a lambda in a function

Luigi Ballabio ballabio at mac.com
Thu Dec 13 04:34:38 EST 2001

At 10:12 PM 12/12/01 +0000, Fred Clare wrote:
>Why does interpreting the five lines:
>def func():
>   x = 1
>   add_one = lambda i: i+x
>   j = add_one(100)
>Give me:
>   Traceback (most recent call last):
>     File "test.py", line 6, in ?
>       func()
>     File "test.py", line 4, in func
>       j = add_one(100)
>     File "test.py", line 3, in <lambda>
>       add_one = lambda i: i+x
>   NameError: There is no variable named 'x'

It's a scope thing. The scope of the lambda doesn't "see" the scope of the 
enclosing function---only the global scope.

You can fix it in two ways: the first is to import x into the lambda scope 
by means of a default parameter; this works in all versions of Python:

def f():
     x = 1
     add_one = lambda i,x=x: i+x
     j = add_one(100)

The second works for Python 2.1 and above: the interpreted lets you declare 
that you want the scope of the function to be visible from the lambda. The 
directive you have to add is:

from __future__ import nested_scopes

def f():
     x = 1
     add_one = lambda i: i+x       # it will work now
     j = add_one(100)

You'll be happy to know that starting from Python 2.2 the above is the 
default behavior---i.e., had you used Python 2.2b2, your code would have 
ran as expected.


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