Number of occurrences of substring in string in Swift - regex

Question

My main string is "hello Swift Swift and Swift" and substring is Swift.
I need to get the number of times the substring "Swift" occurs in the mentioned string.
This code can determine whether the pattern exists.
var string = "hello Swift Swift and Swift"
if string.rangeOfString("Swift") != nil {
println("exists")
}
Now I need to know the number of occurrence.

Answer 1

A simple approach would be to split on "Swift", and subtract 1 from the number of parts:
let s = "hello Swift Swift and Swift"
let tok = s.components(separatedBy:"Swift")
print(tok.count-1)
This code prints 3.
Edit: Before Swift 3 syntax the code looked like this:
let tok = s.componentsSeparatedByString("Swift")

Answer 2

Should you want to count characters rather than substrings:
extension String {
func count(of needle: Character) -> Int {
return reduce(0) {
$1 == needle ? $0 + 1 : $0
}
}
}

Answer 3

Optimising dwsolbergs solution to count faster. Also faster than componentsSeparatedByString.
extension String {
/// stringToFind must be at least 1 character.
func countInstances(of stringToFind: String) -> Int {
assert(!stringToFind.isEmpty)
var count = 0
var searchRange: Range<String.Index>?
while let foundRange = range(of: stringToFind, options: [], range: searchRange) {
count += 1
searchRange = Range(uncheckedBounds: (lower: foundRange.upperBound, upper: endIndex))
}
return count
}
}
Usage:
// return 2
"aaaa".countInstances(of: "aa")
If you want to ignore accents, you may replace options: [] with options: .diacriticInsensitive like dwsolbergs did.
If you want to ignore case, you may replace options: [] with options: .caseInsensitive like ConfusionTowers suggested.
If you want to ignore both accents and case, you may replace options: [] with options: [.caseInsensitive, .diacriticInsensitive] like ConfusionTowers suggested.
If, on the other hand, you want the fastest comparison possible and you can guarantee some canonical form for composed character sequences, then you may consider option .literal and it will only perform exact matchs.

Answer 4

Swift 5 Extension
extension String {
func numberOfOccurrencesOf(string: String) -> Int {
return self.components(separatedBy:string).count - 1
}
}
Example use
let string = "hello Swift Swift and Swift"
let numberOfOccurrences = string.numberOfOccurrencesOf(string: "Swift")
// numberOfOccurrences = 3

Answer 5

I'd recommend an extension to string in Swift 3 such as:
extension String {
func countInstances(of stringToFind: String) -> Int {
var stringToSearch = self
var count = 0
while let foundRange = stringToSearch.range(of: stringToFind, options: .diacriticInsensitive) {
stringToSearch = stringToSearch.replacingCharacters(in: foundRange, with: "")
count += 1
}
return count
}
}
It's a loop that finds and removes each instance of the stringToFind, incrementing the count on each go-round. Once the searchString no longer contains any stringToFind, the loop breaks and the count returns.
Note that I'm using .diacriticInsensitive so it ignore accents (for example résume and resume would both be found). You might want to add or change the options depending on the types of strings you want to find.

Answer 6

I needed a way to count substrings that may contain the start of the next matched substring. Leveraging dwsolbergs extension and Strings range(of:options:range:locale:) method I came up with this String extension
extension String
{
/**
Counts the occurrences of a given substring by calling Strings `range(of:options:range:locale:)` method multiple times.
- Parameter substring : The string to search for, optional for convenience
- Parameter allowOverlap : Bool flag indicating whether the matched substrings may overlap. Count of "🐼🐼" in "🐼🐼🐼🐼" is 2 if allowOverlap is **false**, and 3 if it is **true**
- Parameter options : String compare-options to use while counting
- Parameter range : An optional range to limit the search, default is **nil**, meaning search whole string
- Parameter locale : Locale to use while counting
- Returns : The number of occurrences of the substring in this String
*/
public func count(
occurrencesOf substring: String?,
allowOverlap: Bool = false,
options: String.CompareOptions = [],
range searchRange: Range<String.Index>? = nil,
locale: Locale? = nil) -> Int
{
guard let substring = substring, !substring.isEmpty else { return 0 }
var count = 0
let searchRange = searchRange ?? startIndex..<endIndex
var searchStartIndex = searchRange.lowerBound
let searchEndIndex = searchRange.upperBound
while let rangeFound = range(of: substring, options: options, range: searchStartIndex..<searchEndIndex, locale: locale)
{
count += 1
if allowOverlap
{
searchStartIndex = index(rangeFound.lowerBound, offsetBy: 1)
}
else
{
searchStartIndex = rangeFound.upperBound
}
}
return count
}
}

Answer 7

why not just use some length maths??
extension String {
func occurences(of search:String) -> Int {
guard search.count > 0 else {
preconditionFailure()
}
let shrunk = self.replacingOccurrences(of: search, with: "")
return (self.count - shrunk.count)/search.count
}
}

Answer 8

Try this
var mainString = "hello Swift Swift and Swift"
var count = 0
mainString.enumerateSubstrings(in: mainString.startIndex..<mainString.endIndex, options: .byWords) { (subString, subStringRange, enclosingRange, stop) in
if case let s? = subString{
if s.caseInsensitiveCompare("swift") == .orderedSame{
count += 1
}
}
}
print(count)

Answer 9

For the sake of completeness – and because there is a regex tag – this is a solution with Regular Expression
let string = "hello Swift Swift and Swift"
let regex = try! NSRegularExpression(pattern: "swift", options: .caseInsensitive)
let numberOfOccurrences = regex.numberOfMatches(in: string, range: NSRange(string.startIndex..., in: string))
The option .caseInsensitive is optional.

Answer 10

My solution, maybe it will be better to use String.Index instead of Int range but I think in such way it is a bit easier to read.
extension String {
func count(of char: Character, range: (Int, Int)? = nil) -> Int {
let range = range ?? (0, self.count)
return self.enumerated().reduce(0) {
guard ($1.0 >= range.0) && ($1.0 < range.1) else { return $0 }
return ($1.1 == char) ? $0 + 1 : $0
}
}
}

Answer 11

Solution which uses a higher order functions
func subStringCount(str: String, substr: String) -> Int {
{ $0.isEmpty ? 0 : $0.count - 1 } ( str.components(separatedBy: substr))
}
Unit Tests
import XCTest
class HigherOrderFunctions: XCTestCase {
func testSubstringWhichIsPresentInString() {
XCTAssertEqual(subStringCount(str: "hello Swift Swift and Swift", substr: "Swift"), 3)
}
func testSubstringWhichIsNotPresentInString() {
XCTAssertEqual(subStringCount(str: "hello", substr: "Swift"), 0)
}
}

Answer 12

Another way using RegexBuilder in iOS 16+ & swift 5.7+.
import RegexBuilder
let text = "hello Swift Swift and Swift"
let match = text.matches(of: Regex{"Swift"})
print(match.count) // prints 3
Using this as a function
func countSubstrings(string : String, subString : String)-> Int{
return string.matches(of: Regex{subString}).count
}
print(countSubstrings(string: text, subString: "Swift")) //prints 3
Using this as an Extension
extension String {
func countSubstrings(subString : String)-> Int{
return self.matches(of: Regex{subString}).count
}
}
print(text.countSubstrings(subString: "Swift")) // prints 3