Array", code: {
var resultingArrays = [[String]]()
for value in array {
let ar = array.filter({ $0 == value })
if !resultingArrays.contains(where: {$0 == ar}) {
resultingArrays.append(ar)
}
}
})
You could try something like this:
var someArray = ["1", "1", "2"]
var resultingArrays = [[String]]()
for value in someArray {
let array = someArray.filter({ $0 == value })
if !resultingArrays.contains(where: {$0 == array}) {
resultingArrays.append(array)
}
}
You can try this one :
let arrM = ["1","3","4","6","1","1","3"]
let arrSrtd = Array(Set(arrM))
for ele in arrSrtd{
let a = arrM.filter( {$0 == ele})
print(a)
}
I would like to know how can i fill my label from an Array
func metaDataUpdated(metaData : NSString){
var listItems: NSArray = [metaData.componentsSeparatedByString(";")]
if ([listItems.count] > 0){
println([listItems.objectAtIndex(0)])
titleSong.text = [listItems.objectAtIndex(0)]
}
}
I don't really know how to convert an array to string.
Direct conversion to Swift:
func metaDataUpdated(metaData : String) {
let listItems = metaData.componentsSeparatedByString(";")
if listItems.count > 0 {
print(listItems[0])
titleSong.text = listItems[0]
}
}
Nicer Swift:
func metaDataUpdated(metaData : String) {
let listItems = metaData.componentsSeparatedByString(";")
if let first = listItems.first {
print(first)
titleSong.text = first
}
}
Even nicer Swift, without using Foundation and without the function needing to get every component separated by ";", but only the first one (recommended):
func metaDataUpdated(metaData : String) {
if let index = metaData.characters.indexOf(";") {
let first = metaData[metaData.startIndex ..< index]
print(first)
titleSong.text = first
}
}
you cannot assign NSArray to NSString therefore you need to cast the value of this first index into a string
change this
titleSong.text = [listItems.objectAtIndex(0)]
to
titleSong.text = "\(listItems.objectAtIndex(0))"
or
titleSong.text = listItems[0] as! String
and also change this line to ([listItems.count > 0]) to (listItems.count > 0)
your code will look like this:
Note this not obj-c so remove all []
func metaDataUpdated(metaData : NSString){
var listItems: NSArray = metaData.componentsSeparatedByString(";")
if (listItems.count > 0)
{
println(listItems.objectAtIndex(0))
titleSong.text = listItems.objectAtIndex(0) as! String
}
}
Better use Swift types and objects now: Array instead of NSArray, Dictionary instead of NSDictionary, etc.
func metaDataUpdated(metaData : NSString) {
var listItems = metaData.componentsSeparatedByString(";")
if listItems.count > 0 {
print(listItems[0])
titleSong.text = listItems[0]
}
}
Here componentsSeparatedByString returns an array of strings: [String]. We then use simple index subscripting to retrieve its first value.
Note: I suppose you were trying to adapt code from Objective-C because your example was ridden with [] everywhere...
Put your to string item in "\\()".
For instance:
titleSong.text = "\\([listItems.objectAtIndex(0)])"
Not sure you need the [] brackets though
I am new to Swift. I have been doing Java programming. I have a scenario to code for in Swift.
The following code is in Java. I need to code in Swift for the following scenario
// With String array - strArr1
String strArr1[] = {"Some1","Some2"}
String strArr2[] = {"Somethingelse1","Somethingelse2"}
for( int i=0;i< strArr1.length;i++){
System.out.println(strArr1[i] + " - "+ strArr2[i]);
}
I have a couple of arrays in swift
var strArr1: [String] = ["Some1","Some2"]
var strArr2: [String] = ["Somethingelse1","Somethingelse2"]
for data in strArr1{
println(data)
}
for data in strArr2{
println(data)
}
// I need to loop over in single for loop based on index.
Could you please provide your help on the syntaxes for looping over based on index
You can use zip(), which creates
a sequence of pairs from the two given sequences:
let strArr1 = ["Some1", "Some2"]
let strArr2 = ["Somethingelse1", "Somethingelse2"]
for (e1, e2) in zip(strArr1, strArr2) {
print("\(e1) - \(e2)")
}
The sequence enumerates only the "common elements" of the given sequences/arrays. If they have different length then the additional
elements of the longer array/sequence are simply ignored.
With Swift 5, you can use one of the 4 following Playground codes in order to solve your problem.
