Lambda evaluation
Joshua Ginsberg
listspam at flowtheory.net
Thu Oct 6 16:18:15 EDT 2005
So this part makes total sense to me:
>>> d = {}
>>> for x in [1,2,3]:
... d[x] = lambda y: y*x
...
>>> d[1](3)
9
Because x in the lambda definition isn't evaluated until the lambda is
executed, at which point x is 3.
Is there a way to specifically hard code into that lambda definition the
contemporary value of an external variable? In other words, is there a
way to rewrite the line "d[x] = lambda y: y*x" so that it is always the
case that d[1](3) = 3?
Thanks!
-jag
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