python equivalent of filter() getting two output lists (i.e. partition of a list)

If you don’t have duplicate element in your list you can definitely use set:

>>> a = [1,4,12,7,42]
>>> b = filter(lambda x: x > 10, [1,4,12,7,42])
>>> no_b = set(a) - set(b)
set([1, 4, 7])

or you can do by a list comprehensible:

>>> no_b = [i for i in a if i not in b]

N.B: it’s not a function but just knowing the first fitler() result you can deduce the element that didn’t much your filter criterion .

from itertools import ifilterfalse

def filter2(predicate, iterable):
    return filter(predicate, iterable), list(ifilterfalse(predicate, iterable))

>>> def partition(l, p):
...     return reduce(lambda x, y: (x[0]+[y], x[1]) if p(y) else (x[0], x[1]+[y]), l,  ([], []))
... 
>>> partition([1, 2, 3, 4, 5], lambda x: x < 3)
([1, 2], [3, 4, 5])

and a little uglier but faster version of the above code:

def partition(l, p):
    return reduce(lambda x, y: x[0].append(y) or x if p(y) else x[1].append(y) or x, l,  ([], []))

This is second edit, but I think it matters:

 def partition(l, p):
     return reduce(lambda x, y: x[not p(y)].append(y) or x, l, ([], []))

The second and the third are as quick as the iterative one upper but are less code.

I think groupby might be more relevant here:

http://docs.python.org/library/itertools.html#itertools.groupby

For example, splitting a list into odd and even numbers (or could be an arbitrary number of groups):

>>> l=range(6)
>>> key=lambda x: x % 2 == 0
>>> from itertools import groupby
>>> {k:list(g) for k,g in groupby(sorted(l,key=key),key=key)}
    {False: [1, 3, 5], True: [0, 2, 4]}

Plenty of good answers already. I like to use this:

def partition( pred, iterable ):
    def _dispatch( ret, v ):
        if ( pred( v ) ):
            ret[0].append( v )
        else:
            ret[1].append( v )
        return ret
    return reduce( _dispatch, iterable, ( [], [] ) )

if ( __name__ == '__main__' ):
    import random
    seq = range( 20 )
    random.shuffle( seq )
    print( seq )
    print( partition( lambda v : v > 10, seq ) )

I just had exactly this requirement. I’m not keen on the itertools recipe since it involves two separate passes through the data. Here’s my implementation:

def filter_twoway(test, data):
    "Like filter(), but returns the passes AND the fails as two separate lists"
    collected = {True: [], False: []}
    for datum in data:
        collected[test(datum)].append(datum)
    return (collected[True], collected[False])

Everyone seems to think that their solution is the best, so I decided to use timeit to test all of them. I used “def is_odd(x): return x & 1” as my predicate function, and “xrange(1000)” as the iterable. Here is my version of Python:

Python 2.7.3 (v2.7.3:70274d53c1dd, Apr  9 2012, 20:52:43) 
[GCC 4.2.1 (Apple Inc. build 5666) (dot 3)] on darwin

And here are the results of my testing:

Mark Byers
1000 loops, best of 3: 325 usec per loop

cldy
1000 loops, best of 3: 1.96 msec per loop

Dan S
1000 loops, best of 3: 412 usec per loop

TTimo
1000 loops, best of 3: 503 usec per loop

Those are all comparable to each other. Now, let’s try using the example given in the Python documentation.

import itertools

def partition(pred, iterable,
              # Optimized by replacing global lookups with local variables
              # defined as default values.
              filter=itertools.ifilter,
              filterfalse=itertools.ifilterfalse,
              tee=itertools.tee):
    'Use a predicate to partition entries into false entries and true entries'
    # partition(is_odd, range(10)) --> 0 2 4 6 8   and  1 3 5 7 9
    t1, t2 = tee(iterable)
    return filterfalse(pred, t1), filter(pred, t2)

This seems to be a bit faster.

100000 loops, best of 3: 2.58 usec per loop

The itertools example code beats all comers by a factor of at least 100! The moral is, don’t keep re-inventing the wheel.

You can look at django.utils.functional.partition solution:

def partition(predicate, values):
    """
    Splits the values into two sets, based on the return value of the function
    (True/False). e.g.:

        >>> partition(lambda x: x > 3, range(5))
        [0, 1, 2, 3], [4]
    """
    results = ([], [])
    for item in values:
        results[predicate(item)].append(item)
    return results

In my opinion it’s the most elegant solution presented here.

This part is not documented, only source code can be found on https://docs.djangoproject.com/en/dev/_modules/django/utils/functional/

TL;DR

The accepted, most voted answer [1] by Mark Byers

def partition(pred, iterable):
    trues = []
    falses = []
    for item in iterable:
        if pred(item):
            trues.append(item)
        else:
            falses.append(item)
    return trues, falses

is the simplest and the
fastest.

Benchmarking the different approaches

The different approaches that had been suggested can be classified
broadly in three categories,

1. straightforward list manipulation via lis.append, returning a 2-tuple
   of lists,
2. lis.append mediated by a functional approach, returning a 2-tuple
   of lists,
3. using the canonical recipe given in the itertools fine
   documentation, returning a 2-tuple of, loosely speaking, generators.

Here follows a vanilla implementation of the three techniques, first
the functional approach, then itertools and eventually two different
implementations of direct list manipulation, the alternative being
using the False is zero, True is one trick.

Note that this is Python3 — hence reduce comes from functools —
and that OP request a tuple like (positives, negatives) but my
implementations all return (negatives, positives)…

$ ipython
Python 3.6.2 |Continuum Analytics, Inc.| (default, Jul 20 2017, 13:51:32) 
Type 'copyright', 'credits' or 'license' for more information
IPython 6.1.0 -- An enhanced Interactive Python. Type '?' for help.

