Related
I have readLine() which gives me the character. So, I have this:
223 345 567 and so on. So, when I convert them to Int first thing says these are character so, I researched and found this solution and when I use this:
let size = readLine()
var array = [Int]()
let numbers = readLine()
for number in numbers!
{
if let integer = Int(String(number))
{
array.append(integer)
}
}
print(array)
So, when I am printing the array, I am getting these as [2,2,3,3,4,5,5,6,7] instead of [223,345,567]. Can anyone help?
Here is the extract of new code.
let numbers = readLine()
var array = [Int]()
guard let numberStrings = numbers?.components(separatedBy: " ") else{
fatalError()
}
for number in numberStrings {
if let validNumber = Int(number) {
array.append(validNumber)
}
}
print(array)
You need to split the string and find every number string and then convert them to Int.
let numbers = readLine()
var numberArray: [Int] = []
guard let numberStrings = numbers?.components(separatedBy: " ") else {
fatalError()
}
for number in numberStrings {
if let validNumber = Int(number) {
numberArray.append(validNumber)
}
}
for number in numbers!
This line of code extracts each character in the string (223 345 567) and tries to convert it to int. That's why it converts each valid number in string and left the spaces.
You need to first split the string in to array of string numbers then iterate through the array to convert them.
Split it, further iterate through strNumArray and convert them to integers
var array = [Int]()
if let strNumArray = numbers?.components(separatedBy: " ") {
for number in strNumArray {
if let integer = Int(String(number)) {
array.append(integer)
}
}
}
print(array)
And in more swifty way
var arrayInt = numbers.split(separator: " ").map{Int($0)}
I know how to programmatically do it, but I'm sure there's a built-in way...
Every language I've used has some sort of default textual representation for a collection of objects that it will spit out when you try to concatenate the Array with a string, or pass it to a print() function, etc. Does Apple's Swift language have a built-in way of easily turning an Array into a String, or do we always have to be explicit when stringifying an array?
If the array contains strings, you can use the String's join method:
var array = ["1", "2", "3"]
let stringRepresentation = "-".join(array) // "1-2-3"
In Swift 2:
var array = ["1", "2", "3"]
let stringRepresentation = array.joinWithSeparator("-") // "1-2-3"
This can be useful if you want to use a specific separator (hypen, blank, comma, etc).
Otherwise you can simply use the description property, which returns a string representation of the array:
let stringRepresentation = [1, 2, 3].description // "[1, 2, 3]"
Hint: any object implementing the Printable protocol has a description property. If you adopt that protocol in your own classes/structs, you make them print friendly as well
In Swift 3
join becomes joined, example [nil, "1", "2"].flatMap({$0}).joined()
joinWithSeparator becomes joined(separator:) (only available to Array of Strings)
In Swift 4
var array = ["1", "2", "3"]
array.joined(separator:"-")
With Swift 5, according to your needs, you may choose one of the following Playground sample codes in order to solve your problem.
Turning an array of Characters into a String with no separator:
let characterArray: [Character] = ["J", "o", "h", "n"]
let string = String(characterArray)
print(string)
// prints "John"
Turning an array of Strings into a String with no separator:
let stringArray = ["Bob", "Dan", "Bryan"]
let string = stringArray.joined(separator: "")
print(string) // prints: "BobDanBryan"
Turning an array of Strings into a String with a separator between words:
let stringArray = ["Bob", "Dan", "Bryan"]
let string = stringArray.joined(separator: " ")
print(string) // prints: "Bob Dan Bryan"
Turning an array of Strings into a String with a separator between characters:
let stringArray = ["car", "bike", "boat"]
let characterArray = stringArray.flatMap { $0 }
let stringArray2 = characterArray.map { String($0) }
let string = stringArray2.joined(separator: ", ")
print(string) // prints: "c, a, r, b, i, k, e, b, o, a, t"
Turning an array of Floats into a String with a separator between numbers:
let floatArray = [12, 14.6, 35]
