Objective c 跨n个键使用KVC查找最小值
再次卡在Obj-C中的KVCs上 我想使用KVC来找到多个键的最小值 考虑以下数组:Objective c 跨n个键使用KVC查找最小值,objective-c,ios,nsarray,kvc,Objective C,Ios,Nsarray,Kvc,再次卡在Obj-C中的KVCs上 我想使用KVC来找到多个键的最小值 考虑以下数组: NSArray *data = [[NSArray alloc] initWithObjects: [NSDictionary dictionaryWithObjectsAndKeys:[NSNumber numberWithFloat:2.0], @"a", [NSNumber numberWithFloat:5.0], @"b", [NSNumber numberWithF
NSArray *data = [[NSArray alloc] initWithObjects:
[NSDictionary dictionaryWithObjectsAndKeys:[NSNumber numberWithFloat:2.0], @"a", [NSNumber numberWithFloat:5.0], @"b", [NSNumber numberWithFloat:4.0], @"c", nil],
[NSDictionary dictionaryWithObjectsAndKeys:[NSNumber numberWithFloat:3.0], @"a", [NSNumber numberWithFloat:1.0], @"b", [NSNumber numberWithFloat:1.5], @"c", nil],
[NSDictionary dictionaryWithObjectsAndKeys:[NSNumber numberWithFloat:9.0], @"a", [NSNumber numberWithFloat:7.0], @"b", [NSNumber numberWithFloat:6.0], @"c", nil],
nil];
我可以通过以下方法轻松找到“a”、“b”或“c”的最小值:
float minKeyA = [[data valueForKeyPath:@"@min.a"] floatValue]; // value: 2.0
float minKeyB = [[data valueForKeyPath:@"@min.b"] floatValue]; // value: 1.0
float minKeyC = [[data valueForKeyPath:@"@min.c"] floatValue]; // value: 1.5
我想要实现的是获取一个NSSet或NSArray密钥,并在这些密钥池中找到最小值
NSSet *keySet1 = [NSSet setWithObjects:@"a", @"b", nil];
// use to find absolute minimum value across keys 'a' and 'b' --> desired value 1.0
NSSet *keySet2 = [NSSet setWithObjects:@"a", @"c", nil];
// use to find absolute minimum value across keys 'a' and 'c' --> desired value 1.5
NSSet *keySet3 = [NSSet setWithObjects:@"a", @"b", @"c", nil];
// use to find absolute minimum value across keys 'a', 'b', and 'c' --> desired value 1.0
欣赏任何提示:)一个简单的解决方案是首先找到每个键的最小值,然后在第二步中找到这些最小值中的最小值
NSMutableSet *localMinima = [NSMutableSet setWithCapacity:[keySet1 count]];
for (NSString *key in keySet1) {
NSString *keyPath = [NSString stringWithFormat:@"@min.%@", key];
NSNumber *localMin = [data valueForKeyPath:keyPath];
[localMinima addObject:localMin];
}
NSNumber *globalMin = [localMinima valueForKeyPath:@"@min.self"];
我在想,也许这种方法是最简单的。如果第二天没有其他人以更好的方式做出回应,我会认为这是正确的。谢谢:)