list partition

[Moosebumps]

Is there a function that takes a list and predicate, and returns two
lists -- one for which the predicate is true and one for which it is false?

I know I can call filter(p, list) and filter( not p, list) or something like
that, but I assume it would be more efficient to do it in one pass.

Actually I probably can't call "not p", I would have to wrap p in a function
notP or something. Or is there a shortcut for lambda(x) : not p(x)? I
guess it isn't that long : )

MB

[Peter Otten]

Moosebumps wrote:

Is there a function that takes a list and predicate, and returns two
lists -- one for which the predicate is true and one for which it is
false?

I take a slightly different approach, which is not limited to boolean
predicates:

def predicate(v): return v & 1 .... d = {}
for i in range(10): .... d.setdefault(pr edicate(i), []).append(i)
.... d[True] [1, 3, 5, 7, 9] d[False] [0, 2, 4, 6, 8]

Peter

[David M. Cooke]

At some point, "Moosebumps " <Mo********@Moo sebumps.Moosebu mps> wrote:

Is there a function that takes a list and predicate, and returns two
lists -- one for which the predicate is true and one for which it is false?

I know I can call filter(p, list) and filter( not p, list) or something like
that, but I assume it would be more efficient to do it in one pass.
Not too hard to write your own:

def filtersplit(p, iterable):
ptrue = []
pfalse = []
for x in iterable:
if p(x):
ptrue.append(x)
else:
pfalse.append(x )
return ptrue, pfalse
Actually I probably can't call "not p", I would have to wrap p in a function
notP or something. Or is there a shortcut for lambda(x) : not p(x)? I
guess it isn't that long : )

itertools.ifilt erfalse negates the condition for you.

--
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|David M. Cooke
|cookedm(at)phy sics(dot)mcmast er(dot)ca