let stringArray = floatArray.map { String($0) }
let string = stringArray.joined(separator: "-")
print(string)
// prints "12.0-14.6-35.0"
Swift 2.0 Xcode 7.0 beta 6 onwards uses joinWithSeparator() instead of join():
var array = ["1", "2", "3"]
let stringRepresentation = array.joinWithSeparator("-") // "1-2-3"
joinWithSeparator is defined as an extension on SequenceType
extension SequenceType where Generator.Element == String {
/// Interpose the `separator` between elements of `self`, then concatenate
/// the result. For example:
///
/// ["foo", "bar", "baz"].joinWithSeparator("-|-") // "foo-|-bar-|-baz"
#warn_unused_result
public func joinWithSeparator(separator: String) -> String
}
Swift 3
["I Love","Swift"].joined(separator:" ") // previously joinWithSeparator(" ")
Since no one has mentioned reduce, here it is:
[0, 1, 1, 0].map {"\($0)"}.reduce("") { $0 + $1 } // "0110"
In the spirit of functional programming 🤖
In Swift 4
let array:[String] = ["Apple", "Pear ","Orange"]
array.joined(separator: " ")
To change an array of Optional/Non-Optional Strings
//Array of optional Strings
let array : [String?] = ["1",nil,"2","3","4"]
//Separator String
let separator = ","
//flatMap skips the nil values and then joined combines the non nil elements with the separator
let joinedString = array.flatMap{ $0 }.joined(separator: separator)
//Use Compact map in case of **Swift 4**
let joinedString = array.compactMap{ $0 }.joined(separator: separator
print(joinedString)
Here flatMap, compactMap skips the nil values in the array and appends the other values to give a joined string.
Nowadays, in iOS 13+ and macOS 10.15+, we might use ListFormatter:
let formatter = ListFormatter()
let names = ["Moe", "Larry", "Curly"]
if let string = formatter.string(from: names) {
print(string)
}
That will produce a nice, natural language string representation of the list. A US user will see:
Moe, Larry, and Curly
It will support any languages for which (a) your app has been localized; and (b) the user’s device is configured. For example, a German user with an app supporting German localization, would see:
Moe, Larry und Curly
Mine works on NSMutableArray with componentsJoinedByString
var array = ["1", "2", "3"]
let stringRepresentation = array.componentsJoinedByString("-") // "1-2-3"
In Swift 2.2 you may have to cast your array to NSArray to use componentsJoinedByString(",")
let stringWithCommas = (yourArray as NSArray).componentsJoinedByString(",")
let arrayTemp :[String] = ["Mani","Singh","iOS Developer"]
let stringAfterCombining = arrayTemp.componentsJoinedByString(" ")
print("Result will be >>> \(stringAfterCombining)")
Result will be >>> Mani Singh iOS Developer
If you want to ditch empty strings in the array.
["Jet", "Fire"].filter { !$0.isEmpty }.joined(separator: "-")
If you want to filter nil values as well:
["Jet", nil, "", "Fire"].flatMap { $0 }.filter { !$0.isEmpty }.joined(separator: "-")
you can use joined() to get a single String when you have array of struct also.
struct Person{
let name:String
let contact:String
}
You can easily produce string using map() & joined()
PersonList.map({"\($0.name) - \($0.contact)"}).joined(separator: " | ")
output:
Jhon - 123 | Mark - 456 | Ben - 789
if you want convert custom object array to string or comma separated string (csv) you can use
var stringIds = (self.mylist.map{$0.id ?? 0}).map{String($0)}.joined(separator: ",")
credit to : urvish modi
post: Convert an array of Ints to a comma separated string
A separator can be a bad idea for some languages like Hebrew or Japanese.
Try this:
// Array of Strings
let array: [String] = ["red", "green", "blue"]
let arrayAsString: String = array.description
let stringAsData = arrayAsString.data(using: String.Encoding.utf16)
let arrayBack: [String] = try! JSONDecoder().decode([String].self, from: stringAsData!)
For other data types respectively:
// Set of Doubles
let set: Set<Double> = [1, 2.0, 3]
let setAsString: String = set.description
let setStringAsData = setAsString.data(using: String.Encoding.utf16)
let setBack: Set<Double> = try! JSONDecoder().decode(Set<Double>.self, from: setStringAsData!)
You can either use loops for getting this done. Or by using map.
By mapping:
let array = ["one" , "two" , "three"]
array.map({$0}).joined(seperator : ",")
so in separator you can modify the string.
