I have a problem with my sorting algorithm.
My NSArray (here vcd.signals.key) contains values of Strings for example:
"x [0]", "x [18]", "x [15]", "x [1]"...
When I try to sort this the result ends up in
"x [0]", "x [15]", "x [18]", "x [1]"
instead of:
"x [0]", "x [1]", "x [15]", "x [18]"
This is my code:
let sortedKeys = sorted(vcd.signals.keys) {
var val1 = $0 as! String
var val2 = $1 as! String
return val1 < val2
}
Any idea how I can fix this issue?
Your problem come associated with your comparison , for example see what happen when you compare the two following strings:
println("x [15]" < "x [1]") // true
This is because the default lexicography comparer goes character for character, position by position comparing ,and of course 5 in position 3 is less than ] in position 3:
println("5" < "]") // true
For the explained above you need to create you own comparer but , only compare for the numbers inside the [$0]. For achieve this I use regular expressions to match any numbers inside the brackets like in the following way:
func matchesForRegexInText(regex: String!, text: String!) -> [String] {
let regex = NSRegularExpression(pattern: regex,
options: nil, error: nil)!
let nsString = text as NSString
let results = regex.matchesInString(text,
options: nil, range: NSMakeRange(0, nsString.length))
as! [NSTextCheckingResult]
return map(results) { nsString.substringWithRange($0.range)}
}
var keysSorted = keys.sorted() {
var key1 = $0
var key2 = $1
var pattern = "([0-9]+)"
var m1 = self.matchesForRegexInText(pattern, text: key1)
var m2 = self.matchesForRegexInText(pattern, text: key2)
return m1[0] < m2[0]
}
In the above regular expression I assume that the numbers only appears inside the brackets and match any number inside the String, but feel free to change the regular expression if you want to achieve anything more. Then you achieve the following:
println(keysSorted) // [x [0], x [1], x [15], x [18]]
I hope this help you.
The issue you are running into is the closing brace character ']' comes after digits. This means that "18" is less than "1]". As long as all your strings share the form of "[digits]" then you can remove the closing brace, sort the strings, add the closing brace back to your final array. The code below works for Swift 2:
let arr = ["x [0]", "x [18]", "x [15]", "x [1]"]
let sorted = arr.map { $0.substringToIndex($0.endIndex.predecessor()) }.sort().map { $0 + "]" }
print(sorted)
x[0], x[18], x[15],...there areStringeach one?