I have string:
Simple text with spaces
I need regular expression which select:
leading
trailing
more than 1 space
Example:
_ - space
_Simple text __with ___spaces_
My 2ct:
let text = " Simple text with spaces "
let pattern = "^\\s+|\\s+$|\\s+(?=\\s)"
let trimmed = text.stringByReplacingOccurrencesOfString(pattern, withString: "", options: .RegularExpressionSearch)
println(">\(trimmed)<") // >Simple text with spaces<
^\s+ and \s+$ match one or more white space characters at the start/end of the string.
The tricky part is the \s+(?=\s) pattern, which matches one or more
white space characters followed by another white space character which itself is not considered
part of the match (a "look-ahead assertion").
Generally, \s matches all white-space characters such as the space character itself, horizontal tabulation, newline, carriage-return, linefeed or formfeed. If
you want to remove only the (repeated) space characters then replace the pattern by
let pattern = "^ +| +$| +(?= )"
You can keep the regular expression simple by doing the leading/trailing part as a second stage:
let singlySpaced = " Simple text with spaces "
.stringByReplacingOccurrencesOfString("\\s+", withString: " ", options: .RegularExpressionSearch)
.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet())
(assuming you want to strip all kinds of whitespace – you can adjust it down to just do only spaces fairly easily)
There are more complex regexprs that will do it in one shot, but personally I prefer the two-step version over obfuscation (and as #MartinR mentions, performance is very similar between the two, given a trim is a very lightweight operation vs a slower more-complex regex - so it’s really down to which you prefer the look of).
This ought to clean your strings up:
var string : NSString = " hello world. "
while string.rangeOfString(" ").location != NSNotFound { //note the use of two spaces
string = string.stringByReplacingOccurrencesOfString(" ", withString: " ")
}
println(string)
string = string.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceCharacterSet())
println(string)
Some good answers have been provided, but if you want a regex the following should work:
^ |(?<= ) +| $
The | indicates alternatives, the ^ is the beginning of a string, the $ indicates the end of the string. So this matches beginning of the string followed by a space OR one or more spaces preceded by a space OR a space at the end of the string.
Following will remove all the spaces
NSString *spaces =#"hi how are you ";
NSString *withoutSpace= [spaces stringByReplacingOccurrencesOfString:#" "
withString:#" "];
withoutSpace = [withoutSpace stringByTrimmingCharactersInSet:
[NSCharacterSet whitespaceCharacterSet]];
this function will leave one space between words
let accountNameStr = " test test "
let trimmed = accountNameStr.replacingOccurrences(of: "\\s+", with: " ", options: .regularExpression)
let textWithoutSpace = trimmed.trimmingCharacters(in: .whitespaces)
Output : test test
Related
I have a phone number variable that is assigned a value.
let phone: String = "+1 (512) 477-2900"
I want to remove all the spaces and characters including "(", "+", ")" and "-" from the phone number. How can I do that?
// remove the spaces
let formattedPhone = phone.replacingOccurrences(of: " ", with: "")
I don't know how to remove other characters as mentioned earlier.
I assume you want to remove all the characters in the string other than numbers.
let updatedPhone = phone.filter{$0.isNumber}
print(updatedPhone) // 15124772900
I have this code to replace part of a string and remove white spaces:
let str = "باب ".replacingOccurrences(of: "باب", with: "").trimmingCharacters(in: .whitespacesAndNewlines)
print(str.count) /// gives 1 why not 0
But it gives me 1 always while it should be 0. Why?
If it's RTL Mark (and it probably is), that's "\u{200F}" in Swift. If you want to trim it with the whitespace, you'd just add it to your set. That'd be something like:
.trimmingCharacters(in: whitespacesAndNewlines
.union(CharacterSet(charactersIn: "\u{200f}")))
You can also just replace that directly:
.replacingOccurrences(of: "\u{200f}باب", with: "")
Keep in mind the layout rules, since sometimes bidirectional literal strings like this can get confusing in the editor. You may want to separate the Arabic like:
let bab = "باب"
let rtl = "\u{200f}"
string.replacingOccurrences(of: rtl + bab, with: "")
Let's look at the content of your original string:
func hexCharactersArray(_ string: String) -> String {
string.unicodeScalars.map { String(format: "0x%X", $0.value)}.joined(separator: ",")
}
let originalString = "باب "
print(hexCharactersArray(originalString))
The result is [0x628,0x627,0x628,0x20,0x200F]
0x628 - arabic letter beh
0x627 - arabic letter alef
0x628 - arabic letter beh
0x20 - space
0x200F - right-to-left mark
The first three are letters, then some whitespace, but 0x200F is a unicode character in the category of control characters. It's not a letter and it's not whitespace.