Output-> ("one,two,three")
The Swift equivalent to what you're describing is string interpolation. If you're thinking about things like JavaScript doing "x" + array, the equivalent in Swift is "x\(array)".
As a general note, there is an important difference between string interpolation vs the Printable protocol. Only certain classes conform to Printable. Every class can be string interpolated somehow. That's helpful when writing generic functions. You don't have to limit yourself to Printable classes.
You can print any object using the print function
or use \(name) to convert any object to a string.
Example:
let array = [1,2,3,4]
print(array) // prints "[1,2,3,4]"
let string = "\(array)" // string == "[1,2,3,4]"
print(string) // prints "[1,2,3,4]"
Create extension for an Array:
extension Array {
var string: String? {
do {
let data = try JSONSerialization.data(withJSONObject: self, options: [.prettyPrinted])
return String(data: data, encoding: .utf8)
} catch {
return nil
}
}
}
if you have string array list , then convert to Int
let arrayList = list.map { Int($0)!}
arrayList.description
it will give you string value
for any Element type
extension Array {
func joined(glue:()->Element)->[Element]{
var result:[Element] = [];
result.reserveCapacity(count * 2);
let last = count - 1;
for (ix,item) in enumerated() {
result.append(item);
guard ix < last else{ continue }
result.append(glue());
}
return result;
}
}
Try This:
let categories = dictData?.value(forKeyPath: "listing_subcategories_id") as! NSMutableArray
let tempArray = NSMutableArray()
for dc in categories
{
let dictD = dc as? NSMutableDictionary
tempArray.add(dictD?.object(forKey: "subcategories_name") as! String)
}
let joinedString = tempArray.componentsJoined(by: ",")
use this when you want to convert list of struct type into string
struct MyStruct {
var name : String
var content : String
}
let myStructList = [MyStruct(name: "name1" , content: "content1") , MyStruct(name: "name2" , content: "content2")]
and covert your array like this way
let myString = myStructList.map({$0.name}).joined(separator: ",")
will produce ===> "name1,name2"
FOR SWIFT 3:
func textField(textField: UITextField, shouldChangeCharactersInRange range: NSRange, replacementString string: String) -> Bool {
if textField == phoneField
{
let newString = NSString(string: textField.text!).replacingCharacters(in: range, with: string)
let components = newString.components(separatedBy: NSCharacterSet.decimalDigits.inverted)
let decimalString = NSString(string: components.joined(separator: ""))
let length = decimalString.length
let hasLeadingOne = length > 0 && decimalString.character(at: 0) == (1 as unichar)
if length == 0 || (length > 10 && !hasLeadingOne) || length > 11
{
let newLength = NSString(string: textField.text!).length + (string as NSString).length - range.length as Int
return (newLength > 10) ? false : true
}
var index = 0 as Int
let formattedString = NSMutableString()
if hasLeadingOne
{
formattedString.append("1 ")
index += 1
}
if (length - index) > 3
{
let areaCode = decimalString.substring(with: NSMakeRange(index, 3))
formattedString.appendFormat("(%#)", areaCode)
index += 3
}
if length - index > 3
{
let prefix = decimalString.substring(with: NSMakeRange(index, 3))
formattedString.appendFormat("%#-", prefix)
index += 3
}
let remainder = decimalString.substring(from: index)
formattedString.append(remainder)
textField.text = formattedString as String
return false
}
else
{
return true
}
}
If you question is something like this:
tobeFormattedString = ["a", "b", "c"]
Output = "abc"
String(tobeFormattedString)
here's a part of my code :
let separator = anyWord.indexOf("-")!
if (separator >= 0) {
let res = anyWord[anyWord.characters.index(after: separator)..<anyWord.endIndex] //get an error
//some code
}
I get an error in 'let res' part - Type 'String!' has no subscript members
if I do :
let separator = anyWord.indexOf("-")!
