如何从List< Double>在Java中加倍[]？ [英] How to cast from List<Double> 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 */

有一个新的变量类型 double [] 从Java中的高性能变量？

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 对象包装单个 double 值。如果要将所有这些值存储到 double [] 数组中，则 遍历双重实例。不能进行 O（1）映射，这应该是您可以获得的最快的：

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)
 }

---

$ b b

感谢评论中的其他问题）;这里是 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，我没有使用它的第一个答案 - 即使它看起来...很相似：-D）

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

顺便说一下，Marcelos答案的复杂度是O（2n），因为它迭代两次场景）：首先从列表中创建一个 Double [] ，然后打开 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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