Python: split a list based on a condition?

Question

What's the best way, both aesthetically and from a performance perspective, to split a list of items into multiple lists based on a conditional? The equivalent of:

good = [x for x in mylist if x in goodvals]
bad  = [x for x in mylist if x not in goodvals]

is there a more elegant way to do this?

Update: here's the actual use case, to better explain what I'm trying to do:

# files looks like: [ ('file1.jpg', 33L, '.jpg'), ('file2.avi', 999L, '.avi'), ... ]
IMAGE_TYPES = ('.jpg','.jpeg','.gif','.bmp','.png')
images = [f for f in files if f[2].lower() in IMAGE_TYPES]
anims  = [f for f in files if f[2].lower() not in IMAGE_TYPES]

Answer 1

good = [x for x in mylist if x in goodvals]
bad  = [x for x in mylist if x not in goodvals]

is there a more elegant way to do this?

That code is perfectly readable, and extremely clear!

# files looks like: [ ('file1.jpg', 33L, '.jpg'), ('file2.avi', 999L, '.avi'), ... ]
IMAGE_TYPES = ('.jpg','.jpeg','.gif','.bmp','.png')
images = [f for f in files if f[2].lower() in IMAGE_TYPES]
anims  = [f for f in files if f[2].lower() not in IMAGE_TYPES]

Again, this is fine!

There might be slight performance improvements using sets, but it's a trivial difference, and I find the list comprehension far easier to read, and you don't have to worry about the order being messed up, duplicates being removed as so on.

In fact, I may go another step "backward", and just use a simple for loop:

images, anims = [], []

for f in files:
    if f.lower() in IMAGE_TYPES:
        images.append(f)
    else:
        anims.append(f)

The a list-comprehension or using set() is fine until you need to add some other check or another bit of logic - say you want to remove all 0-byte jpeg's, you just add something like..

if f[1] == 0:
    continue

Answer 2

good, bad = [], []
for x in mylist:
    (bad, good)[x in goodvals].append(x)

Answer 3

Here's the lazy iterator approach:

from itertools import tee

def split_on_condition(seq, condition):
    l1, l2 = tee((condition(item), item) for item in seq)
    return (i for p, i in l1 if p), (i for p, i in l2 if not p)

It evaluates the condition once per item and returns two generators, first yielding values from the sequence where the condition is true, the other where it's false.

Because it's lazy you can use it on any iterator, even an infinite one:

from itertools import count, islice

def is_prime(n):
    return n > 1 and all(n % i for i in xrange(2, n))

primes, not_primes = split_on_condition(count(), is_prime)
print("First 10 primes", list(islice(primes, 10)))
print("First 10 non-primes", list(islice(not_primes, 10)))

Usually though the non-lazy list returning approach is better:

def split_on_condition(seq, condition):
    a, b = [], []
    for item in seq:
        (a if condition(item) else b).append(item)
    return a, b

Edit: For your more specific usecase of splitting items into different lists by some key, heres a generic function that does that:

DROP_VALUE = lambda _:_
def split_by_key(seq, resultmapping, keyfunc, default=DROP_VALUE):
    """Split a sequence into lists based on a key function.

        seq - input sequence
        resultmapping - a dictionary that maps from target lists to keys that go to that list
        keyfunc - function to calculate the key of an input value
        default - the target where items that don't have a corresponding key go, by default they are dropped
    """
    result_lists = dict((key, []) for key in resultmapping)
    appenders = dict((key, result_lists[target].append) for target, keys in resultmapping.items() for key in keys)

    if default is not DROP_VALUE:
        result_lists.setdefault(default, [])
        default_action = result_lists[default].append
    else:
        default_action = DROP_VALUE

    for item in seq:
        appenders.get(keyfunc(item), default_action)(item)

    return result_lists

Usage:

def file_extension(f):
    return f[2].lower()

split_files = split_by_key(files, {'images': IMAGE_TYPES}, keyfunc=file_extension, default='anims')
print split_files['images']
print split_files['anims']

Answer 4

Problem with all proposed solutions is that it will scan and apply the filtering function twice. I'd make a simple small function like this:

def split_into_two_lists(lst, f):
    a = []
    b = []
    for elem in lst:
        if f(elem):
            a.append(elem)
        else:
            b.append(elem)
    return a, b

That way you are not processing anything twice and also are not repeating code.

