python - 识别排序列表中不同的连续倍数

Question

我有一个排序列表，我想在该列表中识别连续的多个数字。该列表可以包含不同顺序的连续倍数，这使得它变得更加困难。

一些测试用例:

[1,3,4,5] -> [[1], [3,4,5]]
[1,3,5,6,7] -> [[1], [3], [5,6,7]]
# consecutive multiples of 1 and 2 (or n)
[1,2,3,7,9,11] -> [[1,2,3], [7,9,11]
[1,2,3,7,10,12,14,25] -> [[1,2,3], [7], [10,12,14], [25]]
# overlapping consecutives !!!
[1,2,3,4,6,8,10] -> [[1,2,3,4], [6,8,10]

现在，我不知道自己在做什么。我所做的是按数字之间的距离成对分组，这是一个好的开始，但后来我遇到了很多问题，无法确定每对中的哪个元素去哪里，即

 # initial list    
 [1,3,4,5]
 # pairs of same distance
 [[1,3], [[3,4], [4,5]]
 # algo to get the final result ?
 [[1], [3,4,5]]

非常感谢任何帮助。

编辑:也许提及我想要它的目的会使它更清楚。

我想转换这样的东西:

[1,5,10,11,12,13,14,15,17,20,22,24,26,28,30]

进入

1, 5, 10 to 15 by 1, 17, 20 to 30 by 2

Answer 1

这是包含@Bakuriu 优化的版本:

MINIMAL_MATCH = 3

def find_some_sort_of_weird_consecutiveness(data):
    """
    >>> find_some_sort_of_weird_consecutiveness([1,3,4,5])
    [[1], [3, 4, 5]]
    >>> find_some_sort_of_weird_consecutiveness([1,3,5,6,7])
    [[1, 3, 5], [6], [7]]
    >>> find_some_sort_of_weird_consecutiveness([1,2,3,7,9,11])
    [[1, 2, 3], [7, 9, 11]]
    >>> find_some_sort_of_weird_consecutiveness([1,2,3,7,10,12,14,25])
    [[1, 2, 3], [7], [10, 12, 14], [25]]
    >>> find_some_sort_of_weird_consecutiveness([1,2,3,4,6,8,10])
    [[1, 2, 3, 4], [6, 8, 10]]
    >>> find_some_sort_of_weird_consecutiveness([1,5,10,11,12,13,14,15,17,20,22,24,26,28,30])
    [[1], [5], [10, 11, 12, 13, 14, 15], [17], [20, 22, 24, 26, 28, 30]]
    """
    def pair_iter(series):
        from itertools import tee
        _first, _next = tee(series)
        next(_next, None)
        for i, (f, n) in enumerate(zip(_first, _next), start=MINIMAL_MATCH - 1):
            yield i, f, n

    result = []
    while len(data) >= MINIMAL_MATCH:
        test = data[1] - data[0]
        if (data[2] - data[1]) == test:
            for i, f, n in pair_iter(data):
                if (n - f) != test:
                    i -= 1
                    break
        else:
            i = 1
        data, match = data[i:], data[:i]
        result.append(match)
    for d in data:
        result.append([d])
    return result

if __name__ == '__main__':
    from doctest import testmod
    testmod()

它处理您当前所有的测试用例。如果有新的失败测试用例，请给我。

正如下面评论中提到的，我假设最短序列现在是三个元素，因为两个元素的序列是微不足道的。

参见 http://docs.python.org/2/library/itertools.html对成对迭代器的解释。