I tried to remove some characters from string in Swift 4, says: I have
QC00012345, return 12345
QC00009876, return 9876
QC12345678, return 12345678
removing first two characters and those leading zeros.
I looked around in here, people are just using dropFirst then convert
String into Int then convert it back to String.
Is there any better way?
You could use a regular expression and dropFirst:
let str = "QC00009876"
let clean = str.dropFirst(2).replacingOccurrences(of: "^0*", with: "", options: .regularExpression)
The expression ^0* means "zero or more 0 characters at the start of the string".
Or with just a regular expression:
let str = "QC00009876"
let clean = str.replacingOccurrences(of: "^..0*", with: "", options: .regularExpression)
The expression ^..0* means "any two characters followed by zero or more 0 characters at the start of the string".
You can simply use drop(while:):
let test = "QC00012345"
let num = test.dropFirst(2).drop { $0 == "0"}
You can also create a ClosedRange<String> and use it as predicate if you don't want to drop first two characters manually:
let num = test.drop { !("1"..."9" ~= $0) }
print(num) // 12345
or check if a string from 1 to 9 does not contain the character:
let num = test.drop { !"123456789".contains($0) }
edit/update:
Swift 5 or later
Using the Character property isWholeNumber
let num = test.drop { !$0.isWholeNumber || $0 == "0" }
Related
I want to delete only integer value from any string. Is it possible or not?
let string = "kjgd5676idbh123456"
i want to result this string = kjgd5676idbh
I want to delete only the last integer value. How to remove the last integer value?
If you would like to delete the whole value 123456 you can use regex \\d+ to match any number with 1 digit or more and anchor it to the end of the string using the dolar sign $
let str = "kjgd5676idbh123456"
if let index = str.range(of: "\\d+$", options: .regularExpression)?.lowerBound {
print("Last Number:", str[index...]) // 123456
print("Prefix:", str[..<index]) // kjgd5676idbh
}
If you need to remove a single digit suffix you just need to remove the plus sign of your regex. "\\d$"
Search backwards for the first non-digit character and drop the trailing digits
let string = "kjgd5676idbh123456"
let result : String
if let range = string.rangeOfCharacter(from: CharacterSet.decimalDigits.inverted, options: .backwards) {
result = String(string[..<range.upperBound])
} else {
result = Int(string) == nil ? string : ""
}
Code to get the string before a certain character:
let string = "Hello World"
if let range = string.range(of: "World") {
let firstPart = string[string.startIndex..<range.lowerBound]
print(firstPart) // print Hello
}
To begin with, I have a program that converts Hex float to a Binary float and I want to remove all "0" from Binary string answer until first "1". Example:
Any ideas?
You can use Regular Expression:
var str = "001110111001"
str = str.replacingOccurrences(of: "0", with: "", options: [.anchored], range: nil)
The anchored option means search for 0s at the start of the string only.
how can I substring the next 2 characters of a string after a certian character. For example I have a strings str1 = "12:34" and other like str2 = "12:345. I want to get the next 2 characters after : the colons.
I want a same code that will work for str1 and str2.
Swift's substring is complicated:
let str = "12:345"
if let range = str.range(of: ":") {
let startIndex = str.index(range.lowerBound, offsetBy: 1)
let endIndex = str.index(startIndex, offsetBy: 2)
print(str[startIndex..<endIndex])
}
It is very easy to use str.index() method as shown in #MikeHenderson's answer, but an alternative to that, without using that method is iterating through the string's characters and creating a new string for holding the first two characters after the ":", like so:
var string1="12:458676"
var nr=0
var newString=""
for c in string1.characters{
if nr>0{
newString+=String(c)
nr-=1
}
if c==":" {nr=2}
}
print(newString) // prints 45
Hope this helps!
A possible solution is Regular Expression,
The pattern checks for a colon followed by two digits and captures the two digits:
let string = "12:34"
let pattern = ":(\\d{2})"
let regex = try! NSRegularExpression(pattern: pattern, options: [])
if let match = regex.firstMatch(in: string, range: NSRange(location: 0, length: string.characters.count)) {
print((string as NSString).substring(with: match.rangeAt(1)))
}
I'm trying to pull out the parts of a string that are in quotation marks, i.e. in "Rouge One" is an awesome movie I want to extract Rouge One.
