On Feb 5, 2:07 pm, rdmur...@bitdance.com wrote:
> Quoth J Kenneth King <ja...@agentultra.com>:
>
>
>
> > mk <mrk...@gmail.com> writes:
>
> > > Hello everybody,
>
> > > Any better solution than this?
>
> > > def flatten(x):
> > > res = []
> > > for el in x:
> > > if isinstance(el,list):
> > > res.extend(flatten(el))
> > > else:
> > > res.append(el)
> > > return res
>
> > > a = [1, 2, 3, [4, 5, 6], [[7, 8], [9, 10]]]
> > > print flatten(a)
>
> > > [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>
> > > Regards,
> > > mk
>
> >http://mail.python.org/pipermail/python-list/2005-July/330367.html
>
> That's worth reading. I'm not sure why I'm finding this fun, but who
> cares. I tried a couple of other functions after reading that article,
> and it looks like a generator that scans the nested lists is actually
> the winner, and also is in many ways the most elegant implementation.
> Of course, as noted in the emails following above article, the test data
> is really inadequate for proper optimization testing ;)
>
> -----------------------------------------------------------------
> from __future__ import print_function
> from timeit import Timer
> from itertools import chain
>
> # This is the one from the article quoted above.
> def flatten6(seq):
> i = 0
> while (i != len(seq)):
> while hasattr(seq[i], '__iter__'):
> seq[i:i+1] = seq[i]
> i = i + 1
> return seq
>
> #This is my favorite from a readability standpoint out of
> #all the things I tried. It also performs the best.
> def flatten8(seq):
> for x in seq:
> if not hasattr(x, '__iter__'): yield x
> else:
> for y in flatten8(x):
> yield y
>
> l = [[1, 2, 3], 5, [7, 8], 3, [9, [10, 11, 12], 4, [9, [10, 5, 7]], 1,
> [[[[[5, 4], 3], 4, 3], 3, 1, 45], 9], 10]]
>
> if __name__=="__main__":
> print(l)
> print('flatten6', flatten6(l))
> print('flatten8', list(flatten8(l)))
> print('flatten6', Timer("flatten6(l)", "from temp3 import flatten6,
> l").timeit())
> print('flatten8', Timer("list(flatten8(l))", "from temp3 import flatten8,
> l").timeit())
>
> -----------------------------------------------------------------
>
> >src/python/Python-3.0/python temp3.py
>
> [[1, 2, 3], 5, [7, 8], 3, [9, [10, 11, 12], 4, [9, [10, 5, 7]], 1, [[[[[5,
> 4], 3], 4, 3], 3, 1, 45], 9], 10]]
> flatten6 [1, 2, 3, 5, 7, 8, 3, 9, 10, 11, 12, 4, 9, 10, 5, 7, 1, 5, 4, 3, 4,
> 3, 3, 1, 45, 9, 10]
> flatten8 [1, 2, 3, 5, 7, 8, 3, 9, 10, 11, 12, 4, 9, 10, 5, 7, 1, 5, 4, 3, 4,
> 3, 3, 1, 45, 9, 10]
> flatten6 32.8386368752
> flatten8 30.7509689331
>
> >python temp3.py
>
> [[1, 2, 3], 5, [7, 8], 3, [9, [10, 11, 12], 4, [9, [10, 5, 7]], 1, [[[[[5,
> 4], 3], 4, 3], 3, 1, 45], 9], 10]]
> flatten6 [1, 2, 3, 5, 7, 8, 3, 9, 10, 11, 12, 4, 9, 10, 5, 7, 1, 5, 4, 3, 4,
> 3, 3, 1, 45, 9, 10]
> flatten8 [1, 2, 3, 5, 7, 8, 3, 9, 10, 11, 12, 4, 9, 10, 5, 7, 1, 5, 4, 3, 4,
> 3, 3, 1, 45, 9, 10]
> flatten6 34.730714798
> flatten8 32.3252940178
>
> --RDM
I think the generator is clearer with a try statement, like in Magnus
Lie Hetland's solution from Beginning Python:
def flatten(nested):
try:
# Don't iterate over string-like objs.
try: nested + ''
except TypeError: pass
else: raise TypeError
for sub in nested:
for elem in flatten(sub):
yield elem
except TypeError:
# The for doesn't work for single elements.
yield nested
You can't iterate over string-like objs because the all strings are
built of infinite empty lists at the beginning, leading to infinite
recursion.
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