so I am filtering data (all strings), and want to show the following:
The search words where the first letters contain your search (example. when searching 'Br' I want to see all the words that begin with 'Br' like Break, Broke,...)
The search words where the string contains the word (example. When searching 'Br' is shows all words containing 'br' in the word like 'groundbreaker').
I've got both of them to work separately (see let found and let foundMore), but now I want to merge both of them (first showing the search words where the first letter contains your search, after the ones where the string contains the word). Tried using the addition sign, but it gives the following error
Binary operator '+' cannot be applied to two 'Range?' operands
extension SearchResultsController : UISearchResultsUpdating {
func updateSearchResultsForSearchController(searchController: UISearchController) {
self.tableView.reloadData()
let searchBar = searchController.searchBar
let target = searchBar.text!
self.filteredData = self.originalData.filter {
s in
let options = NSStringCompareOptions.AnchoredSearch
let found = s.rangeOfString(target, options: options)
let optionsMore = NSStringCompareOptions.CaseInsensitiveSearch
let foundMore = s.rangeOfString(target, options: optionsMore)
let allTogether = found + foundMore
return (allTogether != nil)
}
self.tableView.reloadData()
}
}
Ranges cannot be added together, because the result may not be a proper range when the two ranges do not intersect.
However, you do not need to add this, because in the end you compare the overall range to nil. This means that you can compare the two sides to nil individually:
return s.rangeOfString(target, options: NSStringCompareOptions.AnchoredSearch) != nil
|| s.rangeOfString(target, options: NSStringCompareOptions.CaseInsensitiveSearch) != nil
check 1. and 2., should be viewed like this, not a mix of both
Then you should filter twice, and append the second array to first one:
var filtered1 = self.originalData.filter { s in
return s.rangeOfString(target, options: NSStringCompareOptions.AnchoredSearch) != nil
}
let filtered2 = self.originalData.filter { s in
// Exclude results of the first search
return s.rangeOfString(target, options: NSStringCompareOptions.AnchoredSearch) == nil
&& s.rangeOfString(target, options: NSStringCompareOptions.CaseInsensitiveSearch) != nil
}
filtered1 += filtered2
self.filteredData = filtered1
For your case you can replace + with ?? so that if either range is non-nil it will be set as the value of allTogether. This is minimal code, but isn't necessarily the most obvious or easy to read. An explicit logical OR nil test is more obvious.
If possible, can you can try this:
let indexSet = NSMutableIndexSet()
indexSet.addIndexesInRange(NSMakeRange(0, 5))
indexSet.addIndexesInRange(NSMakeRange(10, 4))
indexSet.addIndex(5)
println(indexSet)
Then you can do your activity here:
indexSet.enumerateIndexesUsingBlock { (index, stop) -> Void in
println(index)
}
Related
Currently, I'm having a problem comparing an array with a string. I have 2 arrays and want to find out if the elements in those 2 arrays are in the string
let resultString = "STEREON10.000 4ailthameGrinreD NOCHIMINNICHNUÖC-LOINHÀ GIAIDACBIET2ty UnOMMOSTCRShitConDONG FlimChineCrJ045 Dòketquásoan: XSHCM goi 7181 8186-8110°593364THUBAY6A7 05-6-2021teIntaiKNInTaiChínhTP.HCM"
let code_province:[String] = ["xsag", "xsbd", "xsbdi", "xsbl","xsbp",
"xsbt", "xsbth", "xscm", "xsct", "xsdl",
"xsdlk", "xsdn", "xsdng", "xsdno", "xsdt",
"xsgl", "xshcm", "xshg", "xskg", "xskh",
"xskt", "xsla", "xsmb", "xsnt", "xspy",
"xsqb", "xsqng", "xsqnm", "xsqt", "xsst",
"xstg", "xstn", "xstth", "xstv", "xsvl",
"xsvt", "xsbri",]
let name_Province:[String] = ["angiang","binhduong","binhdinh","baclieu", "binhphuoc","bentre", "binhthuan", "camau", "cantho", "dalat","daklak", "dongnai", "daNang", "daknong", "dongthap","gialai", "hcm", "haugiang", "kiengiang", "khanhhoa","kontum", "longan", "mienbac", "ninhthuan", "phuyen","quangbinh", "quangNgai", "quangnam", "quangtri", "soctrang","tiengiang", "tayninh", "thuat.hue", "travinh", "vinhlong","vungtau","baria"]
Here is one way:
let f: (String) -> String? = { resultString.localizedStandardContains($0) ? $0 : nil }
let provincesInResult = code_province.compactMap(f)
let namesInResult = name_Province.compactMap(f)
We map the list of things to search for from a list of strings, to nil if not found and the string if found. Then we compact the result to leave us with just a list of the found ones. That may be 0, 1 or more, so consider those possibilities.
