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up vote 0 down vote favorite

I have some code that I find clumsy.

Given:

sample_lists = [(u'1', u'penguin'), (u'2', u'kelp'), (u'5', u'egg')], 
               [(u'3', u'otter'), (u'4', u'clam')]

I want the result: ['penguin', 'kelp', 'otter', 'clam', 'egg'] (ordered by number). You can assume that lists always contains '1', and each sub-list is in ascending order, and that all the numbers in sample_lists are consecutive integers, if that helps.

Currently, the most pythonic/concise way I can think of doing this is:

sample_list = sample_lists[0]+sample_lists[1]
for i in xrange(len(sample_list)):
    sample_list[0] = i+1
return zip(*sorted(sample_list, key=operator.itemgetter(0)))[1]

Is there a better way? I feel like this is quite clumsy. The problem here is sorting by the strings, which I need to convert over to ints one-by-one. A hunch would be some lambda function, but I'm not terribly well versed in that syntax. If this is the best way, I'm still curious as to how to do this using lambda functions, if possible.

Note that the direct string comparison one-liner:

print zip(*sorted(sample_lists[0]+sample_lists[1], key=operator.itemgetter(0)))[1]

Does not work, as it fails if sample_lists contains '11' or '20'.

A python 3 answer is okay, but python 2.7 is preferred.

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2 Answers 2

up vote 3 down vote accepted

itemgetter is cool, but doesn't let you convert to int. In this case a lambda function is better as it lets you do the itemgetting and int converting together

>>> from itertools import chain
>>> [x[1] for x in sorted(chain.from_iterable(sample_lists), key=lambda x:int(x[0]))]
[u'penguin', u'kelp', u'otter', u'clam', u'egg']
share|improve this answer
    
What is the advantage of using chain.from_iterable as opposed to sample_lists[0]+sample_lists[1]? –  JDong Jul 22 '13 at 13:48
1  
@JDong, what if sample_lists has more than two elements? –  John La Rooy Jul 22 '13 at 14:06
up vote 1 down vote

Nice answer, gnibbler! The only thing that I think is missing is the conversion from the sample_lists to your sample_list, which could be done like so:

sample_list = [item for lst in sample_lists for item in lst]

share|improve this answer
1  
Yes, I had missed that. You can also use chain.from_iterable and do it inside the sort –  John La Rooy Jul 15 '13 at 4:36
    
@gnibbler: True, that would probably be more efficient, but I'm not that familiar with chain yet ;-) –  FriendFX Jul 15 '13 at 4:37

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