Roel Schroeven <r...@roelschroeven.net> writes: > Cecil Westerhof via Python-list schreef op 27/07/2022 om 17:43: >> "Michael F. Stemper" <michael.stem...@gmail.com> writes: >> >> > This is orthogonal to your question, but might be of some use to you: >> > >> > The combination of using len(to_try) as an argument to randint() and >> > saving the output to a variable named "index" suggests that you might >> > be setting up to select a random element from to_try, as in: >> > something = to_try[index] >> > >> > If that is the case, you might want to consider using random.choice() >> > instead: >> > >> > >>> from random import choice >> > >>> to_try = [2,3,5,7,11,13,"seventeen",19] >> > >>> choice(to_try) >> > 2 >> > >>> choice(to_try) >> > 'seventeen' >> > >>> choice(to_try) >> > 13 >> > >>> choice(to_try) >> > 5 >> > >>> >> >> Yes, I try to select a random element, but it has also to be removed, >> because an element should not be used more as once. >> This is the code I use: >> # index = randbelow(len(to_try)) >> index = randrange(len(to_try)) >> found = permutation[to_try.pop(index)] > Do you know in advance how many items you'll need, or maybe an upper > limit on the amount? In that case it might be more efficient to use > random.sample(to_try, k=nr_items_needed).
Something else to try. :-) And yes: I will be using half of the list. -- Cecil Westerhof Senior Software Engineer LinkedIn: http://www.linkedin.com/in/cecilwesterhof -- https://mail.python.org/mailman/listinfo/python-list