I use this method for patterning the phone number in UITextField at the .editingChange event
But the delete key only removes the numbers
extension String{
func applyPatternOnNumbers(pattern: String) -> String {
let replacmentCharacter: Character = "#"
let pureNumber = self.replacingOccurrences( of: "[^۰-۹0-9]", with: "", options: .regularExpression)
var result = ""
var pureNumberIndex = pureNumber.startIndex
for patternCharacter in pattern {
if patternCharacter == replacmentCharacter {
guard pureNumberIndex < pureNumber.endIndex else { return result }
result.append(pureNumber[pureNumberIndex])
pureNumber.formIndex(after: &pureNumberIndex)
} else {
result.append(patternCharacter)
}
}
return result
}
}
use at the editingChange event
let pattern = "+# (###) ###-####"
let mobile = textField.text.substring(to: pattern.count-1)
textfield.text = mobile.applyPatternOnNumbers(pattern: pattern)
// print(textfield.text) +1 (800) 666-8888
the problem is space & - , ( , ) chars can not to be removed
The RegEx you are trying is to not consider digits only:
[^۰-۹0-9]
I'm not sure, but you may change it to:
[^۰-۹0-9\s-\(\)]
and it may work. You might just add a \ before your special chars inside [] and you can any other chars into it that you do not need to be replaced.
Or you may simplify it to
[^\d\s-\(\)]
and it might work.
Method 2
You may use this RegEx which is an exact match to the phone number format you are having:
\+\d+\s\(\d{3}\)\s\d{3}-\d{4}
You may remove the first +, if it is unnecessary
\d+\s\(\d{3}\)\s\d{3}-\d{4}
This question already has answers here:
Refactored Solution In Swift
(2 answers)
Closed 6 years ago.
I'm attempting to solve HackerRank's Hash Table Ransom Note challenge. There are 19 test cases and I'm passing all but two of time due to timeout on larger data sets (10,000-30,000 entries).
I'm given:
1) an array of words contained in a magazine and
2) an array of words for a ransom note. My objective is to determine if the words in the magazine can be used to construct a ransom note.
I need to have enough unique elements in the magazineWords to satisfy the quantity needed by noteWords.
I'm using this code to make that determination...and it takes FOREVER...
for word in noteWordsSet {
// check if there are enough unique words in magazineWords to put in the note
if magazineWords.filter({$0==word}).count < noteWords.filter({$0==word}).count {
return "No"
}
}
What is a faster way to accomplish this task?
Below is my complete code for the challenge:
import Foundation
var magazineWords = // Array of 1 to 30,000 strings
var noteWords = // Array of 1 to 30,000 strings
enum RegexString: String {
// Letters a to z, A to Z, 1 to 5 characters long
case wordCanBeUsed = "([a-zA-Z]{1,5})"
}
func matches(for regexString: String, in text: String) -> [String] {
// Hat tip MartinR for this
do {
let regex = try NSRegularExpression(pattern: regexString)
let nsString = text as NSString
let results = regex.matches(in: text, range: NSRange(location: 0, length: nsString.length))
return results.map { nsString.substring(with: $0.range)}
} catch let error {
print("invalid regex: \(error.localizedDescription)")
return []
}
}
func canCreateRansomNote(from magazineWords: [String], for noteWords: [String]) -> String {
// figure out what's unique
let magazineWordsSet = Set(magazineWords)
let noteWordsSet = Set(noteWords)
let intersectingValuesSet = magazineWordsSet.intersection(noteWordsSet)
// constraints specified in challenge
guard magazineWords.count >= 1, noteWords.count >= 1 else { return "No" }
guard magazineWords.count <= 30000, noteWords.count <= 30000 else { return "No" }
// make sure there are enough individual words to work with
guard magazineWordsSet.count >= noteWordsSet.count else { return "No" }
guard intersectingValuesSet.count == noteWordsSet.count else { return "No" }
// check if all the words can be used. assume the regex method works perfectly
guard noteWords.count == matches(for: RegexString.wordCanBeUsed.rawValue, in: noteWords.joined(separator: " ")).count else { return "No" }
// FIXME: this is a processor hog. I'm timing out when I get to this point
// need to make sure there are enough magazine words to write the note
// compare quantity of word in magazine with quantity of word in note
for word in noteWordsSet {
// check if there are enough unique words in magazineWords to put in the note
if magazineWords.filter({$0==word}).count < noteWords.filter({$0==word}).count {
return "No"
}
}
return "Yes"
}
print(canCreateRansomNote(from: magazineWords, for: noteWords))
I don't know how to read from the test case on the contest website or what frameworks you are allowed. If Foundation is allowed, you can use NSCountedSet
import Foundation
let fileContent = try! String(contentsOf: URL(fileURLWithPath: "/path/to/file.txt"))
let scanner = Scanner(string: fileContent)
var m = 0
var n = 0
scanner.scanInt(&m)
scanner.scanInt(&n)
var magazineWords = NSCountedSet(capacity: m)
var ransomWords = NSCountedSet(capacity: n)
for i in 0..<(m+n) {
var word: NSString? = nil
scanner.scanUpToCharacters(from: .whitespacesAndNewlines, into: &word)
if i < m {
magazineWords.add(word!)
