Question

public static float myAverage(List<Float> elems){
    return /* Answer with no loop */; 
}

如何在一行中获得List的平均值; 没有额外的库像apache，没有循环请帮帮我

我无法制作其他方法和类。

-----原始---

//The answer must have balanced parentesis, and not containing "exerciseAverage"
import java.util.*;
public class Exercise{
public static float myAverage(List<Float> elems){[???]}
//compute the average, 0f on empty lists
public static float exerciseAverage(List<Float> elems){
    /*omitted*/
    /*omitted*/
/*omitted*/
}
public static void main(String [] arg){
List<List<Float>> tests=Arrays.asList(
    Arrays.asList(1f,2f,3f,4f,5f,6f,7f),
    Arrays.asList(-1f,-2f,-3f,-7f,-13f),
    Arrays.asList(-0.5f,0f,0f,0f,30f,-100f,200f,23f,42f),
    Arrays.asList(-12f/3f,32f,12f,45f,-1000f,99f),
    Arrays.asList(-13f/0f,-100f/0f),
    Arrays.asList(7f,6f,4f,3f,2f,1f,0f,-100f),
    Arrays.asList(7f,6f,4f,3f,2f,1f,0f,1f,2f,3f,4f,5f,6f,7f,8f)
    );
    for(List<Float> ls:tests)
assert myAverage(ls)==exerciseAverage(ls): "error on"+ls+" "+myAverage(ls)+" " +exerciseAverage(ls);
    }
}

---我找到了！

return (elems.size() == 0 ) ? 0 : (myAverage(elems.subList(0, elems.size()-1) ) * (elems.size()-1) + elems.get(elems.size()-1))/(float)elems.size();

Answer 1

使用Java 8：

return elems.stream().reduce(0F, Float::sum) / elems.size();

或者

return (float) elems.stream().mapToDouble(Float::doubleValue).average().getAsDouble();

Answer 2

使用递归：

public static float myAverage(List<Float> elems){
    final int N = elems.size();  // For brevity, inline if you wish
    return (N == 0) ? 0 : (myAverage(elems.subList(1,N)) * (N-1) + elems.get(0)) / N;
}

实例：http://ideone.com/V7OVT3。

然而，请注意，这将遇到严重的数字问题;你不应该将它用于真正的数学工作。

如何在没有循环的情况下获得java中列表的平均值

2 个答案: