On 27/07/2022 18:24, Cecil Westerhof via Python-list wrote:
MRAB <pyt...@mrabarnett.plus.com> writes:
On 27/07/2022 16:43, Cecil Westerhof via Python-list wrote:
"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)]
When you pop an element from the last, the elements after it need to be
moved down, which takes time.
Try shuffling the list and then popping the now randomly-ordered
elements off the end.
Would shuffling not be a lot more expensive? Especially because I do
not eat the whole list.
You won't know until you time it.
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