This is what I have so far but can't figure out where to go from here: I create a copy of the text so that I can remove the first quotation mark so that I can get the index of the second.
if text.contains("\"") {
guard let firstQuoteMarkIndex = text.range(of: "\"") else {return}
var textCopy = text
let textWithoutFirstQuoteMark = textCopy.replacingCharacters(in: firstQuoteMarkIndex, with: "")
let secondQuoteMarkIndex = textCopy.range(of: "\"")
let stringBetweenQuotes = text.substring(with: Range(start: firstQuoteMarkIndex, end: secondQuoteMarkIndex))
}
There is no need to create copies or to replace substrings for this task.
Here is a possible approach:
Use text.range(of: "\"") to find the first quotation mark.
Use text.range(of: "\"", range:...) to find the second quotation mark, i.e. the first one after the range found in step 1.
Extract the substring between the two ranges.
Example:
let text = " \"Rouge One\" is an awesome movie"
if let r1 = text.range(of: "\""),
let r2 = text.range(of: "\"", range: r1.upperBound..<text.endIndex) {
let stringBetweenQuotes = text.substring(with: r1.upperBound..<r2.lowerBound)
print(stringBetweenQuotes) // "Rouge One"
}
Another option is a regular expression search with "positive lookbehind" and "positive lookahead" patterns:
if let range = text.range(of: "(?<=\\\").*?(?=\\\")", options: .regularExpression) {
let stringBetweenQuotes = text.substring(with: range)
print(stringBetweenQuotes)
}
var rouge = "\"Rouge One\" is an awesome movie"
var separated = rouge.components(separatedBy: "\"") // ["", "Rouge One", " is an awesome movie"]
separated.dropFirst().first
I would use .components(separatedBy:)
let stringArray = text.components(separatedBy: "\"")
Check if stringArray count is > 2 (there is at least 2 quotes).
Check if stringArray count is odd, aka count % 2 == 1.
If it is odd, all the even indices are between 2 quotes and they are what you want.
If it is even, all the even indices - 1 are between 2 quotes (the last one doesn't have an end quote).
This will allow you to also capture multiple sets of quoted strings, like:
"Rogue One" is a "Star Wars" movie.
Another option is to use regular expressions to find pairs of quotes:
let pattern = try! NSRegularExpression(pattern: "\\\"([^\"]+)\\\"")
// Small helper methods making it easier to work with enumerateMatches(in:...)
extension String {
subscript(utf16Range range: Range<Int>) -> String? {
get {
let start = utf16.index(utf16.startIndex, offsetBy: range.lowerBound)
let end = utf16.index(utf16.startIndex, offsetBy: range.upperBound)
return String(utf16[start..<end])
}
}
var fullUTF16Range: NSRange {
return NSRange(location: 0, length: utf16.count)
}
}
// Loop through *all* quoted substrings in the original string.
let str = "\"Rogue One\" is an awesome movie"
pattern.enumerateMatches(in: str, range: str.fullUTF16Range) { (result, flags, stop) in
// rangeAt(1) is the range representing the characters in the 1st
// capture group of the regular expression: ([^"]+)
if let result = result, let range = result.rangeAt(1).toRange() {
print("This was in quotes: \(str[utf16Range: range] ?? "<bad range>")")
}
}
This question already has answers here:
Filter non-digits from string
(12 answers)
Closed 6 years ago.
How to get numbers characters from a string? I don't want to convert in Int.
var string = "string_1"
var string2 = "string_20_certified"
My result have to be formatted like this:
newString = "1"
newString2 = "20"
Pattern matching a String's unicode scalars against Western Arabic Numerals
You could pattern match the unicodeScalars view of a String to a given UnicodeScalar pattern (covering e.g. Western Arabic numerals).
