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_profilerputting a @profile decorator on the above main() call, and running with kernprof produces the following output:Line #Hits TimePer Hit % TimeLine Contents ============================================================== 12 @profile 13 def main(lines): 14 144.00.0stack = [] 15 2000001 9495990.5 11.5for line in lines: 16 200000011269440.6 13.7meth = line[:3] 17 2000000 9746350.5 11.8if meth == b'pus': 18 100000010027331.0 12.2stack.append(int(line[5:])) 19 1000000 4787560.55.8elif meth == b'pop': 20999999 5971140.67.2stack.pop() 21 else: 22 166.00.0parts = line[15:].split() 23 122.00.0end = len(stack)-1 24 111.00.0amount = int(parts[1]) 25500001 2412270.52.9for x in range(int(parts[0])): 26500000 2734770.53.3index = end - x 27500000 3090330.63.7stack[index] += amount 28 200000022958031.1 27.8print(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.SteveOn Wed, Jun 7, 2017 at 7:34 PM Jonathan Hartley <tart...@tartley.com <mailto:tart...@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 <tart...@tartley.com <mailto:tart...@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 hartleytart...@tartley.com <mailto:tart...@tartley.com> http://tartley.comMade out of meat.+1 507-513-1101 <tel:%28507%29%20513-1101> twitter/skype: tartley _______________________________________________ python-uk mailing list python-uk@python.org <mailto:python-uk@python.org> https://mail.python.org/mailman/listinfo/python-uk _______________________________________________ python-uk mailing list python-uk@python.org <mailto:python-uk@python.org> https://mail.python.org/mailman/listinfo/python-uk-- Jonathan hartleytart...@tartley.com <mailto:tart...@tartley.com> http://tartley.comMade out of meat.+1 507-513-1101 <tel:%28507%29%20513-1101> twitter/skype: tartley _______________________________________________ python-uk mailing list python-uk@python.org <mailto:python-uk@python.org> https://mail.python.org/mailman/listinfo/python-uk _______________________________________________ python-uk mailing list python-uk@python.org https://mail.python.org/mailman/listinfo/python-uk
-- Jonathan Hartley tart...@tartley.com http://tartley.com Made out of meat. +1 507-513-1101 twitter/skype: tartley
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