使用纯 Java 步进的 Python 类范围 [英] Python like range with stepping in pure Java

问题描述

In [1]: range(-100, 100, 20)
Out[1]: [-100, -80, -60, -40, -20, 0, 20, 40, 60, 80]

使用 Java 的标准库而不是编写自己的函数来创建上述 Array 的最简单方法是什么?

What's the easiest way to create Array like above using standard libraries of Java instead of writting own function?

有IntStream.range(-100, 100)，但是step被硬编码为1.

There is IntStream.range(-100, 100), but step is hardcoded to 1.

这不是 Java:等效的重复Python 的 range(int, int)?，因为我需要在数字之间有一个 step (偏移量)，并且想要使用 java 内置库而不是 3rd 方库.在添加我自己的问题和答案之前，我已经检查了该问题和答案.差异很微妙，但很重要.

THIS IS NOT A DUPLICATE of Java: Equivalent of Python's range(int, int)?, because I need a step (offset) between numbers and want to use java built-in libraries instead of 3rd-party libraries. I've checked that question and answers before adding my own. The difference is subtle but essential.

推荐答案

使用 IntStream::range 应该可以工作(对于 20 的特殊步骤).

IntStream.range(-100, 100).filter(i -> i % 20 == 0);

允许负面步骤的一般实现可能如下所示:

A general implementation allowing negative steps could look like this:

/**
 * Generate a range of {@code Integer}s as a {@code Stream<Integer>} including
 * the left border and excluding the right border.
 * 
 * @param fromInclusive left border, included
 * @param toExclusive   right border, excluded
 * @param step          the step, can be negative
 * @return the range
 */
public static Stream<Integer> rangeStream(int fromInclusive,
        int toExclusive, int step) {
    // If the step is negative, we generate the stream by reverting all operations.
    // For this we use the sign of the step.
    int sign = step < 0 ? -1 : 1;
    return IntStream.range(sign * fromInclusive, sign * toExclusive)
            .filter(i -> (i - sign * fromInclusive) % (sign * step) == 0)
            .map(i -> sign * i)
            .boxed();
}

参见https://gist.github.com/lutzhorn/9338f3c43b249a6180285cc4b249a6180285版本.

See https://gist.github.com/lutzhorn/9338f3c43b249a618285ccb2028cc4b5 for a detailed version.

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