如何在NSArray中找到NSNumbers的中值？

Question

我正在尝试计算NSArray中一组（小）NSNumbers的中位数。 NSArray中的每个对象都是NSNumber。

这是我正在尝试的，但它不起作用：

NSNumber *median = [smallNSArray valueForKeyPath:@"@median.floatValue"];

Answer 1

NSArray *sorted = [smallNSArray sortedArrayUsingSelector:@selector(compare:)];    // Sort the array by value
NSUInteger middle = [sorted count] / 2;                                           // Find the index of the middle element
NSNumber *median = [sorted objectAtIndex:middle];                                   // Get the middle element

你可以变得更加漂亮。例如，具有偶数个数的集合的中值在技术上是中间两个数字的平均值。您还可以将其打包成NSArray上的类别中的整齐单行方法：

@interface NSArray (Statistics)
- (id)median;
@end

@implementation NSArray (Statistics)

- (id)median
{
    return [[self sortedArrayUsingSelector:@selector(compare:)] objectAtIndex:[self count] / 2];
}

@end

Answer 2

对于那些对此功能有不寻常需求的人，这里有一个NSArray上的类别方法，它可以同时使用奇数和偶数元素：

NSARRAY CATEGORY METHOD

- (float)median {
    if (self.count == 1) return [self[0] floatValue];

    float result = 0;
    NSUInteger middle;

    NSArray * sorted = [self sortedArrayUsingSelector:@selector(compare:)];
    if (self.count % 2 != 0) {  //odd number of members
        middle = (sorted.count / 2);
        result = [[sorted objectAtIndex:middle] floatValue];
    }
    else {
        middle = (sorted.count / 2) - 1;
        result = [[@[[sorted objectAtIndex:middle], [sorted objectAtIndex:middle + 1]] valueForKeyPath:@"@avg.self"] floatValue];
    }
    return result;
}

TEST

NSArray * singleElement = @[@1];
NSArray * oddNumberOfElements = @[@3, @5, @7, @12, @13, @14, @19, @20, @21, @22, @23, @29, @39, @40, @56];
NSArray * evenNumberOfElements = @[@3, @5, @7, @12, @13, @14, @19, @20, @21, @22, @23, @29, @40, @56];

NSLog(
    @"oddNumberOfElements: %f, evenNumberOfElements: %f singleElement: %f",
    [oddNumberOfElements median], [evenNumberOfElements median], [singleElement median]
);
//oddNumberOfElements: 20.000000, evenNumberOfElements: 19.500000 singleElement: 1.000000

Answer 3

快速扩展

extension Array where Element: Comparable {

  var median: Element {
    return self.sort(<)[self.count / 2]
  }

}