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 :)
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By ZeD
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