iOS版 - 最有效的方式找到词汇出现在一个字符串数(iOS - Most efficient wa

给定一个字符串，我需要获得出现在该字符串的每个单词的计数。要做到这一点，我提取的字符串为数组，通过文字和搜索这种方式，但我的感觉是搜索字符串直接是更理想的。下面是我最初写来解决这个问题的代码。我为更好的解决方法的建议虽然。

NSMutableDictionary *sets = [[NSMutableDictionary alloc] init];

NSString *paragraph = [[NSString alloc] initWithContentsOfFile:[[NSBundle mainBundle] pathForResource:@"text" ofType:@"txt"] encoding:NSUTF8StringEncoding error:NULL];

NSMutableArray *words = [[[paragraph lowercaseString] componentsSeparatedByString:@" "] mutableCopy];

while (words.count) {
    NSMutableIndexSet *indexSet = [[NSMutableIndexSet alloc] init];
    NSString *search = [words objectAtIndex:0];
    for (unsigned i = 0; i < words.count; i++) {
        if ([[words objectAtIndex:i] isEqualToString:search]) {
            [indexSet addIndex:i];
        }
    }
    [sets setObject:[NSNumber numberWithInt:indexSet.count] forKey:search];
    [words removeObjectsAtIndexes:indexSet];
}

NSLog(@"%@", sets);

例：

开始字符串：
“这是一个考验。这只是一个测试。”

结果：

“这” - 2
“是” - 2
“a2
“测试” - 2
“只有1个

Answer 1:

这是一个什么NSCountedSet是。

您需要打散串入字（iOS系统是不够好，给我们一个功能，这样我们就不必担心标点符号），只是他们每个人添加到数集，这使一些轨道次每个对象出现在组：

NSString     *string     = @"This is a test. This is only a test.";
NSCountedSet *countedSet = [NSCountedSet new];

[string enumerateSubstringsInRange:NSMakeRange(0, [string length])
                           options:NSStringEnumerationByWords | NSStringEnumerationLocalized
                        usingBlock:^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *stop){

                            // This block is called once for each word in the string.
                            [countedSet addObject:substring];

                            // If you want to ignore case, so that "this" and "This" 
                            // are counted the same, use this line instead to convert
                            // each word to lowercase first:
                            // [countedSet addObject:[substring lowercaseString]];
                        }];

NSLog(@"%@", countedSet);

// Results:  2012-11-13 14:01:10.567 Testing App[35767:fb03] 
// <NSCountedSet: 0x885df70> (a [2], only [1], test [2], This [2], is [2])

Answer 2:

如果我猜的话，我会说NSRegularExpression了点。像这样：

NSUInteger numberOfMatches = [regex numberOfMatchesInString:string
                                                    options:0
                                                      range:NSMakeRange(0, [string length])];

这片段是取自这里。

编辑1.0：

基于什么爵士直到说：

NSString *string = @"This is a test, so it is a test";

NSMutableDictionary *dictionary = [NSMutableDictionary dictionary];
NSArray *arrayOfWords = [string componentsSeparatedByCharactersInSet:[NSCharacterSet whitespaceCharacterSet]];
for (NSString *word in arrayOfWords)
{
    if ([dictionary objectForKey:word])
    {
        NSNumber *numberOfOccurences = [dictionary objectForKey:word];
        NSNumber *increment = [NSNumber numberWithInt:(1 + [numberOfOccurences intValue])];
        [dictionary setValue:increment forKey:word];
    }
    else
    {
        [dictionary setValue:[NSNumber numberWithInt:1] forKey:word];
    }
}

你应该小心：