iOS-查找字符串中单词出现次数的最有效方法 [英] iOS - Most efficient way to find word occurrence count in a string

问题描述

给出一个字符串，我需要获取该字符串中出现的每个单词的计数.为此，我按字词将字符串提取到数组中，然后进行搜索，但是我觉得直接搜索字符串更为理想.下面是我最初为解决该问题而编写的代码.不过，我正在寻求有关更好解决方案的建议.

Given a string, I need to obtain a count of each word that appears in that string. To do so, I extracted the string into an array, by word, and searched that way, but I have the feeling that searching the string directly is more optimal. Below is the code that I originally wrote to solve the problem. I'm up for suggestions on better solutions though.

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);

示例:

起始字符串:
这是一个测试.这只是一个测试."

Starting string:
"This is a test. This is only a test."

结果:

• 此"-2
• 是"-2
• "a"-2
• 测试"-2
• 仅"-1

推荐答案

这正是NSCountedSet的用途.

您需要将字符串分解成单词(iOS足够好，可以为我们提供一个功能，这样我们就不必担心标点符号了)，只需将它们中的每一个添加到计数集中即可，每个对象出现在集合中的次数:

You need to break the string apart into words (which iOS is nice enough to give us a function for so that we don't have to worry about punctuation) and just add each of them to the counted set, which keeps track of the number of times each object appears in the set:

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])

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