When you do:
let replacedString = originalString.replacingOccurrences(of: "باب", with: "").trimmingCharacters(in: .whitespacesAndNewlines)
print(hexCharactersArray(replacedString))
you get [0x200F]
Because you've replaced the letters and trimmed out the whitespace, but you've left behind a control character.
If you want to trim that out too, use:
let replacedString = originalString.replacingOccurrences(of: "باب", with: "").trimmingCharacters(in: .whitespacesAndNewlines.union(.controlCharacters))
I would like to remove an exact combination of words from a string in Swift or Objective-C without removing portions of a word.
You can remove a single word from a string by converting the strings into arrays:
NSString *str = #"Did the favored horse win the race?";
NSString *toRemove = #"horse";
NSMutableArray *mutArray = [str componentsSeparatedByString:#" "];
NSArray *removeArray = [toRemove componentsSeparatedByString:#" "];
[mutarr removeObjectsInArray:removeArr];
You can also remove a two word string from another string if you don't care about whole words using:
str = [str stringByReplacingOccurrencesOfString:#"favored horse " withString:#""];
although you have to work around the spacing issue.
This would fail, however, on a string such as:
str = [str stringByReplacingOccurrencesOfString:#"red horse " withString:#""];
Which would give "Did the favo horse win the race"
How can you remove a multiple word term cleanly without removing partial words leaving fragments?
Thanks for any suggestions.
// Convert string to array of words
let words = string.components(separatedBy: " ")
// Do the same for your search words
let wordsToRemove = "red horse".components(separatedBy: " ")
// remove only the full matching words, and reform the string
let result = words.filter { !wordsToRemove.contains($0) }.joined(separator: " ")
// result = "Did the favored win the race?"
The caveat to this method is that it will remove those exact words anywhere in your original string. If you want the result to only remove the words where they appear in that exact order, then just use a space at the front of the parameter for replacingOccurrencesOf.
If you want to remove some words try to use this extension:
extension String{
func replace(_ dictionary: [String: String]) -> String{
var result = String()
var i = -1
for (of , with): (String, String)in dictionary{
i += 1
if i<1{
result = self.replacingOccurrences(of: of, with: with)
}else{
result = result.replacingOccurrences(of: of, with: with)
}
}
return result
}
}
how to use:
let str = "Did the favored horse win the race?"
let dictionary = ["horse ": "", "the ": ""]
let result = str.replace(dictionary)
print("result: \(result)")
Output:
result: Did favored win race?
For one word:
let str = "Did the favored horse win the race?"
let result = str.replacingOccurrences(of: "horse ", with: "", options: .literal, range:nil)
print("result: \(result)")
Output:
result: Did the favored win the race?
don't forget to include the space in the word that you want to remove... Hope this help
You could also take account of the leading space and replace the whole match with a single space:
str = [str stringByReplacingOccurrencesOfString:#" red horse " withString:#" "];
Or, and you may need to tweak this example, you could just use a Regular Expression - this is the sort of thing they are designed for, and the syntax is nice in Swift
So here is the string s:
"Hi! How are you? I'm fine. It is 6 p.m. Thank you! That's it."
I want them to be separated to a array as:
["Hi", "How are you", "I'm fine", "It is 6 p.m", "Thank you", "That's it"]
Which means the separators should be ". " + "? " + "! "
I've tried:
let charSet = NSCharacterSet(charactersInString: ".?!")
let array = s.componentsSeparatedByCharactersInSet(charSet)
But it will separate p.m. to two elements too. Result:
["Hi", " How are you", " I'm fine", " It is 6 p", "m", " Thank you", " That's it"]
I've also tried
let array = s.componentsSeparatedByString(". ")
It works well for separating ". " but if I also want to separate "? ", "! ", it become messy.
So any way I can do it? Thanks!
There is a method provided that lets you enumerate a string. You can do so by words or sentences or other options. No need for regular expressions.
let s = "Hi! How are you? I'm fine. It is 6 p.m. Thank you! That's it."
var sentences = [String]()
s.enumerateSubstringsInRange(s.startIndex..<s.endIndex, options: .BySentences) {
substring, substringRange, enclosingRange, stop in
sentences.append(substring!)
