对不基于字符串的字符串 nsarray 进行排序

Question

所以我有一个从网络服务中检索的数组，没有特定的顺序

示例：

0 => x large, 
1 => large, 
2 => XX large, 
3 => small,
4 => medium, 
5 => x small

我需要对它们进行排序：首先基于特定的 - 这可能是反向字母顺序：

small
medium
large

其次我需要根据它们的“x”对应部分对它们进行排序：

x small
small
medium
large
x large
xx large

我知道我可以通过强力字符串匹配来做到这一点，但我真的很想知道如何整齐地做到这一点，也许是正则表达式或更优雅的东西？

Answer 1

使用 NSComparator 块语法。 。

NSArray * sizes = [NSArray arrayWithObjects:  @"x small",@"small",@"medium",@"large",@"x large", nil];

NSArray *sortedBySizes =[array sortedArrayUsingComparator:^NSComparisonResult(id obj1, id obj2) {
    if ([sizes indexOfObject:[obj1 size]]> [sizes indexOfObject:[obj2 size]])
        return (NSComparisonResult)NSOrderedAscending;
    if ([sizes indexOfObject:[obj1 size]]< [sizes indexOfObject:[obj2 size]])
        return (NSComparisonResult)NSOrderedDescending;
    return (NSComparisonResult)NSOrderedSame;
}];

---

在第二种方法中，我添加了 Web 服务器发送的数字和 x 大小之间的映射 现在 [obj size]; 假设返回一个 NSNumber 对象。

NSArray * sizesStrings = [NSArray arrayWithObjects:  @"x small",@"small",
                                                     @"medium",@"large",
                                                     @"x large",@"xx large", 
                                                     nil];
NSArray * sizesNumbers = [NSArray arrayWithObjects:[NSNumber numberWithInt:5],
                                                   [NSNumber numberWithInt:3],
                                                   [NSNumber numberWithInt:4],
                                                   [NSNumber numberWithInt:1],
                                                   [NSNumber numberWithInt:0],
                                                   [NSNumber numberWithInt:2], 
                                                   nil];

NSDictionary *sizes = [NSDictionary dictionaryWithObjects:sizesStrings 
                                                   forKeys:sizesNumbers];

NSArray *sortedBySizes = [array sortedArrayUsingComparator:^NSComparisonResult(id obj1, id obj2) {
    NSString *sizeObj1String = [sizes objectForKey:[obj1 size]];
    NSString *sizeObj2String = [sizes objectForKey:[obj1 size]];

    int i1 = [sizesStrings indexOfObject:sizeObj1String];
    int i2 = [sizesStrings indexOfObject:sizeObj2String];

    if (i1 > i2)
        return (NSComparisonResult)NSOrderedAscending;
    if (i2 > i1)
        return (NSComparisonResult)NSOrderedDescending;
    return (NSComparisonResult)NSOrderedSame;
}];

---

该问题的第二个任务——分组为小型、中型、大型——可以这样完成：

NSDictionary *groups = [NSDictionary dictionaryWithObjects:[NSArray arrayWithObjects:[NSMutableArray array],[NSMutableArray array],[NSMutableArray array], nil] 
                                    forKeys:[NSArray arrayWithObjects:@"small",@"medium",@"large",nil]
                        ];
[array enumerateObjectsUsingBlock:^(id obj, NSUInteger idx, BOOL *stop) {
    int i = [[obj size] intValue];
    if (i == 5 || i == 3) 
        [[groups objectForKey:@"small"] addObject:obj];
    else if (i == 2 || i == 0 || i == 1)
        [[groups objectForKey:@"large"] addObject:obj];
    else
        [[groups objectForKey:@"medium"] addObject:obj];

}];

Answer 2

首先，为了理智起见，向 NSString 添加一个类别

@implementation NSString (NSStringContainsCategory)

- (BOOL)contains:(NSString*)string;
{
    if(IsEmpty(string)) {
        return NO;
    }

    NSRange range = [self rangeOfString:string options:NSCaseInsensitiveSearch];

    if (range.location != NSNotFound) {
        return YES;
    }

    return NO;
}

@end

然后，

NSMutableArray* sortMe = [NSMutableArray arrayWithArray:theArrayYouWantToSort];

NSComparator smlComparator = ^NSComparisonResult(id string1, id string2) {
    // 1 = small, 2 = medium, 3 = large        
    NSNumber* string1Number = [NSNumber numberWithInt:0];
    NSNumber* string2Number = [NSNumber numberWithInt:0];

    if([string1 contains:@"large"]) {
        string1Number = [NSNumber numberWithInt:3];
    } else if([string1 contains:@"medium"]) {
        string1Number = [NSNumber numberWithInt:2];
    } else if([string1 contains:@"small"]) {
        string1Number = [NSNumber numberWithInt:1];
    }

    if([string2 contains:@"large"]) {
        string2Number = [NSNumber numberWithInt:3];
    } else if([string2 contains:@"medium"]) {
        string2Number = [NSNumber numberWithInt:2];
    } else if([string2 contains:@"small"]) {
        string2Number = [NSNumber numberWithInt:1];
    }

    NSComparisonResult compareSml = [string1Number compare:string2Number];

    if(compareSml == 0) {
        // if they are the same size, just use normal string comparison to sort x xx xxx etc
        return [string1 compare:string2];
    }
    return compareSml;
};

[sortMe sortUsingComparator:smlComparator];

我不会称其为优雅，但这是我能想到的最简单的答案。您可能想颠倒顺序以满足您的需要 - 只需对比较结果应用负数即可。如果这不起作用请评论，我会再看一下。

Answer 3

也许您可以对字符串进行排序，其中“xx s”被“z”替换，“x s”被“v”替换，“x l”被“b”替换，“xx l”被“a”替换。
q