[python-uk] A stack with better performance than using a list
Jonathan Hartley
tartley at tartley.com
Tue Jun 13 09:41:25 EDT 2017
Very interesting! Thanks for digging deeper and sharing.
I was thinking about horrible complicated structures like storing the
'add_to_first_n' params in parallel to the stack, to apply them at 'pop'
time, which doesn't work at all. As is so often the case with these
things, your solution of pushing those markers onto the stack makes me
feel silly for not realising sooner.
Thanks to everyone for the interesting discussion.
Jonathan
On 2017-06-08 13:17, Stestagg wrote:
> I tracked down the challenge on the site, and have a working solution
> (I won't share for obvious reasons). Basically the timeouts were being
> caused by 'add_to_first_n' being called in horrible ways in the test
> cases.
>
> Because add_to_first_n alters the bottom of the stack, you can just
> push a marker onto the stack rather than iterating and mutating each
> entry, doing this made those test cases pass
>
> Personally, I think it's not a well-described problem, because it's
> expecting you to tune the algo to specific shapes of data without
> allowing any visibility on the data, or a description of what to code
> for. An algo junkie may jump straight to the optimized version, but a
> pragmatic developer would, in my opinion, hesitate to do that without
> any actual evidence that the problem required it.
>
> Steve
>
> On Thu, Jun 8, 2017 at 5:27 PM Jonathan Hartley <tartley at tartley.com>
> wrote:
>
>> Yep, that's a great elimination of the suspicious small overheads.
>>
>> line_profiler is beautiful, I'll definitely be adding it to my
>> toolbox, thanks for that!
>>
>> I tried a variant of accumulating the output and printing it all as
>> a single string, but of course this didn't help, printing is already
>> buffered.
>>
>> Jonathan
>>
>> On 6/8/2017 03:54, Stestagg wrote:
>>
>> I honestly can't see a way to improve this in python. My best
>> solution is:
>>
>> def main(lines):
>> stack = []
>> sa = stack.append
>> sp = stack.pop
>> si = stack.__getitem__
>> for line in lines:
>> meth = line[:3]
>> if meth == b'pus':
>> sa(int(line[5:]))
>> elif meth == b'pop':
>> sp()
>> else:
>> parts = line[15:].split()
>> end = len(stack)-1
>> amount = int(parts[1])
>> for x in range(int(parts[0])):
>> index = end - x
>> stack[index] += amount
>> print(stack[-1] if stack else None)
>>
>> which comes out about 25% faster than your solution.
>>
>> One tool that's interesting to use here is: line_profiler:
>> https://github.com/rkern/line_profiler
>>
>> putting a @profile decorator on the above main() call, and running
>> with kernprof produces the following output:
>>
>> Line # Hits Time Per Hit % Time Line Contents
>>
>> ==============================================================
>>
>> 12 @profile
>>
>> 13 def main(lines):
>>
>> 14 1 4 4.0 0.0 stack = []
>>
>> 15 2000001 949599 0.5 11.5 for line in
>> lines:
>>
>> 16 2000000 1126944 0.6 13.7 meth =
>> line[:3]
>>
>> 17 2000000 974635 0.5 11.8 if meth ==
>> b'pus':
>>
>> 18 1000000 1002733 1.0 12.2
>> stack.append(int(line[5:]))
>>
>> 19 1000000 478756 0.5 5.8 elif meth
>> == b'pop':
>>
>> 20 999999 597114 0.6 7.2
>> stack.pop()
>>
>> 21 else:
>>
>> 22 1 6 6.0 0.0 parts =
>> line[15:].split()
>>
>> 23 1 2 2.0 0.0 end =
>> len(stack)-1
>>
>> 24 1 1 1.0 0.0 amount
>> = int(parts[1])
>>
>> 25 500001 241227 0.5 2.9 for x
>> in range(int(parts[0])):
>>
>> 26 500000 273477 0.5 3.3
>> index = end - x
>>
>> 27 500000 309033 0.6 3.7
>> stack[index] += amount
>>
>> 28 2000000 2295803 1.1 27.8
>> print(stack[-1])
>>
>> which shows that there's no obvious bottleneck (line by line) here
>> (for my sample data).
>>
>> Note the print() overhead dominates the runtime, and that's with me
>> piping the output to /dev/null directly.
>>
>> I had a go at using arrays, deques, and numpy arrays in various ways
>> without luck, but we're getting fairly close to the native python
>> statement execution overhead here (hence folding it all into one
>> function).
