检测列表中的连续整数

Question

我有一个包含数据的列表:

[1, 2, 3, 4, 7, 8, 10, 11, 12, 13, 14]

我想打印出连续整数的范围:

1-4, 7-8, 10-14

是否有内置/快速/有效的方法来做到这一点？

Answer 1

来自文档:

>>> from itertools import groupby
>>> from operator import itemgetter
>>> data = [ 1, 4,5,6, 10, 15,16,17,18, 22, 25,26,27,28]
>>> for k, g in groupby(enumerate(data), lambda (i, x): i-x):
...     print map(itemgetter(1), g)
...
[1]
[4, 5, 6]
[10]
[15, 16, 17, 18]
[22]
[25, 26, 27, 28]

您可以相当容易地调整它以获得一组打印范围.

Answer 2

一个简短的解决方案，无需额外导入即可使用。它接受任何迭代，对未排序的输入进行排序，并删除重复项：

def ranges(nums):
    nums = sorted(set(nums))
    gaps = [[s, e] for s, e in zip(nums, nums[1:]) if s+1 < e]
    edges = iter(nums[:1] + sum(gaps, []) + nums[-1:])
    return list(zip(edges, edges))

例：

>>> ranges([2, 3, 4, 7, 8, 9, 15])
[(2, 4), (7, 9), (15, 15)]

>>> ranges([-1, 0, 1, 2, 3, 12, 13, 15, 100])
[(-1, 3), (12, 13), (15, 15), (100, 100)]

>>> ranges(range(100))
[(0, 99)]

>>> ranges([0])
[(0, 0)]

>>> ranges([])
[]

这与@dansalmo的解决方案相同，我发现了惊人的解决方案，尽管有点难以阅读和应用（因为它没有作为函数给出）。

请注意，可以很容易地对其进行修改以吐出“传统的”开放范围[start, end)，例如通过更改return语句：

    return [(s, e+1) for s, e in zip(edges, edges)]

Answer 3

这将完全按照您指定的方式打印：

>>> nums = [1, 2, 3, 4, 7, 8, 10, 11, 12, 13, 14]
>>> ranges = sum((list(t) for t in zip(nums, nums[1:]) if t[0]+1 != t[1]), [])
>>> iranges = iter(nums[0:1] + ranges + nums[-1:])
>>> print ', '.join([str(n) + '-' + str(next(iranges)) for n in iranges])
1-4, 7-8, 10-14

如果列表具有任何单个数字范围，则它们将显示为nn：

>>> nums = [1, 2, 3, 4, 5, 7, 8, 9, 12, 15, 16, 17, 18]
>>> ranges = sum((list(t) for t in zip(nums, nums[1:]) if t[0]+1 != t[1]), [])
>>> iranges = iter(nums[0:1] + ranges + nums[-1:])
>>> print ', '.join([str(n) + '-' + str(next(iranges)) for n in iranges])
1-5, 7-9, 12-12, 15-18