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android - 使用 nextAfter(双开始,双向);在安卓系统中

转载 作者:行者123 更新时间:2023-11-29 21:43:29 26 4
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我需要帮助。我将计算一个测量变量,然后取前 100 个值并取它们的平均值。请记住,过去 6 周我一直在自学,对某些人来说显而易见的东西对我来说不一定显而易见。

本质上,说“double x”是变量,我有很多接近的值。我需要的是一种计算这些值的前 100 个值的总和(然后是平均值)的方法。

在我的研究中,我能看到最适合我需要的东西是'nextAfter(double start, double direction);在此之前,使用 'max' 来确定最大值,这是否是正确的起点:

double xm = max(x);静态 double (xm, x < xm);

我的问题是如何获得前 100 个值(最大值和 99 个 nextAfter 的值)的总和 - 取平均值很容易 - 只需除以 100。

最佳答案

要计算从源中读取的最大 n 值的平均值,您需要至少存储这些值。由于在结束前的任何给定点,您都不知道总体上最大的 n 值中的一些是否会稍后出现,因此您需要跟踪看到的最大 n 值远。

一个简单的方法是将最大值存储在堆或优先级队列中,因为这样可以轻松添加新值并查找(和删除)存储的最小值。默认的 PriorityQueue 非常适合这个任务,因为它使用元素的自然排序,因此 poll ing 删除了最小的存储元素。如果想要计算 n 最小元素的平均值,则需要使用带有自定义 ComparatorPriorityQueue(或者在这个特殊的在这种情况下,简单地否定所有值并使用自然排序也可以。

实现所需的惰性方法(更少的代码)是简单地将每个传入值添加到队列中,如果队列的大小超过 n [那么它必须是 n+1 ] 从队列中移除最小的元素:

// vp is the value provider
while(vp.hasNext()) {
// read the next value and add it to the queue
pq.add(vp.nextValue());
if (pq.size() > topSize) {
pq.poll();
}

稍微复杂一点的方法是先检查是否需要添加新的值,需要时才修改队列,

double newValue = vp.nextValue();
// Check if we have to put the new value in the queue
// that is the case when the queue is not yet full, or the smallest
// stored value is smaller than the new
if (pq.size() < topSize || pq.peek() < newValue) {
// remove the smallest value from the queue only if it is full
if (pq.size() == topSize()) {
pq.poll();
}
pq.add(newValue);
}

这种方式可能更有效,因为向队列添加一个值和删除最小值都是 O(log size) 操作,而与最小存储值比较是 O( 1)。因此,如果有许多值小于之前看到的最大 n,则第二种方法可以节省一些工作。

如果性能很重要,请注意 PriorityQueue 不能存储原始类型,如 double,因此存储(和平均计算的检索)涉及装箱(包装一个double 对象中的 double 值)分别。拆箱(从 Double 对象中提取 double 值),并因此间接从队列的底层数组到实际值。这些成本可以通过自己使用原始 double[] 实现基于堆的优先级队列来避免。 (但这应该很少是必要的,通常,装箱和间接的成本只占整个处理的一小部分。)

一个头脑简单的完整工作示例:

import java.util.PriorityQueue;

/**
* Example class to collect the largest values from a stream and compute their
* average.
*/
public class Average {
// number of values we want to save
private int topSize;
// number of values read so far
private long count = 0;
// priority queue to save the largest topSize values
private PriorityQueue<Double> pq;
// source of read values, could be a file reader, a device reader, or whatever
private ValueProvider vp;

/**
* Construct an <code>Average</code> to sample the largest <code>n</code>
* values from the source.
*
* @param tops Number of values to save for averaging.
* @param v Source of the values to sample.
*
* @throws IllegalArgumentException when the specified number of values is less than one.
*/
public Average(int tops, ValueProvider v) throws IllegalArgumentException {
if (tops < 1) {
throw new IllegalArgumentException("Can't get average of fewer than one values.");
}
topSize = tops;
vp = v;
// Initialise queue to needed capacity; topSize + 1, since we first add
// and then poll. Thus no resizing should ever be necessary.
pq = new PriorityQueue<Double>(topSize+1);
}

/**
* Compute the average of the values stored in the <code>PriorityQueue<Double></code>
*
* @param prio The queue to average.
* @return the average of the values stored in the queue.
*/
public static double average(PriorityQueue<Double> prio) throws IllegalArgumentException {
if (prio == null || prio.size() == 0) {
throw new IllegalArgumentException("Priority queue argument is null or empty.");
}
double sum = 0;
for(Double d : prio) {
sum += d;
}
return sum/prio.size();
}

/**
* Reads values from the provider until exhausted, reporting the average
* of the largest <code>topSize</code> values read so far from time to time
* and when the source is exhausted.
*/
public void collectAverage() {
while(vp.hasNext()) {
// read the next value and add it to the queue
pq.add(vp.nextValue());
++count;
// If the queue was already full, we now have
// topSize + 1 values in it, so we remove the smallest.
// That is, conveniently, what the default PriorityQueue<Double>
// gives us. If we wanted for example the smallest, we'd need
// to use a PriorityQueue with a custom Comparator (or negate
// the values).
if (pq.size() > topSize) {
pq.poll();
}
// Occasionally report the running average of the largest topSize
// values read so far. This may not be desired.
if (count % (topSize*25) == 0 || count < 11) {
System.out.printf("Average of top %d values after collecting %d is %f\n",
pq.size(), count, average(pq));
}
}
// Report final average. Returning the average would be a natural choice too.
System.out.printf("Average of top %d values of %d total is %f\n",
pq.size(), count, average(pq));
}

public static void main(String[] args) {
Average a = new Average(100, new SimpleProvider(123456));
a.collectAverage();
}
}

使用界面

/**
* Interface for a source of <code>double</code>s.
*/
public interface ValueProvider {
/**
* Gets the next value from the source.
*
* @return The next value if there is one.
* @throws RuntimeException if the source is exhausted.
*/
public double nextValue() throws RuntimeException;

/**
* Checks whether the source has more values to deliver.
*
* @return whether there is at least one more value to be obtained from the source.
*/
public boolean hasNext();
}

和实现类

/**
* Simple provider of a stream of <code>double</code>s.
*/
public class SimpleProvider implements ValueProvider {
// State determining which value to return next.
private long state = 0;
// Last allowed state.
private final long end;

/**
* Construct a provider of <code>e</code> values.
*
* @param e the number of values to yield.
*/
public SimpleProvider(long e) {
end = e > 0 ? e : 0;
}

/**
* Default constructor to provide 10000 values.
*/
public SimpleProvider() {
this(10000);
}

public double nextValue() {
++state;
return Math.log(state)*Math.sin(state) + Math.cos(state/2.0);
}

public boolean hasNext() {
return state < end;
}
}

关于android - 使用 nextAfter(双开始,双向);在安卓系统中,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16499343/

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