lambda requirements

[David Goodger]

Lambda expressions don't have access to the local namespace. Declaring
_list2 as global gets around this problem, but in an "unclean" way (note
that _list1 doesn't need to be global, as it's not accessed inside the
lambda). A cleaner way to implement the lambda (without globals, or
"_"-prefixed local variables) is as follows:

    _onlylist1 = filter(lambda x, l=list2: x not in l, list1)

"l" is bound to "list2" when the lambda expression is evaluated (when the
local namespace *is* accessible), which carries through to when the
resulting anonymous function is executed by filter (when
SymmetricDifference's local namespace is *not* accessible).

-- 
David Goodger    dgoodger at bigfoot.com    Open-source projects:
 - The Go Tools Project: http://gotools.sourceforge.net
 (more to come!)


on 2000-07-03 08:41, Wolfgang Thaemelt (wolfgang.thaemelt at orsoft.de) wrote:
> Running the script:
> 
> #----------------------------------------- begin of script
> def SymmetricDifference(list1, list2):
> # global _list1
> # global _list2
> 
>   _list1 = list1
>   _list2 = list2
> 
>   _onlylist1 = filter(lambda x: x not in _list2, _list1)
> 
>   return _onlylist1
> 
> if __name__ == '__main__':
>   a = [1,2,3,4,5]
>   b = [3,4,5,6,7]
>   x = SymmetricDifference(a,b)
> 
>   print x
> #---------------------------------------- end of script
> 
> gives the error:
> 
> Traceback (innermost last):
> File "E:\up\000701\symmdiff.py", line 16, in ?
> x = SymmetricDifference(a,b)
> File "E:\up\000701\symmdiff.py", line 9, in SymmetricDifference
> _onlylist1 = filter(lambda x: x not in _list2, _list1)
> File "E:\up\000701\symmdiff.py", line 9, in <lambda>
> _onlylist1 = filter(lambda x: x not in _list2, _list1)
> NameError: _list2
> 
> If the comment characters "#"  are removed from the "global" statements
> everything works fine.
> 
> Is there some requirement on a lambda violated in the script?