Drowning in a teacup?

Sat Apr 2 02:26:35 EDT 2016

Michael Selik wrote:
>> > I need to scan a list of strings. If one of the elements matches the
>> > beginning of a search keyword, that element needs to snap to the front
>> > of the list.
>>>> I know this post regards the function passing, but, on you specific
>> problem,
>> can't you just ... sort the list with a custom key?
>> If the number of matches is small relative to the size of the list, I'd
> expect the sort would be slower than most of the other suggestions.

umm... I take the liberty to set up a little test
 $ cat test_sort_prefix.py
 #!/usr/bin/env python3
 from sys import argv
 from timeit import timeit
 def sort_in_place(l):
 def key(e):
 return not e.startswith('yes')
 l.sort(key=key)
 def original_solution(mylist):
 for i in range(len(mylist)):
 if(mylist[i].startswith('yes')):
 mylist[:] = [mylist[i]] + mylist[:i] + mylist[i+1:]
 def main():
 if argv[1] == 'sort_in_place':
 f = sort_in_place
 elif argv[1] == 'original_solution':
 f = original_solution
 else:
 raise Exception()
 nomatches = int(argv[2])
 matches = int(argv[3])
 l = ["no_" for _ in range(nomatches)] + ["yes_" for _ in 
range(matches)]
 print(timeit("f(l)", number=100, globals={"f": f, "l": l}))
 if __name__ == '__main__':
 main()
 $ ./test_sort_prefix.py sort_in_place 1000000 3
 33.260575089996564
 $ ./test_sort_prefix.py original_solution 1000000 3
 35.93009542999789
 $
in my tests there is no sensible difference between the two solutions... and 
the number of matches is risible :)
-- 
By ZeD