i'm using GNUstep shell for programming objective-c. I'm able to convert string to a character set. But unable to print the converted character set in the console. Please tell me a way to print it. Thanks in advance.
This will do the first 65536 characters in unicode, which will do for most situations. I believe unicode can go much higher (2^32?), but this would take much longer to log.
+ (void) logCharacterSet:(NSCharacterSet*)characterSet
{
unichar unicharBuffer[20];
int index = 0;
for (unichar uc = 0; uc < (0xFFFF); uc ++)
{
if ([characterSet characterIsMember:uc])
{
unicharBuffer[index] = uc;
index ++;
if (index == 20)
{
NSString * characters = [NSString stringWithCharacters:unicharBuffer length:index];
NSLog(#"%#", characters);
index = 0;
}
}
}
if (index != 0)
{
NSString * characters = [NSString stringWithCharacters:unicharBuffer length:index];
NSLog(#"%#", characters);
}
}
There are some quite fun looking results, for example here is a sample of 20 characters from punctuationCharacterSet.
༅༆༇༈༉༊་༌།༎༏༐༑༒༔༺༻༼༽྅
Related
please help me with this problem.
I want to check if the targetString match the keyword or not. Consider some character may different, but should still return true.
Example:
targetString = #"#ß<"
keyword = #"abc", #"∂B(", #"#Aß<"
result: all must return true.
(Matched.targetString and all keyword are the same.)
Consider me have an array, contains list of character set that can be the same:
NSArray *variants = [NSArray arrayWithObjects:#"aA#∂", #"bBß", #"c©C<(", nil]
So that when matching, with this rule, it can match as the example above.
Here is what i've done so far (using recursion):
- (BOOL) test:(NSString*)aString include:(NSString*) keyWord doTrim:(BOOL)doTrim {
// break recursion.
if([aString length] < [keyWord length]) return false;
// First, loop through each keyword's character
for (NSUInteger i = 0; i < [keyWord length]; i++) {
// Get #"aA#∂", #"bBß", #"c©C<(" or only the character itself.
// like, if the keyword's character is A, return the string #"aA#∂".
// If the character is not in the variants set, eg. P, return #"P"
char c = [keyWord characterAtIndex:i];
NSString *rs = [self variantsWithChar:c];
// Check if rs (#"aA#∂" or #"P") contains aString[i] character
if([rs rangeOfString:[NSString stringWithCharacters:[aString characterAtIndex:i] length:1]].location == NSNotFound) {
// If not the same char, remove first char in targetString (aString), recursion to match again.
return [self test:[aString substringFromIndex:1] include:keyWord doTrim:NO];
}
}
// If all match with keyword, return true.
return true;
}
- (NSString *) variantsWithChar:(char) c {
for (NSString *s in self.variants) {
if ([s rangeOfString:[NSString stringWithFormat:#"%c",c]].location != NSNotFound) {
return s;
}
}
return [NSString stringWithFormat:#"%c", c];
}
The main problem is, variantsWithChar: doesn't return the correct string. I don't know which datatype and which function should I use here. Please help.
For thou who know ruby, here's the example in ruby. It work super fine!
require 'test/unit/assertions'
include Test::Unit::Assertions
class String
def matching?(keyword)
length >= keyword.length && (keyword.chars.zip(chars).all? { |cs| variants(cs[0]).include?(cs[1]) } || slice(1, length - 1).matching?(keyword))
end
private
VARIANTS = ["aA#∂", "bBß", "c©C<("]
def variants(c)
VARIANTS.find { |cs| cs.include?(c) } || c
end
end
assert "abc".matching?("#ß<")
PS: The fact is, it's containt a japanese character set that sounds the same (like あア, いイ... for thou who know japanese)
PS 2: Please feel free to edit this Question, since my engrish is sooo bad. I may not tell all my thought.
PS 3: And, maybe some may comment about the performance. Like, search about 10,000 target words, with nearly 100 variants, each variant have at most 4 more same characters.
So first off, ignore comments about ASCII and stop using char. NSString and CFString use unichar
If what you really want to do is transpose hiragana and katakana you can do that with CFStringTransform()
It wraps the ICU libraries included in OS X and iOS.
It makes it very simple.
