在另一篇SO文章中 ，一位Python用户询问如何对连续数字进行分组，使得任何序列都可以由其开始/结束代表，并且任何散列表将显示为单个项目。 接受的答案出色地适用于连续序列。
我需要能够适应类似的解决方案，但对于可能（并非总是）不同增量的数字序列。 理想情况下，我如何表示，还包括增量（所以他们会知道它是否是每3,4,5，nth）
参考原始问题，用户询问以下输入/输出
[2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20] # input [(2,5), (12,17), 20]我想要的是以下内容（注意：为了清楚起见，我写了一个元组作为输出，但是使用其step变量优先选择xrange）：
[2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20] # input [(2,5,1), (12,17,1), 20] # note, the last element in the tuple would be the step value它也可以处理以下输入
[2, 4, 6, 8, 12, 13, 14, 15, 16, 17, 20] # input [(2,8,2), (12,17,1), 20] # note, the last element in the tuple would be the increment我知道xrange()支持一个步骤，所以甚至可以使用其他用户的答案的变体。 我试图根据他们在解释中写的内容做出一些编辑，但我无法得到我期待的结果。
对于任何不想点击原始链接的用户，最初由Nadia Alramli发布的代码是：
ranges = [] for key, group in groupby(enumerate(data), lambda (index, item): index - item): group = map(itemgetter(1), group) if len(group) > 1: ranges.append(xrange(group[0], group[-1])) else: ranges.append(group[0])In this other SO post, a Python user asked how to group continuous numbers such that any sequences could just be represented by its start/end and any stragglers would be displayed as single items. The accepted answer works brilliantly for continuous sequences.
I need to be able to adapt a similar solution but for a sequence of numbers that have potentially (not always) varying increments. Ideally, how I represent that will also include the increment (so they'll know if it was every 3, 4, 5, nth)
Referencing the original question, the user asked for the following input/output
[2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20] # input [(2,5), (12,17), 20]What I would like is the following (Note: I wrote a tuple as the output for clarity but xrange would be preferred using its step variable):
[2, 3, 4, 5, 12, 13, 14, 15, 16, 17, 20] # input [(2,5,1), (12,17,1), 20] # note, the last element in the tuple would be the step valueAnd it could also handle the following input
[2, 4, 6, 8, 12, 13, 14, 15, 16, 17, 20] # input [(2,8,2), (12,17,1), 20] # note, the last element in the tuple would be the incrementI know that xrange() supports a step so it may be possible to even use a variant of the other user's answer. I tried making some edits based on what they wrote in the explanation but I wasn't able to get the result I was looking for.
For anyone that doesn't want to click the original link, the code that was originally posted by Nadia Alramli is:
ranges = [] for key, group in groupby(enumerate(data), lambda (index, item): index - item): group = map(itemgetter(1), group) if len(group) > 1: ranges.append(xrange(group[0], group[-1])) else: ranges.append(group[0])最满意答案
您可以创建一个迭代器来帮助分组，并尝试从下一个组中取出下一个元素，该组将成为前一组的结尾：
def ranges(lst): it = iter(lst) next(it) # move to second element for comparison grps = groupby(lst, key=lambda x: (x - next(it, -float("inf")))) for k, v in grps: i = next(v) try: step = next(v) - i # catches single element v or gives us a step nxt = list(next(grps)[1]) yield xrange(i, nxt.pop(0), step) # outliers or another group if nxt: yield nxt[0] if len(nxt) == 1 else xrange(nxt[0], next(next(grps)[1]), nxt[1] - nxt[0]) except StopIteration: yield i # no seq它给你：
In [2]: l1 = [2, 3, 4, 5, 8, 10, 12, 14, 13, 14, 15, 16, 17, 20, 21] In [3]: l2 = [2, 4, 6, 8, 12, 13, 14, 15, 16, 17, 20] In [4]: l3 = [13, 14, 15, 16, 17, 18] In [5]: s1 = [i + 10 for i in xrange(0, 11, 2)] In [6]: s2 = [30] In [7]: s3 = [i + 40 for i in xrange(45)] In [8]: l4 = s1 + s2 + s3 In [9]: l5 = [1, 2, 5, 6, 9, 10] In [10]: l6 = {1, 2, 3, 5, 6, 9, 10, 13, 19, 21, 22, 23, 24} In [11]: In [11]: for l in (l1, l2, l3, l4, l5, l6): ....: print(list(ranges(l))) ....: [xrange(2, 5), xrange(8, 14, 2), xrange(13, 17), 20, 21] [xrange(2, 8, 2), xrange(12, 17), 20] [xrange(13, 18)] [xrange(10, 20, 2), 30, xrange(40, 84)] [1, 2, 5, 6, 9, 10] [xrange(1, 3), 5, 6, 9, 10, 13, 19, xrange(21, 24)]当步长为1它不包含在xrange输出中。
You can create an iterator to help grouping and try to pull the next element from the following group which will be the end of the previous group:
def ranges(lst): it = iter(lst) next(it) # move to second element for comparison grps = groupby(lst, key=lambda x: (x - next(it, -float("inf")))) for k, v in grps: i = next(v) try: step = next(v) - i # catches single element v or gives us a step nxt = list(next(grps)[1]) yield xrange(i, nxt.pop(0), step) # outliers or another group if nxt: yield nxt[0] if len(nxt) == 1 else xrange(nxt[0], next(next(grps)[1]), nxt[1] - nxt[0]) except StopIteration: yield i # no seqwhich give you:
In [2]: l1 = [2, 3, 4, 5, 8, 10, 12, 14, 13, 14, 15, 16, 17, 20, 21] In [3]: l2 = [2, 4, 6, 8, 12, 13, 14, 15, 16, 17, 20] In [4]: l3 = [13, 14, 15, 16, 17, 18] In [5]: s1 = [i + 10 for i in xrange(0, 11, 2)] In [6]: s2 = [30] In [7]: s3 = [i + 40 for i in xrange(45)] In [8]: l4 = s1 + s2 + s3 In [9]: l5 = [1, 2, 5, 6, 9, 10] In [10]: l6 = {1, 2, 3, 5, 6, 9, 10, 13, 19, 21, 22, 23, 24} In [11]: In [11]: for l in (l1, l2, l3, l4, l5, l6): ....: print(list(ranges(l))) ....: [xrange(2, 5), xrange(8, 14, 2), xrange(13, 17), 20, 21] [xrange(2, 8, 2), xrange(12, 17), 20] [xrange(13, 18)] [xrange(10, 20, 2), 30, xrange(40, 84)] [1, 2, 5, 6, 9, 10] [xrange(1, 3), 5, 6, 9, 10, 13, 19, xrange(21, 24)]When the step is 1 it is not included in the xrange output.
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