let subStartIndex = anyWord.characters.index(after: separator)
the extension to String :
extension String {
public func indexOf(_ char: Character) -> Int! {
if let idx = self.characters.index(of: char) {
return self.characters.distance(from: self.startIndex, to: idx)
}
return Const.INVALID_VALUE
}
I get error :
String may not be indexed with 'Int", it has variable size elements
You are thinking too complicated, it's pretty easy in Swift 3 using range(of and upperBound
let anyWord = "asdghf-fkgjh"
if let separator = anyWord.range(of:"-") {
let res = anyWord.substring(from: separator.upperBound)
//some code
}
Cast it to string
let test: String
test = "Test-Word"
if let separatorWord = anyWord.range(of: "-")
{
let word = anyWord.substring(from: anyWord.index(after:
separatorWord.lowerBound))
print(word)
}
I like 'pure' swift solution ( no Foundation required )
let anyWord = "asdghf-fkgjh-blablabla--bla"
anyWord.characters.split(separator: "-").forEach { print( String($0)) }
it prints
asdghf
fkgjh
blablabla
bla
How could we convert anyobject to string in swift 3, it's very easy in the older version by using.
var str = toString(AnyObject)
I tried String(AnyObject) but the output is always optional, even when i'm sure that AnyObject is not a optional value.
The compiler suggests that you replace your code with:
let s = String(describing: str)
One other option is available if you have a situation where you want to silently fail with an empty string rather than store something that might not originally be a string as a string.
let s = str as? String ?? ""
else you have the ways of identifying and throwing an error in the answers above/below.
Here's three options for you:
Option 1 - if let
if let b = a as? String {
print(b) // Was a string
} else {
print("Error") // Was not a string
}
Option 2 - guard let
guard let b = a as? String
else {
print("Error") // Was not a string
return // needs a return or break here
}
print(b) // Was a string
Option 3 - let with ?? (null coalescing operator)
let b = a as? String ?? ""
print(b) // Print a blank string if a was not a string
Here's a simple function (repl.it) that will mash any value into a string, with nil becoming an empty string. I found it useful for dealing with JSON that inconsistently uses null, blank, numbers, and numeric strings for IDs.
import Foundation
func toString(_ value: Any?) -> String {
return String(describing: value ?? "")
}
let d: NSDictionary = [
"i" : 42,
"s" : "Hello, World!"
]
dump(toString(d["i"]))
dump(toString(d["s"]))
dump(toString(d["x"]))
Prints:
- "42"
- "Hello, World!"
- ""
Try
let a = "Test" as AnyObject
guard let b = a as? String else { // Something went wrong handle it here }
print(b) // Test
try this -
var str:AnyObject?
str = "Hello, playground" as AnyObject?
if let value = str
{
var a = value as! String
}
OR
var a = str as? String
I know how to programmatically do it, but I'm sure there's a built-in way...
Every language I've used has some sort of default textual representation for a collection of objects that it will spit out when you try to concatenate the Array with a string, or pass it to a print() function, etc. Does Apple's Swift language have a built-in way of easily turning an Array into a String, or do we always have to be explicit when stringifying an array?
If the array contains strings, you can use the String's join method:
var array = ["1", "2", "3"]
let stringRepresentation = "-".join(array) // "1-2-3"
In Swift 2:
var array = ["1", "2", "3"]
let stringRepresentation = array.joinWithSeparator("-") // "1-2-3"
This can be useful if you want to use a specific separator (hypen, blank, comma, etc).
Otherwise you can simply use the description property, which returns a string representation of the array:
let stringRepresentation = [1, 2, 3].description // "[1, 2, 3]"
Hint: any object implementing the Printable protocol has a description property. If you adopt that protocol in your own classes/structs, you make them print friendly as well
In Swift 3
join becomes joined, example [nil, "1", "2"].flatMap({$0}).joined()
joinWithSeparator becomes joined(separator:) (only available to Array of Strings)
In Swift 4
var array = ["1", "2", "3"]
array.joined(separator:"-")
With Swift 5, according to your needs, you may choose one of the following Playground sample codes in order to solve your problem.
