On Wed, 25 Nov 2015 14:51:23 +0100, Antoon Pardon wrote:
> Am I missing something?
The issue is with lambdas rather than with list comprehensions per se.
Python's lambdas capture free variables by reference, not value.
> x = 3
> f = lambda y: x + y
> f(0)
3
> x = 7
> f(0)
7
The same issue applies to nested functions:
> def foo():
= x = 3
= def f(y):
= return x + y
= print f(0)
= x = 7
= print f(0)
=
> foo()
3
7
And also to non-nested functions (but most people expect that):
> x = 3
> def f(y,x=x):
= return x + y
=
> print f(0)
3
> x=7
> print f(0)
3
If you want to capture a variable by value, add a parameter with a default
value using that variable:
> def foo():
= x = 3
= def f(y, x=x):
= return x + y
= print f(0)
= x = 7
= print f(0)
=
> foo()
3
3
This also works for lambdas:
> x = 3
> f = lambda y,x=x: x + y
> f(0)
3
> x = 7
> f(0)
3
Returning to the original expression:
> q = [lambda x: i * x for i in range(4)]
> q[0](1), q[3](1)
(3, 3)
> q = [lambda x,i=i: i * x for i in range(4)]
> q[0](1), q[3](1)
(0, 3)
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