Answer 5

My take on it. I propose a lazy, single-pass, partition function, which preserves relative order in the output subsequences.

1. Requirements

I assume that the requirements are:

• maintain elements' relative order (hence, no sets and dictionaries)
• evaluate condition only once for every element (hence not using (i)filter or groupby)
• allow for lazy consumption of either sequence (if we can afford to precompute them, then the naïve implementation is likely to be acceptable too)

2. split library

My partition function (introduced below) and other similar functions have made it into a small library:

• python-split

It's installable normally via PyPI:

pip install --user split

To split a list base on condition, use partition function:

>>> from split import partition
>>> files = [ ('file1.jpg', 33L, '.jpg'), ('file2.avi', 999L, '.avi') ]
>>> image_types = ('.jpg','.jpeg','.gif','.bmp','.png')
>>> images, other = partition(lambda f: f[-1] in image_types, files)
>>> list(images)
[('file1.jpg', 33L, '.jpg')]
>>> list(other)
[('file2.avi', 999L, '.avi')]

3. partition function explained

Internally we need to build two subsequences at once, so consuming only one output sequence will force the other one to be computed too. And we need to keep state between user requests (store processed but not yet requested elements). To keep state, I use two double-ended queues (deques):

from collections import deque

SplitSeq class takes care of the housekeeping:

class SplitSeq:
    def __init__(self, condition, sequence):
        self.cond = condition
        self.goods = deque([])
        self.bads = deque([])
        self.seq = iter(sequence)

Magic happens in its .getNext() method. It is almost like .next() of the iterators, but allows to specify which kind of element we want this time. Behind the scene it doesn't discard the rejected elements, but instead puts them in one of the two queues:

    def getNext(self, getGood=True):
        if getGood:
            these, those, cond = self.goods, self.bads, self.cond
        else:
            these, those, cond = self.bads, self.goods, lambda x: not self.cond(x)
        if these:
            return these.popleft()
        else:
            while 1: # exit on StopIteration
                n = self.seq.next()
                if cond(n):
                    return n
                else:
                    those.append(n)

The end user is supposed to use partition function. It takes a condition function and a sequence (just like map or filter), and returns two generators. The first generator builds a subsequence of elements for which the condition holds, the second one builds the complementary subsequence. Iterators and generators allow for lazy splitting of even long or infinite sequences.

def partition(condition, sequence):
    cond = condition if condition else bool  # evaluate as bool if condition == None
    ss = SplitSeq(cond, sequence)
    def goods():
        while 1:
            yield ss.getNext(getGood=True)
    def bads():
        while 1:
            yield ss.getNext(getGood=False)
    return goods(), bads()

I chose the test function to be the first argument to facilitate partial application in the future (similar to how map and filter have the test function as the first argument).

Answer 6

I basically like Anders' approach as it is very general. Here's a version that puts the categorizer first (to match filter syntax) and uses a defaultdict (assumed imported).

def categorize(func, seq):
    """Return mapping from categories to lists
    of categorized items.
    """
    d = defaultdict(list)
    for item in seq:
        d[func(item)].append(item)
    return d

Answer 7

First go (pre-OP-edit): Use sets:

mylist = [1,2,3,4,5,6,7]
goodvals = [1,3,7,8,9]

myset = set(mylist)
goodset = set(goodvals)

print list(myset.intersection(goodset))  # [1, 3, 7]
print list(myset.difference(goodset))    # [2, 4, 5, 6]

That's good for both readability (IMHO) and performance.

Second go (post-OP-edit):

Create your list of good extensions as a set:

IMAGE_TYPES = set(['.jpg','.jpeg','.gif','.bmp','.png'])

and that will increase performance. Otherwise, what you have looks fine to me.