} else {
ransomWords.add(word!)
}
}
var canCreate = true
for w in ransomWords {
if ransomWords.count(for: w) > magazineWords.count(for: w) {
canCreate = false
break
}
}
print(canCreate ? "Yes" : "No")
It works by going through the input file one word at a time, counting how many times that word appears in the magazine and in the ransom note. Then if any word appear more frequently in the ransom note than in the magazine, it fails the test immediately. Run the 30,000 words test case in less than 1 second on my iMac 2012.
I have a Swift based iOS app and one of the features allows you to comment on a post. Anyway, users can add "#mentions" in their posts to tag other people. However I want to stop the user from adding a username with a capital letter.
Is there anyway I can convert a string, so that the #usernames are all in lowercase?
For example:
I really enjoy sightseeing with #uSerABC (not allowed)
I really enjoy sightseeing with #userabc (allowed)
I know there is a property for the string in swift called .lowercaseString - but the problem with that, is that it makes the entire string lowercase and thats not what I want. I only want the #username to be in lower case.
Is there any way around this with having to use the .lowercase property.
Thanks for your time, Dan.
This comes from a code I use to detect hashtags, I've modified to detect mentions:
func detectMentionsInText(text: String) -> [NSRange]? {
let mentionsDetector = try? NSRegularExpression(pattern: "#(\\w+)", options: NSRegularExpressionOptions.CaseInsensitive)
let results = mentionsDetector?.matchesInString(text, options: NSMatchingOptions.WithoutAnchoringBounds, range: NSMakeRange(0, text.utf16.count)).map { $0 }
return results?.map{$0.rangeAtIndex(0)}
}
It detects all the mentions in a string by using a regex and returns an NSRange array, by using a range you have the beginning and the end of the "mention" and you can easily replace them with a lower case version.
Split the string into two using the following command -
let arr = myString.componentsSeparatedByString("#")
//Convert arr[1] to lower case
//Append to arr[0]
//Enjoy
Thanks to everyone for their help. In the end I couldn't get any of the solutions to work and after a lot of testing, I came up with this solution:
func correctStringWithUsernames(inputString: String, completion: (correctString: String) -> Void) {
// Create the final string and get all
// the seperate strings from the data.
var finalString: String!
var commentSegments: NSArray!
commentSegments = inputString.componentsSeparatedByString(" ")
if (commentSegments.count > 0) {
for (var loop = 0; loop < commentSegments.count; loop++) {
// Check the username to ensure that there
// are no capital letters in the string.
let currentString = commentSegments[loop] as! String
let capitalLetterRegEx = ".*[A-Z]+.*"
let textData = NSPredicate(format:"SELF MATCHES %#", capitalLetterRegEx)
let capitalResult = textData.evaluateWithObject(currentString)
// Check if the current loop string
// is a #user mention string or not.
if (currentString.containsString("#")) {
// If we are in the first loop then set the
// string otherwise concatenate the string.
if (loop == 0) {
if (capitalResult == true) {
// The username contains capital letters
// so change it to a lower case version.
finalString = currentString.lowercaseString
}
else {
// The username does not contain capital letters.
finalString = currentString
}
}
else {
if (capitalResult == true) {
// The username contains capital letters
// so change it to a lower case version.
finalString = "\(finalString) \(currentString.lowercaseString)"
}
else {
// The username does not contain capital letters.
finalString = "\(finalString) \(currentString)"
}
}
}
else {
// The current string is NOT a #user mention
// so simply set or concatenate the finalString.
if (loop == 0) {
finalString = currentString
}
else {
finalString = "\(finalString) \(currentString)"
}
}
}
}
else {
// No issues pass back the string.
finalString = inputString
}
// Pass back the correct username string.
completion(correctString: finalString)
}
Its certainly not the most elegant or efficient solution around but it does work. If there are any ways of improving it, please leave a comment.
for example I have a String text like : "I have to go to the kitchen"
and If I searched this text using the 'av' phrase I want a way that return me the whole word 'have'
how I can do this in swift
There is very nice solution with filter in swift.You can use rangeOfString method of String with filter to get only filtered string having "av"
var s = "I have to go to the kitchen"
//will return "have"
let abc:[String] = s.componentsSeparatedByString(" ").filter({ $0.rangeOfString("av", options: NSStringCompareOptions.CaseInsensitiveSearch, range: nil, locale: nil) != nil } )
There is an API for that, enumerateSubstrings(in:options:using:)
The byWords option returns all words in the closure
let string = "I have to go to the kitchen"
var found : String?
string.enumerateSubstrings(in: string.startIndex..., options: .byWords) { substring, _, _, stop in
if let word = substring, word.contains("av") {
found = word
stop = true
}
}
print(found ?? "not found")
Split your string into array by space char (" "), and return component that contains your 'av' string.
let words = stringYouWantToSearchIn.componentsSeparatedByString(" ")
for word in words
{
var range = word.rangeOfString(lastWord)
if (range != nil)
{
//you got what do you want in 'word variable'
break
}
}