Turning an array of Characters into a String with no separator:
let characterArray: [Character] = ["J", "o", "h", "n"]
let string = String(characterArray)
print(string)
// prints "John"
Turning an array of Strings into a String with no separator:
let stringArray = ["Bob", "Dan", "Bryan"]
let string = stringArray.joined(separator: "")
print(string) // prints: "BobDanBryan"
Turning an array of Strings into a String with a separator between words:
let stringArray = ["Bob", "Dan", "Bryan"]
let string = stringArray.joined(separator: " ")
print(string) // prints: "Bob Dan Bryan"
Turning an array of Strings into a String with a separator between characters:
let stringArray = ["car", "bike", "boat"]
let characterArray = stringArray.flatMap { $0 }
let stringArray2 = characterArray.map { String($0) }
let string = stringArray2.joined(separator: ", ")
print(string) // prints: "c, a, r, b, i, k, e, b, o, a, t"
Turning an array of Floats into a String with a separator between numbers:
let floatArray = [12, 14.6, 35]
let stringArray = floatArray.map { String($0) }
let string = stringArray.joined(separator: "-")
print(string)
// prints "12.0-14.6-35.0"
Swift 2.0 Xcode 7.0 beta 6 onwards uses joinWithSeparator() instead of join():
var array = ["1", "2", "3"]
let stringRepresentation = array.joinWithSeparator("-") // "1-2-3"
joinWithSeparator is defined as an extension on SequenceType
extension SequenceType where Generator.Element == String {
/// Interpose the `separator` between elements of `self`, then concatenate
/// the result. For example:
///
/// ["foo", "bar", "baz"].joinWithSeparator("-|-") // "foo-|-bar-|-baz"
#warn_unused_result
public func joinWithSeparator(separator: String) -> String
}
Swift 3
["I Love","Swift"].joined(separator:" ") // previously joinWithSeparator(" ")
Since no one has mentioned reduce, here it is:
[0, 1, 1, 0].map {"\($0)"}.reduce("") { $0 + $1 } // "0110"
In the spirit of functional programming 🤖
In Swift 4
let array:[String] = ["Apple", "Pear ","Orange"]
array.joined(separator: " ")
To change an array of Optional/Non-Optional Strings
//Array of optional Strings
let array : [String?] = ["1",nil,"2","3","4"]
//Separator String
let separator = ","
//flatMap skips the nil values and then joined combines the non nil elements with the separator
let joinedString = array.flatMap{ $0 }.joined(separator: separator)
//Use Compact map in case of **Swift 4**
let joinedString = array.compactMap{ $0 }.joined(separator: separator
print(joinedString)
Here flatMap, compactMap skips the nil values in the array and appends the other values to give a joined string.
Nowadays, in iOS 13+ and macOS 10.15+, we might use ListFormatter:
let formatter = ListFormatter()
let names = ["Moe", "Larry", "Curly"]
if let string = formatter.string(from: names) {
print(string)
}
That will produce a nice, natural language string representation of the list. A US user will see:
Moe, Larry, and Curly
It will support any languages for which (a) your app has been localized; and (b) the user’s device is configured. For example, a German user with an app supporting German localization, would see:
Moe, Larry und Curly
Mine works on NSMutableArray with componentsJoinedByString
var array = ["1", "2", "3"]
let stringRepresentation = array.componentsJoinedByString("-") // "1-2-3"
In Swift 2.2 you may have to cast your array to NSArray to use componentsJoinedByString(",")
let stringWithCommas = (yourArray as NSArray).componentsJoinedByString(",")
let arrayTemp :[String] = ["Mani","Singh","iOS Developer"]
let stringAfterCombining = arrayTemp.componentsJoinedByString(" ")
print("Result will be >>> \(stringAfterCombining)")
Result will be >>> Mani Singh iOS Developer
If you want to ditch empty strings in the array.
["Jet", "Fire"].filter { !$0.isEmpty }.joined(separator: "-")
If you want to filter nil values as well:
["Jet", nil, "", "Fire"].flatMap { $0 }.filter { !$0.isEmpty }.joined(separator: "-")
you can use joined() to get a single String when you have array of struct also.
struct Person{
let name:String
let contact:String
}
You can easily produce string using map() & joined()
PersonList.map({"\($0.name) - \($0.contact)"}).joined(separator: " | ")
output:
Jhon - 123 | Mark - 456 | Ben - 789
if you want convert custom object array to string or comma separated string (csv) you can use
var stringIds = (self.mylist.map{$0.id ?? 0}).map{String($0)}.joined(separator: ",")
credit to : urvish modi
post: Convert an array of Ints to a comma separated string
A separator can be a bad idea for some languages like Hebrew or Japanese.
