如何从 List<Double> 投射在Java中加倍[]? [英] How to cast from List&lt;Double&gt; to double[] in Java?

问题描述

我有一个这样的变量:

List<Double> frameList =  new ArrayList<Double>();

/* Double elements has added to frameList */

如何在 Java 中从该变量中获得一个具有 double[] 类型的新变量并具有高性能?

How can I have a new variable has a type of double[] from that variable in Java with high performance?

推荐答案

高性能 - 每个 Double 对象包装一个 double 值.如果要将所有这些值存储到 double[] 数组中，则必须遍历 Double 实例的集合.O(1) 映射是不可能的，这应该是你能得到的最快的:

High performance - every Double object wraps a single double value. If you want to store all these values into a double[] array, then you have to iterate over the collection of Double instances. A O(1) mapping is not possible, this should be the fastest you can get:

 double[] target = new double[doubles.size()];
 for (int i = 0; i < target.length; i++) {
    target[i] = doubles.get(i).doubleValue();  // java 1.4 style
    // or:
    target[i] = doubles.get(i);                // java 1.5+ style (outboxing)
 }

<小时>

感谢您在评论中提出的其他问题；) 这是拟合 ArrayUtils#toPrimitive 方法的源代码:

public static double[] toPrimitive(Double[] array) {
  if (array == null) {
    return null;
  } else if (array.length == 0) {
    return EMPTY_DOUBLE_ARRAY;
  }
  final double[] result = new double[array.length];
  for (int i = 0; i < array.length; i++) {
    result[i] = array[i].doubleValue();
  }
  return result;
}

(相信我，我的第一个答案并没有使用它 - 即使它看起来......非常相似:-D)

(And trust me, I didn't use it for my first answer - even though it looks ... pretty similiar :-D )

顺便说一句，Marcelos 答案的复杂度是 O(2n)，因为它迭代了两次(在幕后):首先从列表中生成一个 Double[]，然后解开double 值.

By the way, the complexity of Marcelos answer is O(2n), because it iterates twice (behind the scenes): first to make a Double[] from the list, then to unwrap the double values.

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