Reputation: 12575
Say I have a string here:
var fullName: String = "First Last"
I want to split the string based on whitespace and assign the values to their respective variables
var fullNameArr = // something like: fullName.explode(" ")
var firstName: String = fullNameArr[0]
var lastName: String? = fullnameArr[1]
Also, sometimes users might not have a last name.
Upvotes: 841
Views: 821315
Reputation: 2091
Update for Swift 5.2 and the simpliest way
let paragraph = "Bob hit a ball, the hit BALL flew far after it was hit. Hello! Hie, How r u?"
let words = paragraph.components(separatedBy: [",", " ", "!",".","?"])
This prints the following
["Bob", "hit", "a", "ball", "", "the", "hit", "BALL", "flew", "far", "after", "it", "was", "hit", "", "Hello", "", "Hie", "", "How", "r", "u", ""]
However, if you want to filter out empty string,
let words = paragraph.components(separatedBy: [",", " ", "!",".","?"]).filter({!$0.isEmpty})
Output,
["Bob", "hit", "a", "ball", "the", "hit", "BALL", "flew", "far", "after", "it", "was", "hit", "Hello", "Hie", "How", "r", "u"]
But make sure, Foundation is imported.
Upvotes: 91
Reputation: 1337
Do not use whitespace to separate words. It will not work well for English. And it will not work at all for languages like Japanese or Chinese.
Here is a sample for tokenizing names from a given string:
import NaturalLanguage
actor Tokenizer {
private let tagger = NLTagger(tagSchemes: [.nameType])
func tokens(for text: String) async -> [String] {
tagger.string = text
let tags = tagger.tags(
in: text.startIndex ..< text.endIndex,
unit: .word,
scheme: .nameType,
options: [
.omitPunctuation,
.omitWhitespace,
.joinNames,
]
)
return tags.map { String(text[$1]) }
}
}
Upvotes: 2
Reputation: 552
Simple way to split a string into array
var fullName: String = "First Last";
var fullNameArr = fullName.componentsSeparatedByString(" ")
var firstName: String = fullNameArr[0]
var lastName: String = fullNameArr[1]
Upvotes: 1
Reputation: 6037
The simplest solution is
let fullName = "First Last"
let components = fullName.components(separatedBy: .whitespacesAndNewlines).compactMap { $0.isEmpty ? nil : $0 }
This will handled multiple white spaces in a row of different types (white space, tabs, newlines etc) and only returns a two element array, you can change the CharacterSet
to include more character you like, if you want to get cleaver you can use Regular Expression Decoder, this lets you write regular expression that can be used to decoded string directly into your own class/struct that implement the Decoding protocol. For something like this is over kill, but if you are using it as an example for more complicate string it may make more sense.
Upvotes: 5
Reputation: 8407
OFFTOP:
For people searching how to split a string with substring (not a character), then here is working solution:
// TESTING
let str1 = "Hello user! What user's details? Here user rounded with space."
let a = str1.split(withSubstring: "user") // <-------------- HERE IS A SPLIT
print(a) // ["Hello ", "! What ", "\'s details? Here ", " rounded with space."]
// testing the result
var result = ""
for item in a {
if !result.isEmpty {
result += "user"
}
result += item
}
print(str1) // "Hello user! What user's details? Here user rounded with space."
print(result) // "Hello user! What user's details? Here user rounded with space."
print(result == str1) // true
/// Extension providing `split` and `substring` methods.