Try this:
// Array of Strings
let array: [String] = ["red", "green", "blue"]
let arrayAsString: String = array.description
let stringAsData = arrayAsString.data(using: String.Encoding.utf16)
let arrayBack: [String] = try! JSONDecoder().decode([String].self, from: stringAsData!)
For other data types respectively:
// Set of Doubles
let set: Set<Double> = [1, 2.0, 3]
let setAsString: String = set.description
let setStringAsData = setAsString.data(using: String.Encoding.utf16)
let setBack: Set<Double> = try! JSONDecoder().decode(Set<Double>.self, from: setStringAsData!)
You can either use loops for getting this done. Or by using map.
By mapping:
let array = ["one" , "two" , "three"]
array.map({$0}).joined(seperator : ",")
so in separator you can modify the string.
Output-> ("one,two,three")
The Swift equivalent to what you're describing is string interpolation. If you're thinking about things like JavaScript doing "x" + array, the equivalent in Swift is "x\(array)".
As a general note, there is an important difference between string interpolation vs the Printable protocol. Only certain classes conform to Printable. Every class can be string interpolated somehow. That's helpful when writing generic functions. You don't have to limit yourself to Printable classes.
You can print any object using the print function
or use \(name) to convert any object to a string.
Example:
let array = [1,2,3,4]
print(array) // prints "[1,2,3,4]"
let string = "\(array)" // string == "[1,2,3,4]"
print(string) // prints "[1,2,3,4]"
Create extension for an Array:
extension Array {
var string: String? {
do {
let data = try JSONSerialization.data(withJSONObject: self, options: [.prettyPrinted])
return String(data: data, encoding: .utf8)
} catch {
return nil
}
}
}
if you have string array list , then convert to Int
let arrayList = list.map { Int($0)!}
arrayList.description
it will give you string value
for any Element type
extension Array {
func joined(glue:()->Element)->[Element]{
var result:[Element] = [];
result.reserveCapacity(count * 2);
let last = count - 1;
for (ix,item) in enumerated() {
result.append(item);
guard ix < last else{ continue }
result.append(glue());
}
return result;
}
}
Try This:
let categories = dictData?.value(forKeyPath: "listing_subcategories_id") as! NSMutableArray
let tempArray = NSMutableArray()
for dc in categories
{
let dictD = dc as? NSMutableDictionary
tempArray.add(dictD?.object(forKey: "subcategories_name") as! String)
}
let joinedString = tempArray.componentsJoined(by: ",")
use this when you want to convert list of struct type into string
struct MyStruct {
var name : String
var content : String
}
let myStructList = [MyStruct(name: "name1" , content: "content1") , MyStruct(name: "name2" , content: "content2")]
and covert your array like this way
let myString = myStructList.map({$0.name}).joined(separator: ",")
will produce ===> "name1,name2"
FOR SWIFT 3:
func textField(textField: UITextField, shouldChangeCharactersInRange range: NSRange, replacementString string: String) -> Bool {
if textField == phoneField
{
let newString = NSString(string: textField.text!).replacingCharacters(in: range, with: string)
let components = newString.components(separatedBy: NSCharacterSet.decimalDigits.inverted)
let decimalString = NSString(string: components.joined(separator: ""))
let length = decimalString.length
let hasLeadingOne = length > 0 && decimalString.character(at: 0) == (1 as unichar)
if length == 0 || (length > 10 && !hasLeadingOne) || length > 11
{
let newLength = NSString(string: textField.text!).length + (string as NSString).length - range.length as Int
return (newLength > 10) ? false : true
}
var index = 0 as Int
let formattedString = NSMutableString()
if hasLeadingOne
{
formattedString.append("1 ")
index += 1
}
if (length - index) > 3
{
let areaCode = decimalString.substring(with: NSMakeRange(index, 3))
formattedString.appendFormat("(%#)", areaCode)
index += 3
}
if length - index > 3
{
let prefix = decimalString.substring(with: NSMakeRange(index, 3))
formattedString.appendFormat("%#-", prefix)
index += 3
}
let remainder = decimalString.substring(from: index)
formattedString.append(remainder)
textField.text = formattedString as String
return false
}
else
{
return true
}
}
If you question is something like this:
tobeFormattedString = ["a", "b", "c"]
Output = "abc"
String(tobeFormattedString)