}
print(sentences)
The result is:
["Hi! ", "How are you? ", "I\'m fine. ", "It is 6 p.m. ", "Thank you! ", "That\'s it."]
rmaddy's answer is correct (+1). A Swift 3 implementation is:
var sentences = [String]()
string.enumerateSubstrings(in: string.startIndex ..< string.endIndex, options: .bySentences) { substring, substringRange, enclosingRange, stop in
sentences.append(substring!)
}
You can also use regular expression, NSRegularExpression, though it's much hairier than rmaddy's .bySentences solution. In Swift 3:
var sentences = [String]()
let regex = try! NSRegularExpression(pattern: "(^|\\s+)(\\w.*?[.!?]+)(?=(\\s+|$))")
regex.enumerateMatches(in: string, range: NSMakeRange(0, string.characters.count)) { match, flags, stop in
sentences.append((string as NSString).substring(with: match!.rangeAt(2)))
}
Or Swift 2:
let regex = try! NSRegularExpression(pattern: "(^|\\s+)(\\w.*?[.!?]+)(?=(\\s+|$))", options: [])
var sentences = [String]()
regex.enumerateMatchesInString(string, options: [], range: NSMakeRange(0, string.characters.count)) { match, flags, stop in
sentences.append((string as NSString).substringWithRange(match!.rangeAtIndex(2)))
}
The [.!?] syntax matches any of those three characters. The | means "or". The ^ matches the start of the string. The $ matches the end of the string. The \\s matches a whitespace character. The \\w matches a "word" character. The * matches zero or more of the preceding character. The + matches one or more of the preceding character. The (?=) is a look-ahead assertion (e.g. see if there's something there, but don't advance through that match).
I've tried to simplify this a bit, and it's still pretty complicated. Regular expressions offer rich text pattern matching, but, admittedly, it is a little dense when you first use it. But this rendition matches (a) repeated punctuation (e.g. "Thank you!!!"), (b) leading spaces, and (c) trailing spaces, too.
If the splitting basis is something a little more esoteric than sentences, this extension could work.
extension String {
public func components(separatedBy separators: [String]) -> [String] {
var output: [String] = [self]
for separator in separators {
output = output.flatMap { $0.components(separatedBy: separator) }
}
return output.map { $0.trimmingCharacters(in: .whitespaces)}
}
}
let artists = "Rihanna, featuring Calvin Harris".components(separated by: [", with", ", featuring"])
I tried to find a regex to solve this too: (([^.!?]+\s)*\S+(\.|!|\?))
Here the explanation from regexper and an example
Well I've found a regex too from here
var pattern = "(?<=[.?!;…])\\s+(?=[\\p{Lu}\\p{N}])"
let s = "Hi! How are you? I'm fine. It is 6 p.m. Thank you! That's it."
let sReplaced = s.stringByReplacingOccurrencesOfString(pattern, withString:"[*-SENTENCE-*]" as String, options:NSStringCompareOptions.RegularExpressionSearch, range:nil)
let array = sReplaced.componentsSeparatedByString("[*-SENTENCE-*]")
Perhaps it's not a good way as it has to first replace and than separate the string. :)
UPDATE:
For regex part, if you also want to match Chinese/Japanese punctuations (which space after each punctuation is not necessary), you can use the following one:
((?<=[.?!;…])\\s+|(?<=[。！？；…])\\s*)(?=[\\p{L}\\p{N}])
I have a string which is "Optional("5")". I need to remove the "" surrounding the 5. I have removed the 'Optional' by doing:
text2 = text2.stringByReplacingOccurrencesOfString("Optional(", withString: "", options: NSStringCompareOptions.LiteralSearch, range: nil)
I am having difficulties removing the " characters as they designate the end of a string in the code.