#1. Using zip(_:_:) function
In the simplest case, you can use zip(_:_:) to create a new sequence of pairs (tuple) of the elements of your initial arrays.
let strArr1 = ["Some1", "Some2", "Some3"]
let strArr2 = ["Somethingelse1", "Somethingelse2"]
let sequence = zip(strArr1, strArr2)
for (el1, el2) in sequence {
print("\(el1) - \(el2)")
}
/*
prints:
Some1 - Somethingelse1
Some2 - Somethingelse2
*/
#2. Using Array's makeIterator() method and a while loop
It is also easy to loop over two arrays simultaneously with a simple while loop and iterators:
let strArr1 = ["Some1", "Some2", "Some3"]
let strArr2 = ["Somethingelse1", "Somethingelse2"]
var iter1 = strArr1.makeIterator()
var iter2 = strArr2.makeIterator()
while let el1 = iter1.next(), let el2 = iter2.next() {
print("\(el1) - \(el2)")
}
/*
prints:
Some1 - Somethingelse1
Some2 - Somethingelse2
*/
#3. Using a custom type that conforms to IteratorProtocol
In some circumstances, you may want to create you own type that pairs the elements of your initials arrays. This is possible by making your type conform to IteratorProtocol. Note that by making your type also conform to Sequence protocol, you can use instances of it directly in a for loop:
struct TupleIterator: Sequence, IteratorProtocol {
private var firstIterator: IndexingIterator<[String]>
private var secondIterator: IndexingIterator<[String]>
init(firstArray: [String], secondArray: [String]) {
self.firstIterator = firstArray.makeIterator()
self.secondIterator = secondArray.makeIterator()
}
mutating func next() -> (String, String)? {
guard let el1 = firstIterator.next(), let el2 = secondIterator.next() else { return nil }
return (el1, el2)
}
}
let strArr1 = ["Some1", "Some2", "Some3"]
let strArr2 = ["Somethingelse1", "Somethingelse2"]
let tupleSequence = TupleIterator(firstArray: strArr1, secondArray: strArr2)
for (el1, el2) in tupleSequence {
print("\(el1) - \(el2)")
}
/*
prints:
Some1 - Somethingelse1
Some2 - Somethingelse2
*/
#4. Using AnyIterator
As an alternative to the previous example, you can use AnyIterator. The following code shows a possible implementation of it inside an Array extension method:
extension Array {
func pairWithElements(of array: Array) -> AnyIterator<(Element, Element)> {
var iter1 = self.makeIterator()
var iter2 = array.makeIterator()
return AnyIterator({
guard let el1 = iter1.next(), let el2 = iter2.next() else { return nil }
return (el1, el2)
})
}
}
let strArr1 = ["Some1", "Some2", "Some3"]
let strArr2 = ["Somethingelse1", "Somethingelse2"]
let iterator = strArr1.pairWithElements(of: strArr2)
for (el1, el2) in iterator {
print("\(el1) - \(el2)")
}
/*
prints:
Some1 - Somethingelse1
Some2 - Somethingelse2
*/
Try This:
zip([0,2,4,6], [1,3,5,7]).forEach {
print($0,$1)
}
zip([0,2,4,6], [1,3,5,7]).forEach {
print($0.0,$0.1)
}
You could also enumerate over one array and used the index to look inside the second array:
Swift 1.2:
for (index, element) in enumerate(strArr1) {
println(element)
println(strArr2[index])
}
Swift 2:
for (index, element) in strArr1.enumerate() {
print(element)
print(strArr2[index])
}
Swift 3:
for (index, element) in strArr1.enumerated() {
print(element)
print(strArr2[index])
}
You could use Range if you still want to use for in.
var strArr1: [String] = ["Some1","Some2"]
var strArr2: [String] = ["Somethingelse1","Somethingelse2"]
for i in Range(start: 0, end: strArr1.count) {
println(strArr1[i] + " - " + strArr2[i])
}
for(var i = 0; i < strArr1.count ; i++)
{
println(strArr1[i] + strArr2[i])
}
That should do it. Never used swift before so make sure to test.
Updated to recent Swift syntax
for i in 0..< strArr1.count {
print(strArr1[i] + strArr2[i])
}
> Incase of unequal count
let array1 = ["some1","some2"]
let array2 = ["some1","some2","some3"]
var iterated = array1.makeIterator()
let finalArray = array2.map({
let itemValue = iterated.next()
return "\($0)\(itemValue != nil ? "-"+itemValue! : EmptyString)" })
// result : ["some1-some1","some2-some2","some3"]