如何进行逆“范围"，即根据一组数字创建一个紧凑的范围? [英] How to do an inverse `range`, i.e. create a compact range based on a set of numbers?

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?

推荐答案

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']

这篇关于如何进行逆“范围"，即根据一组数字创建一个紧凑的范围?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！