Swift uses backslash to escape double quotes. Here is the list of escaped special characters in Swift:
\0 (null character)
\\ (backslash)
\t (horizontal tab)
\n (line feed)
\r (carriage return)
\" (double quote)
\' (single quote)
This should work:
text2 = text2.replacingOccurrences(of: "\\", with: "", options: NSString.CompareOptions.literal, range: nil)
Swift 3 and Swift 4:
text2 = text2.textureName.replacingOccurrences(of: "\"", with: "", options: NSString.CompareOptions.literal, range:nil)
Latest documents updated to Swift 3.0.1 have:
Null Character (\0)
Backslash (\\)
Horizontal Tab (\t)
Line Feed (\n)
Carriage Return (\r)
Double Quote (\")
Single Quote (\')
Unicode scalar (\u{n}), where n is between one and eight hexadecimal digits
If you need more details you can take a look to the official docs here
Here is the swift 3 updated answer
var editedText = myLabel.text?.replacingOccurrences(of: "\"", with: "")
Null Character (\0)
Backslash (\\)
Horizontal Tab (\t)
Line Feed (\n)
Carriage Return (\r)
Double Quote (\")
Single Quote (\')
Unicode scalar (\u{n})
To remove the optional you only should do this
println("\(text2!)")
cause if you dont use "!" it takes the optional value of text2
And to remove "" from 5 you have to convert it to NSInteger or NSNumber easy peasy. It has "" cause its an string.
Replacing for Removing is not quite logical.
String.filter allows to iterate a string char by char and keep only true assertion.
Swift 4 & 5
var aString = "Optional(\"5\")"
aString = aString.filter { $0 != "\"" }
> Optional(5)
Or to extend
var aString = "Optional(\"5\")"
let filteredChars = "\"\n\t"
aString = aString.filter { filteredChars.range(of: String($0)) == nil }
> Optional(5)
I've eventually got this to work in the playground, having multiple characters I'm trying to remove from a string:
var otherstring = "lat\" : 40.7127837,\n"
var new = otherstring.stringByTrimmingCharactersInSet(NSCharacterSet.init(charactersInString: "la t, \n \" ':"))
count(new) //result = 10
println(new)
//yielding what I'm after just the numeric portion 40.7127837
If you want to remove more characters for example "a", "A", "b", "B", "c", "C" from string you can do it this way:
someString = someString.replacingOccurrences(of: "[abc]", with: "", options: [.regularExpression, .caseInsensitive])
As Martin R says, your string "Optional("5")" looks like you did something wrong.
dasblinkenlight answers you so it is fine, but for future readers, I will try to add alternative code as:
if let realString = yourOriginalString {
text2 = realString
} else {
text2 = ""
}
text2 in your example looks like String and it is maybe already set to "" but it looks like you have an yourOriginalString of type Optional(String) somewhere that it wasn't cast or use correctly.
I hope this can help some reader.
Swift 5 (working). Only 1 line code.
For removing single / multiple characters.
trimmingCharacters(in: CharacterSet)
In action:
var yourString:String = "(\"This Is: Your String\")"
yourString = yourString.trimmingCharacters(in: ["("," ",":","\"",")"])
print(yourString)
Output:
ThisIsYourString
You are entering a Set that contains characters you're required to trim.
Let's say you have a string:
var string = "potatoes + carrots"
And you want to replace the word "potatoes" in that string with "tomatoes"
string = string.replacingOccurrences(of: "potatoes", with: "tomatoes", options: NSString.CompareOptions.literal, range: nil)
If you print your string, it will now be: "tomatoes + carrots"
If you want to remove the word potatoes from the sting altogether, you can use:
string = string.replacingOccurrences(of: "potatoes", with: "", options: NSString.CompareOptions.literal, range: nil)
If you want to use some other characters in your sting, use:
Null Character (\0)
Backslash (\)
Horizontal Tab (\t)
Line Feed (\n)
Carriage Return (\r)
Double Quote (\")
Single Quote (\')
Example:
string = string.replacingOccurrences(of: "potatoes", with: "dog\'s toys", options: NSString.CompareOptions.literal, range: nil)
Output: "dog's toys + carrots"
If you are getting the output Optional(5) when trying to print the value of 5 in an optional Int or String, you should unwrap the value first:
if value != nil
{ print(value)
}
or you can use this:
if let value = text {
print(value)
}
or in simple just 1 line answer:
print(value ?? "")
The last line will check if variable 'value' has any value assigned to it, if not it will print empty string
You've instantiated text2 as an Optional (e.g. var text2: String?).
This is why you receive Optional("5") in your string.
take away the ? and replace with:
var text2: String = ""
If you are getting the output Optional(5) when trying to print the value of 5 in an optional Int or String, you should unwrap the value first:
if let value = text {
print(value)
}
Now you've got the value without the "Optional" string that Swift adds when the value is not unwrapped before.