>>
>> My only thoughts would be to see if there were some magic that could
>> be done by offloading the work onto a non-python library somehow.
>>
>> Another thing that might help some situations (hence my previous
>> questions) would be to implement the add_to_first_n as a lazy
>> operator (i.e. have a stack of the add_to_first_n values and
>> dynamically add to the results of pop() but that would proabably be
>> much slow in the average case.
>>
>> Steve
>>
>> On Wed, Jun 7, 2017 at 7:34 PM Jonathan Hartley
>> <tartley at tartley.com> wrote:
>>
>> Hey.
>>
>> Thanks for engaging, but I can't help with the most important of
>> those questions - the large data sets on which my solution failed
>> due to timeout are hidden from candidates. Not unreasonable to
>> assume that they do exercise deep stacks, and large args to
>> add_to_first_n, etc.
>>
>> Yes, the input looks exactly like your example. All args are
>> integers. The question asked for output corresponding to the top of
>> the stack after every operation. I omitted this print from inside
>> the 'for' loop in 'main', thinking it irrelevant.
>>
>> I converted to integers inside 'dispatch'. 'args' must have actually
>> been created with:
>>
>> args = [int(i) for i in tokens[1:]]
>>
>> Where len(tokens) is never going to be bigger than 3.
>>
>> Return values (from 'pop') were unused.
>>
>> On 6/7/2017 13:25, Stestagg wrote:
>>
>> Do you have any more context?
>> For example, is the add_to_first_n likely to be called with very
>> large numbers, or very often? Does the stack get very deep, or stay
>> shallow?
>>
>> I'm assuming that lines look like this:
>>
>> push 1
>> push 2
>> add_to_first_n 2 10
>> pop
>> pop
>>
>> with all arguments as integers, and the final value being returned
>> from main()?
>> How did you convert from string inputs to numeric values?
>> How did you manage return values?
>>
>> :D
>>
>> On Wed, Jun 7, 2017 at 6:51 PM Jonathan Hartley
>> <tartley at tartley.com> wrote:
>>
>> I recently submitted a solution to a coding challenge, in an
>> employment context. One of the questions was to model a simple
>> stack. I wrote a solution which appended and popped from the end of
>> a list. This worked, but failed with timeouts on their last few
>> automated tests with large (hidden) data sets.
>>
>> From memory, I think I had something pretty standard:
>>
>> class Stack:
>>
>> def __init__(self):
>> self.storage = []
>>
>> def push(arg):
>> self.storage.append(arg)
>>
>> def pop():
>> return self.storage.pop() if self.storage else None
>>
>> def add_to_first_n(n, amount):
>> for n in range(n):
>> self.storage[n] += amount
>>
>> def dispatch(self, line)
>> tokens = line.split()
>> method = getattr(self, tokens[0])
>> args = tokens[1:]
>> method(*args)
>>
>> def main(lines):
>> stack = Stack()
>> for line in lines:
>> stack.dispatch(line)
>>
>> (will that formatting survive? Apologies if not)
>>
>> Subsequent experiments have confirmed that appending to and popping
>> from the end of lists are O(1), amortized. So why is my solution too
>> slow?
>>
>> This question was against the clock, 4th question of 4 in an hour.
>> So I wasn't expecting to produce Cython or C optimised code in that
>> timeframe (Besides, my submitted .py file runs on their servers, so
>> the environment is limited.)
>>
>> So what am I missing, from a performance perspective? Are there
>> other data structures in stdlib which are also O(1) but with a
>> better constant?
>>
>> Ah. In writing this out, I have begun to suspect that my slicing of
>> 'tokens' to produce 'args' in the dispatch is needlessly wasting
>> time. Not much, but some.
>>
>> Thoughts welcome,
>>
>> Jonathan
>>
>> --
>> Jonathan Hartley tartley at tartley.com http://tartley.com
>> Made out of meat. +1 507-513-1101 [1] twitter/skype:
>> tartley
>>
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> --
> Jonathan Hartley tartley at tartley.com http://tartley.com
> Made out of meat. +1 507-513-1101 [1] twitter/skype: tartley
>
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> Jonathan Hartley tartley at tartley.com http://tartley.com
> Made out of meat. +1 507-513-1101 [2] twitter/skype: tartley
>
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Jonathan Hartley Made out of meat.
tartley at tartley.com +1 507-513-1101
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