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added 2196 characters in body

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edited Apr 13, 2014 at 23:31

I was thinking about applications with larger amounts of text and I'd rather not have to count my array. For this, I added to the method to get this:

- (NSDictionary *) numberOfLettersNeededFromString:(NSString *)sourceString {

    sourceString = [sourceString stringByReplacingOccurrencesOfString:@"\n" withString:@""];
    sourceString = [sourceString stringByReplacingOccurrencesOfString:@" " withString:@""];
    const char * sourceChars = sourceString.UTF8String;
    NSMutableArray * arr = [NSMutableArray new];
    for (int i = 0; i < sourceString.length; i++) {
        [arr addObject:[NSString stringWithFormat:@"%c", sourceChars[i]]];
    }
 
    static NSString * alphabet = @"abcdefghijklmnopqrstuvwxyz";
    NSMutableDictionary * masterDictionary = [NSMutableDictionary new];
    for (int i = 0; i < alphabet.length; i++) {
        NSString * alphabetLetter = [alphabet substringWithRange:NSMakeRange(i, 1)];
        NSIndexSet * indexes = [arr indexesOfObjectsPassingTest:^BOOL(id obj, NSUInteger idx, BOOL *stop) {
            if ([[(NSString *)obj lowercaseString] isEqualToString:alphabetLetter]) {
                return YES;
            }
            else {
                return NO;
            }
        }];
    
        masterDictionary[alphabetLetter] = @(indexes.count);
    }

    return masterDictionary;
}

Run like:

NSDictionary * lettersNeeded = [self numberOfLettersNeededFromString:@"Hello\nI love cat\nI love dog\nI love mommy\nMommy loves daddy"];
NSLog(@"%@", lettersNeeded);

Will give you:

{ a = 2; b = 0; c = 1; d = 4; e = 5; f = 0; g = 1; h = 1; i = 3; j = 0; k = 0; l = 6; m = 6; n = 0; o = 8; p = 0; q = 0; r = 0; s = 1; t = 1; u = 0; v = 4; w = 0; x = 0; y = 3; z = 0; }

Which I think is better if I had a very large amount of text and I just needed to know how many of each letter I would need.

I was thinking about applications with larger amounts of text and I'd rather not have to count my array. For this, I added to the method to get this:

- (NSDictionary *) numberOfLettersNeededFromString:(NSString *)sourceString {

    sourceString = [sourceString stringByReplacingOccurrencesOfString:@"\n" withString:@""];
    sourceString = [sourceString stringByReplacingOccurrencesOfString:@" " withString:@""];
    const char * sourceChars = sourceString.UTF8String;
    NSMutableArray * arr = [NSMutableArray new];
    for (int i = 0; i < sourceString.length; i++) {
        [arr addObject:[NSString stringWithFormat:@"%c", sourceChars[i]]];
    }
 
    static NSString * alphabet = @"abcdefghijklmnopqrstuvwxyz";
    NSMutableDictionary * masterDictionary = [NSMutableDictionary new];
    for (int i = 0; i < alphabet.length; i++) {
        NSString * alphabetLetter = [alphabet substringWithRange:NSMakeRange(i, 1)];
        NSIndexSet * indexes = [arr indexesOfObjectsPassingTest:^BOOL(id obj, NSUInteger idx, BOOL *stop) {
            if ([[(NSString *)obj lowercaseString] isEqualToString:alphabetLetter]) {
                return YES;
            }
            else {
                return NO;
            }
        }];
    
        masterDictionary[alphabetLetter] = @(indexes.count);
    }

    return masterDictionary;
}

Run like:

NSDictionary * lettersNeeded = [self numberOfLettersNeededFromString:@"Hello\nI love cat\nI love dog\nI love mommy\nMommy loves daddy"];
NSLog(@"%@", lettersNeeded);

Will give you:

{ a = 2; b = 0; c = 1; d = 4; e = 5; f = 0; g = 1; h = 1; i = 3; j = 0; k = 0; l = 6; m = 6; n = 0; o = 8; p = 0; q = 0; r = 0; s = 1; t = 1; u = 0; v = 4; w = 0; x = 0; y = 3; z = 0; }

Which I think is better if I had a very large amount of text and I just needed to know how many of each letter I would need.

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created Apr 13, 2014 at 23:13

Logan

I realize this probably isn't the most efficient answer, but I wanted to try and solve the problem anyways. Here's my ObjC variation:

- (NSArray *) lettersNeededForString:(NSString *)sourceString {
    sourceString = [sourceString stringByReplacingOccurrencesOfString:@"\n" withString:@""];
    sourceString = [sourceString stringByReplacingOccurrencesOfString:@" " withString:@""];
    const char * sourceChars = sourceString.UTF8String;
    NSMutableArray * arr = [NSMutableArray new];
    for (int i = 0; i < sourceString.length; i++) {
        [arr addObject:[NSString stringWithFormat:@"%c", sourceChars[i]]];
    }
    return [arr sortedArrayUsingSelector:@selector(localizedCaseInsensitiveCompare:)];
}

Then you can call it for whatever string:

NSArray * letters = [self lettersNeededForString:@"Hello\nI love cat\nI love dog\nI love mommy\nMommy loves daddy"];
NSLog(@"%@",letters);