Swift: math operation on regex capture numbers - regex

Question

I like to change "I run 12 miles" to "I run 24 miles" using regex. "12" can be any number. I use the function "replace2()" to do the math operation. The capture parameter "$1" in passed to function replace2("$1") is fine. Please help!
func replace( str:String, pattern : String, repl:String)->String?{
let regex = NSRegularExpression .regularExpressionWithPattern(pattern, options: nil,error: nil)
let replacedString = regex.stringByReplacingMatchesInString( str, options: nil,
range: NSMakeRange(0, countElements( str)), withTemplate: repl )
return replacedString
}
replace( "I run 12 miles and walk 12.45 km","\\d+(.\\d+)?", "-")!
func replace2(x: String)->String {
let xx = x.toInt()! * 2 // error : return nil?
return String(format: "\(xx)")
}
replace( "I run 12 miles ","(\\d+)", replace2("$1") )! // error

Answer 1

Try this in a playground:
let str1: NSString = "I run 12 miles"
let str2 = "I run 12 miles"
let match = str1.rangeOfString("\\d+", options: .RegularExpressionSearch)
let finalStr = str1.substringWithRange(match).toInt()
let n: Double = 2.2*Double(finalStr!)
let newStr = str2.stringByReplacingOccurrencesOfString("\\d+", withString: "\(n)", options: NSStringCompareOptions.RegularExpressionSearch, range: nil)
println(newStr) //I run 26.4 miles
//or more simply
let newStr2 = "I run \(n) kilometers"
//yes, I know my conversion is off

Answer 2

If you want to dynamically catch the values and replace them based upon the captured value, you have to use enumerateMatchesInString. You can then then write a loop that does the replacements:
func stringWithDoubleNumbers(string: String) -> String {
// build array of ranges that need replacing
var ranges = [NSRange]()
let regex = NSRegularExpression(pattern: "[\\d.]+", options: nil, error: nil) as NSRegularExpression
regex.enumerateMatchesInString(string, options: nil, range: NSMakeRange(0, countElements(string))) {
match, flags, stop in
ranges.append(match.range)
}
var doubledString = NSMutableString(string: string)
// iterate backwards so that location is valid despite other replacements
for range in reverse(ranges) {
let foundString = doubledString.substringWithRange(range)
if let value = foundString.toInt() {
let numericString = "\(value * 2)"
doubledString.replaceCharactersInRange(range, withString: numericString)
} else {
let value = NSString(string: foundString).doubleValue
let numericString = "\(value * 2.0)"
doubledString.replaceCharactersInRange(range, withString: numericString)
}
}
return doubledString
}
I do some additional logic to handle integer values differently than floating point types, but you could simplify this if you don't want to worry about that complication.

Answer 3

// I experiment with Steve and Rob codes above, I think I can get the codes to solve the problem for simple string. I still get problem with complex string. I try to parse a data record with various multiple (-20) numbers with either integers/doubles not in any sequence like handling both str2 and str3 in a single routine. Anyway thanks to Steve and Rob.
let str2 = "I run 89 miles and 28.576 km"
let str3 = "I run 89.45 miles and 34 km in 34F degree"
var re = NSRegularExpression(pattern: "(\\d+).* (\\d+\\.\\d+)", options: nil, error: nil)
var match = re.firstMatchInString(str2, options: nil, range: NSMakeRange(0, countElements(str2)))
var startIndex = advance(str2.startIndex, match.rangeAtIndex(1).location)
var endIndex = advance(str2.startIndex, match.rangeAtIndex(1).location + match.rangeAtIndex(1).length)
let x1 = str2.substringWithRange(Range(start: startIndex, end: endIndex)).toInt()!
x1
startIndex = advance(str2.startIndex, match.rangeAtIndex(2).location)
endIndex = advance(str2.startIndex, match.rangeAtIndex(2).location + match.rangeAtIndex(2).length)
let x2 = str2.substringWithRange(Range(start: startIndex, end: endIndex))
let x2x = NSString(string: x2).doubleValue
let n1 = x1 * 2
let n2: Double = 2.0 * x2x
str2.stringByReplacingOccurrencesOfString("(\\d+)(.* )(\\d+\\.\\d+)", withString: "\(n1)$2\(n2)", options: NSStringCompareOptions.RegularExpressionSearch, range: nil)
// I run 178 miles and 57.152 km"