extension String {
/// Split given string with substring into array
/// - Parameters:
/// - string: the string
/// - substring: the substring to search
/// - Returns: array of components
func split(withSubstring substring: String) -> [String] {
var a = [String]()
var str = self
while let range = str.range(of: substring) {
let i = str.distance(from: str.startIndex, to: range.lowerBound)
let j = str.distance(from: str.startIndex, to: range.upperBound)
let left = str.substring(index: 0, length: i)
let right = str.substring(index: j, length: str.length - j)
a.append(left)
str = right
}
if !str.isEmpty {
a.append(str)
}
return a
}
/// the length of the string
public var length: Int {
return self.count
}
/// Get substring, e.g. "ABCDE".substring(index: 2, length: 3) -> "CDE"
///
/// - parameter index: the start index
/// - parameter length: the length of the substring
///
/// - returns: the substring
public func substring(index: Int, length: Int) -> String {
if self.length <= index {
return ""
}
let leftIndex = self.index(self.startIndex, offsetBy: index)
if self.length <= index + length {
return String(self[leftIndex..<self.endIndex])
}
let rightIndex = self.index(self.endIndex, offsetBy: -(self.length - index - length))
return String(self[leftIndex..<rightIndex])
}
}
Upvotes: 2
Reputation: 773
Expounding off of Don Vaughn's Answer, I liked the use of Regular Expressions. I'm surprised that this is only the 2nd Regex answer. However, if we could solve this in just one split
method, instead of multiple methods, that would be great.
I was also inspired by Mithra Singam's Answer to exclude all punctuation as well as whitespace. However, having to create a list of disallowed characters didn't vibe with me.
\w
- Regular Expression for a Letter or Number symbol. No punctuation.let foo = "(..# Hello,,(---- World ".split {
String($0).range(of: #"\w"#, options: .regularExpression) == nil
}
print(foo) // Prints "Hello World"
Let's say you aren't comfortable will all of Unicode. How about just ASKII Letters and Numbers?
let bar = "(..# Hello,,(---- World ".split {
!($0.isASCII && ($0.isLetter || $0.isNumber))
}
print(bar) // Prints "Hello World"
Upvotes: 0
Reputation: 1893
Only the split
is the correct answer, here are the difference for more than 2 spaces.
Swift 5
var temp = "Hello world ni hao"
let arr = temp.components(separatedBy: .whitespacesAndNewlines)
// ["Hello", "world", "", "", "", "", "ni", "hao"]
let arr2 = temp.components(separatedBy: " ")
// ["Hello", "world", "", "", "", "", "ni", "hao"]
let arr3 = temp.split(whereSeparator: {$0 == " "})
// ["Hello", "world", "ni", "hao"]
Upvotes: 28
Reputation: 6540
This gives an array of split parts directly
var fullNameArr = fullName.components(separatedBy:" ")
then you can use like this,
var firstName: String = fullNameArr[0]
var lastName: String? = fullnameArr[1]
Upvotes: 9
Reputation: 211
Steps to split a string into an array in Swift 4.
Note: variableName.components(separatedBy: "split keyword")
let fullName: String = "First Last @ triggerd event of the session by session storage @ it can be divided by the event of the trigger."
let fullNameArr = fullName.components(separatedBy: "@")
print("split", fullNameArr)
Upvotes: 9
Reputation: 236488
Swift 4 or later
If you just need to properly format a person name, you can use PersonNameComponentsFormatter.
The PersonNameComponentsFormatter class provides localized representations of the components of a person’s name, as represented by a PersonNameComponents object. Use this class to create localized names when displaying person name information to the user.
// iOS (9.0 and later), macOS (10.11 and later), tvOS (9.0 and later), watchOS (2.0 and later)
let nameFormatter = PersonNameComponentsFormatter()
let name = "Mr. Steven Paul Jobs Jr."
// personNameComponents requires iOS (10.0 and later)
if let nameComps = nameFormatter.personNameComponents(from: name) {
nameComps.namePrefix // Mr.
nameComps.givenName // Steven
nameComps.middleName // Paul
nameComps.familyName // Jobs
nameComps.nameSuffix // Jr.
// It can also be configured to format your names
// Default (same as medium), short, long or abbreviated
nameFormatter.style = .default
nameFormatter.string(from: nameComps) // "Steven Jobs"
nameFormatter.style = .short
nameFormatter.string(from: nameComps) // "Steven"
nameFormatter.style = .long
nameFormatter.string(from: nameComps) // "Mr. Steven Paul Jobs jr."
nameFormatter.style = .abbreviated
nameFormatter.string(from: nameComps) // SJ
// It can also be use to return an attributed string using annotatedString method
nameFormatter.style = .long
nameFormatter.annotatedString(from: nameComps) // "Mr. Steven Paul Jobs jr."