In [1]: import functools
   ...: 
   ...: def partition_fu(p, l, r=functools.reduce):
   ...:     return r(lambda x, y: x[p(y)].append(y) or x, l, ([], []))
   ...: 

In [2]: import itertools
   ...: 
   ...: def partition_it(pred, iterable,
   ...:               filterfalse=itertools.filterfalse,
   ...:               tee=itertools.tee):
   ...:     t1, t2 = tee(iterable)
   ...:     return filterfalse(pred, t1), filter(pred, t2)
   ...: 

In [3]: def partition_li(p, l):
   ...:     a, b = [], []
   ...:     for n in l:
   ...:         if p(n):
   ...:             b.append(n)
   ...:         else:
   ...:             a.append(n)
   ...:     return a, b
   ...: 

In [4]: def partition_li_alt(p, l):
   ...:     x = [], []
   ...:     for n in l: x[p(n)].append(n)
   ...:     return x
   ...: 

We need a predicate to apply to our lists and lists (again, loosely
speaking) on which to operate.

In [5]: p = lambda n:n%2

In [6]: five, ten = range(50000), range(100000)

To overcome the problem in testing the itertools approach, that was
reported by joeln on
Oct 31 ’13 at 6:17

Nonsense. You’ve calculated the time taken to construct the
generators in filterfalse and filter, but you’ve not iterated
through the input or called pred once! The advantage of the
itertools recipe is that it does not materialise any list, or look
further ahead in the input than necessary. It calls pred twice as
often and takes almost twice as long as Byers et al.

I have thought of a void loop that just instantiates all the couples
of elements in the two iterables returned by the different partition
functions.

First we use two fixed lists to have an idea of the
overload implied (using the very convenient IPython’s magic %timeit)

In [7]: %timeit for e, o in zip(five, five): pass
4.21 ms ± 39.4 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Next we use the different implementations, one after the other

In [8]: %timeit for e, o in zip(*partition_fu(p, ten)): pass
53.9 ms ± 112 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [9]: %timeit for e, o in zip(*partition_it(p, ten)): pass
44.5 ms ± 3.84 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [10]: %timeit for e, o in zip(*partition_li(p, ten)): pass
36.3 ms ± 101 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [11]: %timeit for e, o in zip(*partition_li_alt(p, ten)): pass
37.3 ms ± 109 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [12]:

Comments

The plainest of the approaches is also the fastest one.

Using the x[p(n)] trick is, ehm, useless because at every step you
have to index a data structure, giving you a slight penalty — it’s
however nice to know if you want to persuade a survivor of a declining
culture at pythonizing.

The functional approach, that is operatively equivalent to the
alternative append implementation, is ~50% slower, possibly due to
the fact that we have an extra (w/r to predicate evaluation) function
call for each list element.

The itertools approach has the (customary) advantages that ❶ no
potentially large list is instantiated and ❷ the input list is not
entirely processed if you break out of the consumer loop, but when we
use it it is slower because of the need to apply the predicate on both
ends of the tee

Aside

I’ve fallen in love with the object.mutate() or object idiom that
was exposed by Marii
in their answer showing
a functional approach to the problem — I’m afraid that, sooner or later,
I’m going to abuse it.

Footnotes

[1] Accepted and most voted as today, Sep 14 2017 — but of course I have the highest hopes for this answer of mine!

Concise code for appending to target list

    def partition(cond,inputList):
        a,b= [],[]
        for item in inputList:
            target = a if cond(item) else b
            target.append(item)
        return a, b


    >>> a, b= partition(lambda x: x > 10,[1,4,12,7,42])
    >>> a
    [12, 42]
    >>> b
    [1, 4, 7]

The existing answers either partition an iterable into two lists, or inefficiently partition it into two generators. Here is an implementation that efficiently partitions an iterable into two generators, i.e. the predicate function is called at most once for each element in the iterable. One instance where you might want to use this version is if your need to partition a very large (or even infinite) iterable with an expensive to compute predicate.

from collections import deque

def partition(pred, iterable):
    seq = iter(iterable)
    true_buffer = deque()
    false_buffer = deque()
    def true_iter():
        while True:
            while true_buffer:
                yield true_buffer.popleft()
            item = next(seq)
            if pred(item):
                yield item
            else:
                false_buffer.append(item)
    def false_iter():
        while True:
            while false_buffer:
                yield false_buffer.popleft()
            item = next(seq)
            if not pred(item):
                yield item
            else:
                true_buffer.append(item)
    return true_iter(), false_iter()

Basically, this steps through each item in the the iterator, checks the predicate, and either yields it, if the corresponding generator is being used, or puts it in the buffer for the other iterable. Additionally, each generator will first pull items from its buffer before checking the original iterable. Note that each partition has it’s own buffer which grows each time the other partition is iterated, so this implementation may not be suitable for use cases where one partition is iterated much more than the other.

example use case:

from itertools import count
from random import random

odds, evens = partition(lambda n: n % 2, count())
for _ in range(500):
    if random() < 0.5:
        print(next(odds))
    else:
        print(next(evens))

The three top voted answers to an equivalent question propose to use itertools.tee() (as already covered here) and two even simpler approaches as wells.

The collections.defaultdict method is an excellent helper for sorting operations.

“

import collections
input_list = ['a','b','ana','beta','gamma']
filter_key = lambda x: len(x) == 1


## sorting code
cc = collections.defaultdict(list)
for item in input_list: cc[ filter_key(item) ].append( item )

print( cc )

This approach will also work for any number of categories generated by the `filter_key` function.