Search for that function and you will find examples of how to use it.
After a while (a day) working on the code above, I finally get it through. But don't know about the performance. Someone comment and help me improve about performance, please. Thanks.
- (BOOL) test:(NSString*)aString include:(NSString*) keyWord doTrim:(BOOL)doTrim {
// break recursion.
if([aString length] < [keyWord length]) return false;
// First, loop through each keyword's character
for (NSUInteger i = 0; i < [keyWord length]; i++) {
// Get #"aA#∂", #"bBß", #"c©C<(" or only the character itself.
// like, if the keyword's character is A, return the string #"aA#∂".
// If the character is not in the variants set, eg. P, return #"P"
NSString* c = [NSString stringWithFormat:#"%C", [keyWord characterAtIndex:i]];
NSString *rs = [self variantsWithChar:c];
NSString *theTargetChar = [NSString stringWithFormat:#"%C", [aString characterAtIndex:i]];
// Check if rs (#"aA#∂" or #"P") contains aString[i] character
if([rs rangeOfString:theTargetChar].location == NSNotFound) {
// If not the same char, remove first char in targetString (aString), recursion to match again.
return [self test:[aString substringFromIndex:1] include:keyWord doTrim:NO];
}
}
// If all match with keyword, return true.
return true;
}
If you remove all comment, it'll be pretty short...
////////////////////////////////////////
- (NSString *) variantsWithChar:(NSString *) c{
for (NSString *s in self.variants) {
if ([s rangeOfString:c].location != NSNotFound) {
return s;
}
}
return c;
}
You could try comparing ascii values of the japanese characters in the variants's each character's ascii value. These japanese characters aren't treated like usual characters or string. Hence, string functions like rangeOfString won't work on them.
to be more precise: have a look at the following code.
it will search for "∂" in the string "aA#∂"
NSString *string = #"aA#∂";
NSMutableSet *listOfAsciiValuesOfString = [self getListOfAsciiValuesForString:string]; //method definition given below
NSString *charToSearch = #"∂";
NSNumber *ascii = [NSNumber numberWithInt:[charToSearch characterAtIndex:0]];
int countBeforeAdding = [listOfAsciiValuesOfString count],countAfterAdding = 0;
[listOfAsciiValuesOfString addObject:ascii];
countAfterAdding = [listOfAsciiValuesOfString count];
if(countAfterAdding == countBeforeAdding){ //element found
NSLog(#"element exists"); //return string
}else{
NSLog(#"Doesnt exists"); //return char
}
===================================
-(NSMutableSet*)getListOfAsciiValuesForString:(NSString*)string{
NSMutableSet *set = [[NSMutableSet alloc] init];
for(int i=0;i<[string length];i++){
NSNumber *ascii = [NSNumber numberWithInt:[string characterAtIndex:i]];
[set addObject:ascii];
}
return set;
}
I'm trying to use an existing piece of code in an iOS project to alphabetize a list of words in an array (for instance, to make tomato into amoott, or stack into ackst). The code seems to work if I run it on its own, but I'm trying to integrate it into my existing app.
Each word I want it to alphabetize is stored as an NSString inside an array. The issue seems to be that the code takes the word as an array of chars, and I can't get my NSStrings into that format.
If I use string = [currentWord UTFString], I get an error of Array type char[128] is not assignable, and if I try to create the char array inside the loop (const char *string = [curentWord UTF8String]) I get warnings relating to Initializing char with type const char discards qualifiers. Not quite sure how I can get around it – any tips? The method is below, I'll take care of storing the alphabetized versions later.
- (void) alphabetizeWord {
char string[128], temp;
int n, i, j;
for (NSString* currentWord in wordsList) {
n = [currentWord length];
for (i = 0; i < n-1; i++) {
for (j = i+1; j < n; j++) {
if (string[i] > string[j]) {
temp = string[i];
string[i] = string[j];
string[j] = temp;
}
}
}
NSLog(#"The word %# in alphabetical order is %s", currentWord, string);
}
}
This should work :
- (void)alphabetizeWord {
char str[128];
for (NSString *currentWord in wordList)
{
int wordLength = [currentWord length];
for (int i = 0; i < wordLength; i++)
{
str[i] = [currentWord characterAtIndex:i];
}
// Adding the termination char
str[wordLength] = 0;
// Add your word
}
}
EDIT : Sorry, didn't fully understand at first. Gonna check this out.
