如何进行逆“范围"，即根据一组数字创建一个紧凑的范围?
问题描述
Python 有一个 range 方法，它允许使用以下内容:
Python has a range method, which allows for stuff like:
>>> range(1, 6)
[1, 2, 3, 4, 5]
我正在寻找的是相反的:获取一个数字列表，然后返回开始和结束.
What I’m looking for is kind of the opposite: take a list of numbers, and return the start and end.
>>> magic([1, 2, 3, 4, 5])
[1, 5] # note: 5, not 6; this differs from `range()`
对于上面的示例来说，这很容易做到，但是是否也可以允许间隙或多个范围，以类似 PCRE 的字符串格式返回范围? 像这样:
This is easy enough to do for the above example, but is it possible to allow for gaps or multiple ranges as well, returning the range in a PCRE-like string format? Something like this:
>>> magic([1, 2, 4, 5])
['1-2', '4-5']
>>> magic([1, 2, 3, 4, 5])
['1-5']
我正在寻找 Python 解决方案，但我也欢迎使用其他语言的工作示例.它更多地是关于找出一个优雅、高效的算法.额外的问题:是否有任何编程语言对此具有内置方法?
I’m looking for a Python solution, but I welcome working examples in other languages as well. It’s more about figuring out an elegant, efficient algorithm. Bonus question: is there any programming language that has a built-in method for this?
解决方案
简化代码的一个好技巧是查看排序列表的每个元素及其索引的差异:
A nice trick to simplify the code is to look at the difference of each element of the sorted list and its index:
a = [4, 2, 1, 5]
a.sort()
print [x - i for i, x in enumerate(a)]
打印
[1, 1, 2, 2]
每个相同数字的运行对应于 a 中的一系列连续数字.我们现在可以使用 itertools.groupby() 来提取这些运行.这是完整的代码:
Each run of the same number corresponds to a run of consecutive numbers in a. We can now use itertools.groupby() to extract these runs. Here's the complete code:
from itertools import groupby
def sub(x):
return x[1] - x[0]
a = [5, 3, 7, 4, 1, 2, 9, 10]
ranges = []
for k, iterable in groupby(enumerate(sorted(a)), sub):
rng = list(iterable)
if len(rng) == 1:
s = str(rng[0][1])
else:
s = "%s-%s" % (rng[0][1], rng[-1][1])
ranges.append(s)
print ranges
打印
['1-5', '7', '9-10']
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