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

Fri Feb 19 11:47:54 EST 2016

LOL

vulnerable to square brackets injection attacks! [1, 2, [1, 2, 3, [1, 2, ‘[9,1,9]', 4], 5], 4, 5]

On Feb 19, 2016, at 10:39 AM, JS Irick <hundredpercentjuice at gmail.com<mailto:hundredpercentjuice at gmail.com>> wrote:

I think we can agree that there is only one true solution:

>>> my_list

[1, 2, [1, 2, 3, [1, 2, 3, 4], 5], 4, 5]

>>> json.loads("["+re.sub('[\[\]]','',json.dumps(my_list))+"]")

[1, 2, 1, 2, 3, 1, 2, 3, 4, 5, 4, 5]

On Fri, Feb 19, 2016 at 10:05 AM, Aaron Elmquist <elmq0022 at umn.edu<mailto:elmq0022 at umn.edu>> wrote:
That's still potentially a lot of list copying though, isn't it?

On Fri, Feb 19, 2016 at 9:40 AM, Aaron Elmquist <elmq0022 at umn.edu<mailto:elmq0022 at umn.edu>> wrote:
Brad, that's a really cool approach and very readable.  Thanks!

On Fri, Feb 19, 2016 at 9:34 AM, Brad Martsberger <bradley.marts at gmail.com<mailto:bradley.marts at gmail.com>> wrote:
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<mailto: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<mailto:mgraves87 at gmail.com>> wrote:
Doug,

Also, I didn't see your question get answered.

"The" answer to why is recursion expensive vs iteration is stack traces.  See Guido's answer here<https://t.yesware.com/tt/6640a48a14dbdef70b47105ac6b72156559fc5a6/5ba2375237a9fdc8efa681b19014981f/d6c3025efb0710ebe9f6fa425f843d2c/plus.google.com/115212051037621986145/posts/HajXHPGN752> or 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.

On Thu, Feb 18, 2016 at 7:55 PM, Adam Forsyth <adam at adamforsyth.net<mailto:adam at adamforsyth.net>> 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.

Adam

On Feb 18, 2016 19:22, "Robare, Phillip (TEKSystems)" <proba at allstate.com<mailto: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 data structure that I got from loading a JSON string.

Phil Robare

<snip/>

On Thu, Feb 18, 2016 at 1:43 PM, Aaron Elmquist <elmq0022 at umn.edu<mailto: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<mailto: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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