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POConvert List of Numbers to String Ranges
 

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    <p>I'd like to know if there is a simple (or already created) way of doing the opposite of this: <a href="http://code.activestate.com/recipes/577279-generate-list-of-numbers-from-hyphenated-and-comma/" rel="nofollow noreferrer">Generate List of Numbers from Hyphenated...</a>. This link could be used to do:</p> <pre><code>&gt;&gt; list(hyphen_range('1-9,12,15-20,23')) [1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15, 16, 17, 18, 19, 20, 23]: </code></pre> <p>I'm looking to do the opposite (note that 10 and 21 are included so it would be compatible with the range function, where range(1,10)=[1,2,3,4,5,6,7,8,9]):</p> <pre><code>&gt;&gt; list_to_ranges([1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15, 16, 17, 18, 19, 20, 23]) '1-10,12,15-21,23' </code></pre> <p>Eventually, I would like to have the output also incorporate a step where the last number of the output indicates the step:</p> <pre><code>&gt;&gt; list_to_ranges([1, 3, 5, 7, 8, 9, 10, 11]) '1-13:2,8,10' </code></pre> <p>Essentially, this would end up being kind of like an "inverse" range function</p> <pre><code>&gt;&gt; tmp = list_to_ranges([1, 3, 5]) &gt;&gt; print tmp '1-7:2' &gt;&gt; range(1, 7, 2) [1, 3, 5] </code></pre> <p>My guess is that there is no really easy/simple way to do this, but I thought I would ask on here before I go make some brute force, long method.</p> <p><strong>EDIT</strong></p> <p>Using the code from an answer to <a href="https://stackoverflow.com/questions/3149440/python-splitting-list-based-on-missing-numbers-in-a-sequence">this post</a> as an example, I came up with a simple way to do the first part. But I think that identifying the patterns to do steps would be a bit harder.</p> <pre><code>from itertools import groupby from operator import itemgetter data = [ 1, 4,5,6, 10, 15,16,17,18, 22, 25,26,27,28] print data, '\n' str_list = [] for k, g in groupby(enumerate(data), lambda (i,x):i-x): ilist = map(itemgetter(1), g) print ilist if len(ilist) &gt; 1: str_list.append('%d-%d' % (ilist[0], ilist[-1]+1)) else: str_list.append('%d' % ilist[0]) print '\n', ','.join(str_list) </code></pre> <p><strong>EDIT 2</strong></p> <p>Here is my attempt at including the step size...it is pretty close, but the first numbers get repeated. I think that with a little bit of tweaking of this, it will be close to what I want - or at least good enough.</p> <pre><code>import numpy as np from itertools import groupby def list_to_ranges(data): data = sorted(data) diff_data = np.diff(data).tolist() ranges = [] i = 0 for k, iterable in groupby(diff_data, None): rng = list(iterable) step = rng[0] if len(rng) == 1: ranges.append('%d' % data[i]) elif step == 1: ranges.append('%d-%d' % (data[i], data[i+len(rng)]+step)) else: ranges.append('%d-%d:%d' % (data[i], data[i+len(rng)]+step, step)) i += len(rng) return ','.join(ranges) data = [1, 3, 5, 6, 7, 11, 13, 15, 16, 17, 18, 19, 22, 25, 28] print data data_str = list_to_ranges(data) print data_str _list = [] for r in data_str.replace('-',':').split(','): r = [int(a) for a in r.split(':')] if len(r) == 1: _list.extend(r) elif len(r) == 2: _list.extend(range(r[0], r[1])) else: _list.extend(range(r[0], r[1], r[2])) print _list print list(set(_list)) </code></pre>
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    1. COThe brute force method you talk about, doesn't necessarily have to be that long at all..
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    2. COI agree. You're going to have to parse the list in order to indentify patterns, especially if you want to add non-unity step recognition as well.
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      1. USJoel Cornett
    3. COThere is some ambiguity here: `1-13:2,8,10` is the same as `1-7:2,7-11`. Before we can really look at an algorithm, you'll have to give a more precise definition of what you want.
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      1. USWinston Ewert
 

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