# [Chicago] Resolving lists within lists within lists within .....

Fri Feb 19 10:34:27 EST 2016

```Aaron, Thanks for your example. One thing to point out is that popping from
the front of a list is expensive because the entire list has to be copied.
Some options are to flatten the list from the back (popping off the end of
the list is cheap), or copying the list into a deque (from collections
import deque).

Here is another example of a non recursive version of flatten. It's not
nearly as elegant as the recursive version. It's longer than Aaron's
iterative version, but avoids hand manipulating the iteration over the
lists (no popping or inserting).

def press(lst):
"""
Flattens nested lists one level

Returns a tuple (new_list, changed) where changed is a boolean
indicating
whether new_list is different from lst.
"""
changed = False
new_list = []
for element in lst:
if isinstance(element, list):
new_list.extend(element)
changed = True
else:
new_list.append(element)

return new_list, changed

def flatten(lst):
"""
Fully flattens nested lists into a list with no sublists
"""
new_list = lst
changed = True
while changed:
new_list, changed = press(new_list)
return new_list

On Fri, Feb 19, 2016 at 6:59 AM, Aaron Elmquist <elmq0022 at umn.edu> wrote:

> Here's one last approach that is stack based.  There is some clean up to
> do here for sure (I'm mutating the original list for one), but the point is
> to illustrate an approach that is not recursive.
>
> def flatten_big_list(lst):
>     stack = []
>     while(lst):
>         top = lst.pop(0)
>         while(isinstance(top,list)):
>             temp = top.pop(0)
>             if top:
>                 lst.insert(0,top)
>             top = temp
>         stack.append(top)
>     return stack
>
>
> def flatten_big_list_gen(lst):
>     while(lst):
>         top = lst.pop(0)
>         while(isinstance(top,list)):
>             temp = top.pop(0)
>             if top:
>                 lst.insert(0,top)
>             top = temp
>         yield top
>
>
> print(flatten_big_list([1, [2, [3, [4, 5]]]]))
> print(list(flatten_big_list_gen([1, [2, [3, [4, 5]]]])))
>
> Feedback is always welcome.
>
>
> On Thu, Feb 18, 2016 at 9:29 PM, Mark Graves <mgraves87 at gmail.com> wrote:
>
>> Doug,
>>
>>
>> "The" answer to why is recursion expensive vs iteration is stack traces.
>> try it yourself as mentioned here
>> <http://t.yesware.com/tt/6640a48a14dbdef70b47105ac6b72156559fc5a6/5ba2375237a9fdc8efa681b19014981f/dda1509570b2b5d9d162e6293a1b3f07/stackoverflow.com/questions/22893139/why-is-a-function-method-call-in-python-expensive>
>> .
>>
>> Recursion means creating more functions / stack traces.
>>
>> wrote:
>>
>>> Phil,
>>>
>>> That's generally true, but one small correction. Aaron's solution won't
>>> actually won't flatten strings, as they don't have "__iter__" methods. They
>>> implement iteration because they take sequential numeric indexes starting
>>> at 0, and raise an IndexError after the index passed is too large.
>>>
>>> On Feb 18, 2016 19:22, "Robare, Phillip (TEKSystems)" <
>>> proba at allstate.com> wrote:
>>>
>>>> Aaron, unlike Massimo’s elegant one-liner you don’t check that what you
>>>> are iterating over is a list.  Since Python will happily iterate over
>>>> strings, dictionaries, and much more you quickly get into problems when the
>>>> list includes more types than lists and numbers.  I recount this from
>>>> experience when I tried to throw together a flatten routine and pass it a
>>>>
>>>>
>>>>
>>>> Phil Robare
>>>>
>>>>
>>>>
>>>> *<snip/>*
>>>>
>>>>
>>>>
>>>> On Thu, Feb 18, 2016 at 1:43 PM, Aaron Elmquist <elmq0022 at umn.edu>
>>>> wrote:
>>>>
>>>> Douglas,
>>>>
>>>> Here's one more version for you and the rest of the list. It's based on
>>>> Brad's code.  I will let you think about why this version might be better
>>>> or worse.  Also, recursion is great.  It's just too bad it's not one of
>>>> python's strong points.
>>>>
>>>>
>>>> def flatten(lst):
>>>>     for item1 in lst:
>>>>         if hasattr(item1, '__iter__'):
>>>>             for item2 in flatten(item1):
>>>>                 yield item2
>>>>         else:
>>>>             yield item1
>>>>
>>>> print([x for x in flatten([1, [2,3,[4,5,6,[7,8,9]]]]) if x%2 == 1])
>>>>
>>>> y = flatten([1, [2,3,[4,5,6,[7,8,9]]]])
>>>>
>>>> print(next(y))
>>>> print(next(y))
>>>> print(next(y))
>>>> .
>>>> .
>>>> .
>>>> <snip/>
>>>>
>>>> On Wed, Feb 17, 2016 at 9:48 PM, DiPierro, Massimo <
>>>> MDiPierro at cs.depaul.edu> wrote:
>>>>
>>>> here is a one liner:
>>>>
>>>> def flatten(x):
>>>>     return [z for y in x for z in flatten(y)] if isinstance(x,list)
>>>> else [x]
>>>>
>>>>
>>>>
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>>>>
>>>>
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>>>
>>
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>
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