filter with dynamic value
aleax at aleax.it
Fri Sep 14 15:28:26 CEST 2001
"Achim Domma" <achim.domma at syynx.de> wrote in message
news:9nsuqg$tmk$00$1 at news.t-online.com...
> I have something like this:
> ids = [(1,'abc'),(2,'abc'),(3,'xyz'),(4,'xyz')]
> comp = 'xyz'
> now I want to use filter to remove all items from ids which don't have
> as second value. I think the following should work:
> filter(lambda x: x=='xyz')
No, because filter takes two arguments and you're calling
it with just one. I suspect you mean something like:
xyzids = filter(lambda x: x=='xyz', ids)
> but how can I do this with an dynamic value ? If I try the following I get
> problems with the scope of comp :
> filter(lamdba x: x==comp)
Use Python 2.2 (when it comes out -- it's now in alpha 3),
or, in Python 2.1, have as the first statement of your
from __future__ import nested_scope
Or use list comprehensions instead, Python 2.0 and later:
xyzids = [x for x in ids if x==comp]
although this may be slightly slower than the filter (if
this is a bottleneck for your performance you'll need to
check this carefully).
Or use the good old default value trick, in any Python:
xyzids = filter(lambda x, comp=comp: x==comp, ids)
the default-value comp is evaluated once, when the lambda
is being executed to create the anonymous function, and in
the scope of the caller -- it's then bound to a name comp
that is an argument of the lambda and thus local to it.
There are more than these three solutions, of course, but
thse three are the simplest ones!
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