The Hexa-Tri-Decimal number is 0-9 and A-Z. I know I can covert from hex with a NSScanner but not sure how to go about converting Hexa-Tri-Decimal.
For example I have a NSString with "0XPM" the int value should be 43690, "1BLC" would be 61680.
Objective C is built on top of C, and luckily enough you can use the functions there to accomplish the conversion. What you're looking for is strtol or one of it's sibling functions. If I recall correctly strtol handles up to base36 (the hexa-tri-decimal you refer to).
http://www.cplusplus.com/reference/clibrary/cstdlib/strtol/
I can only think to do this using C strings, as they offer easier access to individual characters.
This seemed like an interesting problem to solve, so I had a go at writing it:
int parseBase36Number(NSString *input)
{
const char *inputCString = [[input lowercaseString] UTF8String];
size_t inputLength = [input length];
int orderOfMagnitudeMultiplier = 1;
int result = 0;
// iterate backward through the number
for (int i = inputLength - 1; i >= 0; i--)
{
char inputChar = inputCString[i];
int charNumericValue;
if (isdigit(inputChar))
{
charNumericValue = inputChar - '0';
}
else if (islower(inputChar))
{
charNumericValue = inputChar - 'a' + 10;
}
else
{
// unhanded character, throw error
}
result += charNumericValue * orderOfMagnitudeMultiplier;
orderOfMagnitudeMultiplier *= 36;
}
return result;
}
NOTE: I've not tested this at all, so take care and let me know how it goes!
What would be the most efficient way to convert a string like "ThisStringIsJoined" to "This String Is Joined" in objective-c?
I receive strings like this from a web service thats out of my control and I would like to present the data to the user, so I would just like to tidy it up a bit by adding spaces infront of each uppercase word. The strings are always formatted with each word beginning in an uppercase letter.
I'm quite new to objective-c so cant really figure this one out.
Thanks
One way of achieving this is as follows:
NSString *string = #"ThisStringIsJoined";
NSRegularExpression *regexp = [NSRegularExpression
regularExpressionWithPattern:#"([a-z])([A-Z])"
options:0
error:NULL];
NSString *newString = [regexp
stringByReplacingMatchesInString:string
options:0
range:NSMakeRange(0, string.length)
withTemplate:#"$1 $2"];
NSLog(#"Changed '%#' -> '%#'", string, newString);
The output in this case would be:
'ThisStringIsJoined' -> 'This String Is Joined'
You might want to tweak the regular expression to you own needs. You might want to make this into a category on NSString.
NSRegularExpressions are the way to go, but as trivia, NSCharacterSet can also be useful:
- (NSString *)splitString:(NSString *)inputString {
int index = 1;
NSMutableString* mutableInputString = [NSMutableString stringWithString:inputString];
while (index < mutableInputString.length) {
if ([[NSCharacterSet uppercaseLetterCharacterSet] characterIsMember:[mutableInputString characterAtIndex:index]]) {
[mutableInputString insertString:#" " atIndex:index];
index++;
}
index++;
}
return [NSString stringWithString:mutableInputString];
}
Here's a category on NSString that will do what you want. This will handle non-ASCII letters. It will also split "IDidAGoodThing" properly.
#implementation NSString (SeparateCapitalizedWords)
-(NSString*)stringBySeparatingCapitalizedWords
{
static NSRegularExpression * __regex ;
static dispatch_once_t onceToken;
dispatch_once(&onceToken, ^{
NSError * error = nil ;
__regex = [ NSRegularExpression regularExpressionWithPattern:#"[\\p{Uppercase Letter}]" options:0 error:&error ] ;
if ( error ) { #throw error ; }
});
NSString * result = [ __regex stringByReplacingMatchesInString:self options:0 range:(NSRange){ 1, self.length - 1 } withTemplate:#" $0" ] ;
return result ;
}
#end
Here is Swift Code (objective c code by webstersx), Thanks !