}
edit/update:
Swift 5 or later
For just splitting a string by non letter characters we can use the new Character property isLetter
:
let fullName = "First Last"
let components = fullName.split{ !$0.isLetter }
print(components) // "["First", "Last"]\n"
Upvotes: 72
Reputation: 2408
Swift 4, Xcode 10 and iOS 12 Update 100% working
let fullName = "First Last"
let fullNameArr = fullName.components(separatedBy: " ")
let firstName = fullNameArr[0] //First
let lastName = fullNameArr[1] //Last
See the Apple's documentation here for further information.
Upvotes: 18
Reputation: 17932
In Swift 4.2 and Xcode 10
//This is your str
let str = "This is my String" //Here replace with your string
Option 1
let items = str.components(separatedBy: " ")//Here replase space with your value and the result is Array.
//Direct single line of code
//let items = "This is my String".components(separatedBy: " ")
let str1 = items[0]
let str2 = items[1]
let str3 = items[2]
let str4 = items[3]
//OutPut
print(items.count)
print(str1)
print(str2)
print(str3)
print(str4)
print(items.first!)
print(items.last!)
Option 2
let items = str.split(separator: " ")
let str1 = String(items.first!)
let str2 = String(items.last!)
//Output
print(items.count)
print(items)
print(str1)
print(str2)
Option 3
let arr = str.split {$0 == " "}
print(arr)
Option 4
let line = "BLANCHE: I don't want realism. I want magic!"
print(line.split(separator: " "))
// Prints "["BLANCHE:", "I", "don\'t", "want", "realism.", "I", "want", "magic!"]"
let line = "BLANCHE: I don't want realism. I want magic!"
print(line.split(separator: " "))
// Prints "["BLANCHE:", "I", "don\'t", "want", "realism.", "I", "want", "magic!"]"
print(line.split(separator: " ", maxSplits: 1))//This can split your string into 2 parts
// Prints "["BLANCHE:", " I don\'t want realism. I want magic!"]"
print(line.split(separator: " ", maxSplits: 2))//This can split your string into 3 parts
print(line.split(separator: " ", omittingEmptySubsequences: false))//array contains empty strings where spaces were repeated.
// Prints "["BLANCHE:", "", "", "I", "don\'t", "want", "realism.", "I", "want", "magic!"]"
print(line.split(separator: " ", omittingEmptySubsequences: true))//array not contains empty strings where spaces were repeated.
print(line.split(separator: " ", maxSplits: 4, omittingEmptySubsequences: false))
print(line.split(separator: " ", maxSplits: 3, omittingEmptySubsequences: true))
Upvotes: 38
Reputation: 512
This is for String and CSV file for swift 4.2 at 20181206 1610
var dataArray : [[String]] = []
let path = Bundle.main.path(forResource: "csvfilename", ofType: "csv")
let url = URL(fileURLWithPath: path!)
do {
let data = try Data(contentsOf: url)
let content = String(data: data, encoding: .utf8)
let parsedCSV = content?.components(separatedBy: "\r\n").map{ $0.components(separatedBy: ";") }
for line in parsedCSV!
{
dataArray.append(line)
}
}
catch let jsonErr {
print("\n Error read CSV file: \n ", jsonErr)
}
print("\n MohNada 20181206 1610 - The final result is \(dataArray) \n ")
Upvotes: -2
Reputation: 70
You can use this common function and add any string which you want to separate
func separateByString(String wholeString: String, byChar char:String) -> [String] {
let resultArray = wholeString.components(separatedBy: char)
return resultArray
}
var fullName: String = "First Last"
let array = separateByString(String: fullName, byChar: " ")
var firstName: String = array[0]
var lastName: String = array[1]
print(firstName)
print(lastName)
Upvotes: -2
Reputation: 12218
I was looking for loosy split, such as PHP's explode
where empty sequences are included in resulting array, this worked for me:
"First ".split(separator: " ", maxSplits: 1, omittingEmptySubsequences: false)
Output:
["First", ""]
Upvotes: 3
Reputation: 12044
Swift 4
let string = "loremipsum.dolorsant.amet:"
let result = string.components(separatedBy: ".")
print(result[0])
print(result[1])
print(result[2])
print("total: \(result.count)")
Output
loremipsum
dolorsant
amet:
total: 3
Upvotes: 6
Reputation: 241
Swift 4 makes it much easier to split characters, just use the new split function for Strings.
