Is it possible to "dynamically" create local variables in Python?
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Chapters
00:00 Is It Possible To &Quot;Dynamically&Quot; Create Local Variables In Python?
00:33 Accepted Answer Score 9
00:53 Answer 2 Score 5
01:17 Thank you
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Full question
https://stackoverflow.com/questions/8799...
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Tags
#python #variables #memory #local #instantiation
#avk47
Rise to the top 3% as a developer or hire one of them at Toptal: https://topt.al/25cXVn
--------------------------------------------------
Music by Eric Matyas
https://www.soundimage.org
Track title: Dreaming in Puzzles
--
Chapters
00:00 Is It Possible To &Quot;Dynamically&Quot; Create Local Variables In Python?
00:33 Accepted Answer Score 9
00:53 Answer 2 Score 5
01:17 Thank you
--
Full question
https://stackoverflow.com/questions/8799...
--
Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...
--
Tags
#python #variables #memory #local #instantiation
#avk47
ACCEPTED ANSWER
Score 9
If you really want to do this, you could use exec:
print 'iWantAVariableWithThisName' in locals()
junkVar = 'iWantAVariableWithThisName'
exec(junkVar + " = 1")
print 'iWantAVariableWithThisName' in locals()
Of course, anyone will tell you how dangerous and hackish using exec is, but then so will be any implementation of this "trickery."
ANSWER 2
Score 5
You can play games and update locals() manually, which will sometimes work, but you shouldn't. It's specifically warned against in the docs. If I had to do this, I'd probably use exec:
>>> 'iWantAVariableWithThisName' in locals()
False
>>> junkVar = 'iWantAVariableWithThisName'
>>> exec(junkVar + '= None')
>>> 'iWantAVariableWithThisName' in locals()
True
>>> print iWantAVariableWithThisName
None
But ninety-three times out of one hundred you really want to use a dictionary instead.