filter with dynamic value

Alex Martelli aleax at
Fri Sep 14 15:28:26 CEST 2001

"Achim Domma" <achim.domma at> wrote in message
news:9nsuqg$tmk$00$1 at
> Hi,
> 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[1]=='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[1]=='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[1]==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[1]==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[1]==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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