extension String {
var westernArabicNumeralsOnly: String {
let pattern = UnicodeScalar("0")..."9"
return String(unicodeScalars
.flatMap { pattern ~= $0 ? Character($0) : nil })
}
}
Example usage:
let str1 = "string_1"
let str2 = "string_20_certified"
let str3 = "a_1_b_2_3_c34"
let newStr1 = str1.westernArabicNumeralsOnly
let newStr2 = str2.westernArabicNumeralsOnly
let newStr3 = str3.westernArabicNumeralsOnly
print(newStr1) // 1
print(newStr2) // 20
print(newStr3) // 12334
Extending to matching any of several given patterns
The unicode scalar pattern matching approach above is particularly useful extending it to matching any of a several given patterns, e.g. patterns describing different variations of Eastern Arabic numerals:
extension String {
var easternArabicNumeralsOnly: String {
let patterns = [UnicodeScalar("\u{0660}")..."\u{0669}", // Eastern Arabic
"\u{06F0}"..."\u{06F9}"] // Perso-Arabic variant
return String(unicodeScalars
.flatMap { uc in patterns.contains{ $0 ~= uc } ? Character(uc) : nil })
}
}
This could be used in practice e.g. if writing an Emoji filter, as ranges of unicode scalars that cover emojis can readily be added to the patterns array in the Eastern Arabic example above.
Why use the UnicodeScalar patterns approach over Character ones?
A Character in Swift contains of an extended grapheme cluster, which is made up of one or more Unicode scalar values. This means that Character instances in Swift does not have a fixed size in the memory, which means random access to a character within a collection of sequentially (/contiguously) stored character will not be available at O(1), but rather, O(n).
Unicode scalars in Swift, on the other hand, are stored in fixed sized UTF-32 code units, which should allow O(1) random access. Now, I'm not entirely sure if this is a fact, or a reason for what follows: but a fact is that if benchmarking the methods above vs equivalent method using the CharacterView (.characters property) for some test String instances, its very apparent that the UnicodeScalar approach is faster than the Character approach; naive testing showed a factor 10-25 difference in execution times, steadily growing for growing String size.
Knowing the limitations of working with Unicode scalars vs Characters in Swift
Now, there are drawbacks using the UnicodeScalar approach, however; namely when working with characters that cannot represented by a single unicode scalar, but where one of its unicode scalars are contained in the pattern to which we want to match.
E.g., consider a string holding the four characters "Café". The last character, "é", is represented by two unicode scalars, "e" and "\u{301}". If we were to implement pattern matching against, say, UnicodeScalar("a")...e, the filtering method as applied above would allow one of the two unicode scalars to pass.
extension String {
var onlyLowercaseLettersAthroughE: String {
let patterns = [UnicodeScalar("1")..."e"]
return String(unicodeScalars
.flatMap { uc in patterns.contains{ $0 ~= uc } ? Character(uc) : nil })
}
}
let str = "Cafe\u{301}"
print(str) // Café
print(str.onlyLowercaseLettersAthroughE) // Cae
/* possibly we'd want "Ca" or "Caé"
as result here */
In the particular use case queried by from the OP in this Q&A, the above is not an issue, but depending on the use case, it will sometimes be more appropriate to work with Character pattern matching over UnicodeScalar.
Edit: Updated for Swift 4 & 5
Here's a straightforward method that doesn't require Foundation:
let newstring = string.filter { "0"..."9" ~= $0 }
or borrowing from #dfri's idea to make it a String extension:
extension String {
var numbers: String {
return filter { "0"..."9" ~= $0 }
}
}
print("3 little pigs".numbers) // "3"
print("1, 2, and 3".numbers) // "123"
import Foundation
let string = "a_1_b_2_3_c34"
let result = string.components(separatedBy: CharacterSet.decimalDigits.inverted).joined(separator: "")
print(result)
Output:
12334
Here is a Swift 2 example:
let str = "Hello 1, World 62"
let intString = str.componentsSeparatedByCharactersInSet(
NSCharacterSet
.decimalDigitCharacterSet()
.invertedSet)
.joinWithSeparator("") // Return a string with all the numbers
This method iterate through the string characters and appends the numbers to a new string:
class func getNumberFrom(string: String) -> String {
var number: String = ""
for var c : Character in string.characters {
if let n: Int = Int(String(c)) {
if n >= Int("0")! && n < Int("9")! {
number.append(c)
}
}
}
return number
}
For example with regular expression
let text = "string_20_certified"
let pattern = "\\d+"
let regex = try! NSRegularExpression(pattern: pattern, options: [])
if let match = regex.firstMatch(in: text, options: [], range: NSRange(location: 0, length: text.characters.count)) {
let newString = (text as NSString).substring(with: match.range)
print(newString)
}
If there are multiple occurrences of the pattern use matches(in..
let matches = regex.matches(in: text, options: [], range: NSRange(location: 0, length: text.characters.count))
for match in matches {
let newString = (text as NSString).substring(with: match.range)
print(newString)
}