var str: NSMutableString = "iLoveSwiftCode"
var str2: NSMutableString = NSMutableString()
for var i:NSInteger = 0 ; i < str.length ; i++ {
var ch:NSString = str.substringWithRange(NSMakeRange(i, 1))
if(ch .rangeOfCharacterFromSet(NSCharacterSet.uppercaseLetterCharacterSet()).location != NSNotFound) {
str2 .appendString(" ")
}
str2 .appendString(ch)
}
println("\(str2.capitalizedString)")
}
Output : I Love Swift Code
For anyone who came here looking for the similar question answered in Swift:
Perhaps a cleaner (adding to Sankalp's answer), and more 'Swifty' approach:
func addSpaces(to givenString: String) -> String{
var string = givenString
//indexOffset is needed because each time replaceSubrange is called, the resulting count is incremented by one (owing to the fact that a space is added to every capitalised letter)
var indexOffset = 0
for (index, character) in string.characters.enumerated(){
let stringCharacter = String(character)
//Evaluates to true if the character is a capital letter
if stringCharacter.lowercased() != stringCharacter{
guard index != 0 else { continue } //"ILoveSwift" should not turn into " I Love Swift"
let stringIndex = string.index(string.startIndex, offsetBy: index + indexOffset)
let endStringIndex = string.index(string.startIndex, offsetBy: index + 1 + indexOffset)
let range = stringIndex..<endStringIndex
indexOffset += 1
string.replaceSubrange(range, with: " \(stringCharacter)")
}
}
return string
}
You call the function like so:
var string = "iLoveSwiftCode"
addSpaces(to: string)
//Result: string = "i Love Swift Code"
Alternatively, if you prefer extensions:
extension String{
mutating func seperatedWithSpaces(){
//indexOffset is needed because each time replaceSubrange is called, the resulting count is incremented by one (owing to the fact that a space is added to every capitalised letter)
var indexOffset = 0
for (index, character) in characters.enumerated(){
let stringCharacter = String(character)
if stringCharacter.lowercased() != stringCharacter{
guard index != 0 else { continue } //"ILoveSwift" should not turn into " I Love Swift"
let stringIndex = self.index(self.startIndex, offsetBy: index + indexOffset)
let endStringIndex = self.index(self.startIndex, offsetBy: index + 1 + indexOffset)
let range = stringIndex..<endStringIndex
indexOffset += 1
self.replaceSubrange(range, with: " \(stringCharacter)")
}
}
}
}
Call the method from a string:
var string = "iLoveSwiftCode"
string.seperatedWithSpaces()
//Result: string = "i Love Swift Code"
You could try making a new string that is a lowercase copy of the original string. Then compare the two strings and insert spaces wherever the characters are different.
Use the NSString method to turn to lowercase.
- (NSString *)lowercaseString
suppose I have string like "this str1ng for test" now i want to check if character at position [i-1] and [i+1] are both alphabet but character at [i] is number, like this example in word "str1ng" then character at position [i] replaced by appropriate alphabet.
or vice versa.
I need this for post processing for output of OCR. TQ
You might have an easier time using Regular Expressions.
NSString are immutable, so you'll have to create a new NSMutableString from it, and mutate this copy, or to allocate a unichar* buffer, copy data from the NSString, perform the correction, and then recreate a new NSString from the result. Once you're working on a mutable copy of the string, you can use whatever algorithm you want.
So you'll need to have a function like that:
- (NSString*)correctOCRErrors:(NSString*)string
{
BOOL hasError = NO;
for (int i = 0; i < [string length]; ++ i)
{
if (isIncorrect([string characterAtIndex:i]))
{
hasError = YES;
break;
}
}
if (hasError)
{
unichar* buffer = (unichar*)malloc([string length]);
for (int i = 0; i < [string length]; ++ i)
{
unichar chr = [string characterAtIndex:i];
if (isIncorrect(chr))
chr = correctChar(chr);
buffer[i] = chr;
}
string = [[[NSString alloc] initWithCharactersNoCopy:buffer length:[string length] freeWhenDone:YES] autorelease];
}
return string;
}
You can access character in a NSString by passing a message charAtIndex:(NSUInteger)index.
And now you can get the ascii value at the particular index you are interested in and change it according to your requirement.
NSString Ref
Hope this is helpful !