Example:
let s = "hi, hello"
let a = s.split(separator: ",")
print(a)
Now you got an array with 'hi' and ' hello'.
Upvotes: 23
Reputation: 363
Let's say you have a variable named "Hello World" and if you want to split it and store it into two different variables you can use like this:
var fullText = "Hello World"
let firstWord = fullText.text?.components(separatedBy: " ").first
let lastWord = fullText.text?.components(separatedBy: " ").last
Upvotes: 4
Reputation: 36660
I had a scenario where multiple control characters can be present in the string I want to split. Rather than maintain an array of these, I just let Apple handle that part.
The following works with Swift 3.0.1 on iOS 10:
let myArray = myString.components(separatedBy: .controlCharacters)
Upvotes: 10
Reputation: 1200
var fullName = "James Keagan Michael"
let first = fullName.components(separatedBy: " ").first?.isEmpty == false ? fullName.components(separatedBy: " ").first! : "John"
let last = fullName.components(separatedBy: " ").last?.isEmpty == false && fullName.components(separatedBy: " ").last != fullName.components(separatedBy: " ").first ? fullName.components(separatedBy: " ").last! : "Doe"
Upvotes: 0
Reputation: 22245
Generally, people reinvent this problem and bad solutions over and over. Is this a space? " " and what about "\n", "\t" or some unicode whitespace character that you've never seen, in no small part because it is invisible. While you can get away with
import Foundation
let pieces = "Mary had little lamb".componentsSeparatedByString(" ")
If you ever need to shake your grip on reality watch a WWDC video on strings or dates. In short, it is almost always better to allow Apple to solve this kind of mundane task.
The way to do this correctly, IMHO, is to use NSCharacterSet
since as stated earlier your whitespace might not be what you expect and Apple has provided a whitespace character set. To explore the various provided character sets check out Apple's NSCharacterSet developer documentation and then, only then, augment or construct a new character set if it doesn't fit your needs.
Returns a character set containing the characters in Unicode General Category Zs and CHARACTER TABULATION (U+0009).
let longerString: String = "This is a test of the character set splitting system"
let components = longerString.components(separatedBy: .whitespaces)
print(components)
Upvotes: 40
Reputation: 6255
Swift 4
let words = "these words will be elements in an array".components(separatedBy: " ")
Upvotes: 52
Reputation: 421
Swift 3
let line = "AAA BBB\t CCC"
let fields = line.components(separatedBy: .whitespaces).filter {!$0.isEmpty}
AAA
, BBB
and CCC
.whitespaces
with .whitespacesAndNewlines
Upvotes: 22
Reputation: 8588
The easiest method to do this is by using componentsSeparatedBy:
For Swift 2:
import Foundation
let fullName : String = "First Last";
let fullNameArr : [String] = fullName.componentsSeparatedByString(" ")
// And then to access the individual words:
var firstName : String = fullNameArr[0]
var lastName : String = fullNameArr[1]
For Swift 3:
import Foundation
let fullName : String = "First Last"
let fullNameArr : [String] = fullName.components(separatedBy: " ")
// And then to access the individual words:
var firstName : String = fullNameArr[0]
var lastName : String = fullNameArr[1]
Upvotes: 215
Reputation: 1829
let fullName : String = "Steve.Jobs"
let fullNameArr : [String] = fullName.components(separatedBy: ".")
var firstName : String = fullNameArr[0]
var lastName : String = fullNameArr[1]
Upvotes: 0
Reputation: 3877
I haven't found the solution that would handle names with 3 or more components and support older iOS versions.
