排序nsarray的字符串，而不是基于字符串 [英] Sorting an nsarray of strings not string based

问题描述

所以我有一个数组，我从一个Web服务检索没有特定的顺序

So i have an array that i retrieve from a web service in no particular order

例如：

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：首先根据特定的 - 可以颠倒字母： p>

I need to sort them: firstly based on specific - which could be reverse alphabetic:

small
medium
large

其次，我需要根据他们的x计数器部分对它们进行排序：

Secondly i need to sort them based on their 'x' counter parts:

x small
small
medium
large
x large
xx large

$ b b

我知道我可以用暴力字符串匹配，但我真的想建议如何做这个整洁，也许是一个正则表达式或更优雅的东西？

I know i can do this with brute force string matching but i would really like a suggestion on how to do this tidily, perhaps a regex or something more elegant?

推荐答案

使用 NSComparator 块语法。像

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，由web服务器发送的数字和x尺寸。现在 [obj size]; 是假设返回一个NSNumber对象。

---

In the second approach I added a mapping between the numbers send by the web server and the x-sizes. Now [obj size]; is suppose to return a NSNumber object.

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;
}];

---

问题的第二个任务 - 小，中，大 - 可以这样做：

---

The second task of the question — the grouping into small, medium, large — could be done like this:

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];

}];

^{注意：直接键入的未经测试的代码。 sup>}

^{Note: Codes untested, as directly typed.}

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