struct NameComponentsSplitter {
static func split(fullName: String) -> (String?, String?) {
guard !fullName.isEmpty else {
return (nil, nil)
}
let components = fullName.components(separatedBy: .whitespacesAndNewlines)
let lastName = components.last
let firstName = components.dropLast().joined(separator: " ")
return (firstName.isEmpty ? nil : firstName, lastName)
}
}
Passed test cases:
func testThatItHandlesTwoComponents() {
let (firstName, lastName) = NameComponentsSplitter.split(fullName: "John Smith")
XCTAssertEqual(firstName, "John")
XCTAssertEqual(lastName, "Smith")
}
func testThatItHandlesMoreThanTwoComponents() {
var (firstName, lastName) = NameComponentsSplitter.split(fullName: "John Clark Smith")
XCTAssertEqual(firstName, "John Clark")
XCTAssertEqual(lastName, "Smith")
(firstName, lastName) = NameComponentsSplitter.split(fullName: "John Clark Jr. Smith")
XCTAssertEqual(firstName, "John Clark Jr.")
XCTAssertEqual(lastName, "Smith")
}
func testThatItHandlesEmptyInput() {
let (firstName, lastName) = NameComponentsSplitter.split(fullName: "")
XCTAssertEqual(firstName, nil)
XCTAssertEqual(lastName, nil)
}
Upvotes: 0
Reputation: 2812
Swift Dev. 4.0 (May 24, 2017)
A new function split
in Swift 4 (Beta).
import Foundation
let sayHello = "Hello Swift 4 2017";
let result = sayHello.split(separator: " ")
print(result)
Output:
["Hello", "Swift", "4", "2017"]
Accessing values:
print(result[0]) // Hello
print(result[1]) // Swift
print(result[2]) // 4
print(result[3]) // 2017
Xcode 8.1 / Swift 3.0.1
Here is the way multiple delimiters with array.
import Foundation
let mathString: String = "12-37*2/5"
let numbers = mathString.components(separatedBy: ["-", "*", "/"])
print(numbers)
Output:
["12", "37", "2", "5"]
Upvotes: 164
Reputation: 9276
String handling is still a challenge in Swift and it keeps changing significantly, as you can see from other answers. Hopefully things settle down and it gets simpler. This is the way to do it with the current 3.0 version of Swift with multiple separator characters.
Swift 3:
let chars = CharacterSet(charactersIn: ".,; -")
let split = phrase.components(separatedBy: chars)
// Or if the enums do what you want, these are preferred.
let chars2 = CharacterSet.alphaNumerics // .whitespaces, .punctuation, .capitalizedLetters etc
let split2 = phrase.components(separatedBy: chars2)
Upvotes: 3
Reputation: 15153
Most of these answers assume the input contains a space - not whitespace, and a single space at that. If you can safely make that assumption, then the accepted answer (from bennett) is quite elegant and also the method I'll be going with when I can.
When we can't make that assumption, a more robust solution needs to cover the following siutations that most answers here don't consider:
\n
) and Windows (\r\n
) newline charactersTo cover these cases this solution uses regex to convert all whitespace (including recurring and Windows newline characters) to a single space, trims, then splits by a single space:
Swift 3:
let searchInput = " First \r\n \n \t\t\tMiddle Last "
let searchTerms = searchInput
.replacingOccurrences(
of: "\\s+",
with: " ",
options: .regularExpression
)
.trimmingCharacters(in: .whitespaces)
.components(separatedBy: " ")
// searchTerms == ["First", "Middle", "Last"]
Upvotes: 11
Reputation: 624
Xcode 8.0 / Swift 3
let fullName = "First Last"
var fullNameArr = fullName.components(separatedBy: " ")
var firstname = fullNameArr[0] // First
var lastname = fullNameArr[1] // Last
Long Way:
var fullName: String = "First Last"
fullName += " " // this will help to see the last word
var newElement = "" //Empty String
var fullNameArr = [String]() //Empty Array
for Character in fullName.characters {
if Character == " " {
fullNameArr.append(newElement)
newElement = ""
} else {
newElement += "\(Character)"
}
}
var firsName = fullNameArr[0] // First
var lastName = fullNameArr[1] // Last
Upvotes: 14