Answer 8

itertools.groupby almost does what you want, except it requires the items to be sorted to ensure that you get a single contiguous range, so you need to sort by your key first (otherwise you'll get multiple interleaved groups for each type). eg.

def is_good(f):
    return f[2].lower() in IMAGE_TYPES

files = [ ('file1.jpg', 33L, '.jpg'), ('file2.avi', 999L, '.avi'), ('file3.gif', 123L, '.gif')]

for key, group in itertools.groupby(sorted(files, key=is_good), key=is_good):
    print key, list(group)

gives:

False [('file2.avi', 999L, '.avi')]
True [('file1.jpg', 33L, '.jpg'), ('file3.gif', 123L, '.gif')]

Similar to the other solutions, the key func can be defined to divide into any number of groups you want.

Answer 9

good.append(x) if x in goodvals else bad.append(x)

This elegant and concise answer by @dansalmo showed up buried in the comments, so I'm just reposting it here as an answer so it can get the prominence it deserves, especially for new readers.

Complete example:

good, bad = [], []
for x in my_list:
    good.append(x) if x in goodvals else bad.append(x)

Answer 10

If you want to make it in FP style:

good, bad = [ sum(x, []) for x in zip(*(([y], []) if y in goodvals else ([], [y])
                                        for y in mylist)) ]

Not the most readable solution, but at least iterates through mylist only once.

Answer 11

Personally, I like the version you cited, assuming you already have a list of goodvals hanging around. If not, something like:

good = filter(lambda x: is_good(x), mylist)
bad = filter(lambda x: not is_good(x), mylist)

Of course, that's really very similar to using a list comprehension like you originally did, but with a function instead of a lookup:

good = [x for x in mylist if is_good(x)]
bad  = [x for x in mylist if not is_good(x)]

In general, I find the aesthetics of list comprehensions to be very pleasing. Of course, if you don't actually need to preserve ordering and don't need duplicates, using the intersection and difference methods on sets would work well too.

Answer 12

Sometimes, it looks like list comprehension is not the best thing to use !

I made a little test based on the answer people gave to this topic, tested on a random generated list. Here is the generation of the list (there's probably a better way to do, but it's not the point) :

good_list = ('.jpg','.jpeg','.gif','.bmp','.png')

import random
import string
my_origin_list = []
for i in xrange(10000):
    fname = ''.join(random.choice(string.lowercase) for i in range(random.randrange(10)))
    if random.getrandbits(1):
        fext = random.choice(good_list)
    else:
        fext = "." + ''.join(random.choice(string.lowercase) for i in range(3))

    my_origin_list.append((fname + fext, random.randrange(1000), fext))

And here we go

# Parand
def f1():
    return [e for e in my_origin_list if e[2] in good_list], [e for e in my_origin_list if not e[2] in good_list]

# dbr
def f2():
    a, b = list(), list()
    for e in my_origin_list:
        if e[2] in good_list:
            a.append(e)
        else:
            b.append(e)
    return a, b

# John La Rooy
def f3():
    a, b = list(), list()
    for e in my_origin_list:
        (b, a)[e[2] in good_list].append(e)
    return a, b

# Ants Aasma
def f4():
    l1, l2 = tee((e[2] in good_list, e) for e in my_origin_list)
    return [i for p, i in l1 if p], [i for p, i in l2 if not p]

# My personal way to do
def f5():
    a, b = zip(*[(e, None) if e[2] in good_list else (None, e) for e in my_origin_list])
    return list(filter(None, a)), list(filter(None, b))

# BJ Homer
def f6():
    return filter(lambda e: e[2] in good_list, my_origin_list), filter(lambda e: not e[2] in good_list, my_origin_list)

Using the cmpthese function, the best result is the dbr answer :

f1     204/s  --    -5%   -14%   -15%   -20%   -26%
f6     215/s     6%  --    -9%   -11%   -16%   -22%
f3     237/s    16%    10%  --    -2%    -7%   -14%
f4     240/s    18%    12%     2%  --    -6%   -13%
f5     255/s    25%    18%     8%     6%  --    -8%
f2     277/s    36%    29%    17%    15%     9%  --

Answer 13

Elegant and Fast

Inspired by DanSalmo's comment, here is a solution that is concise, elegant, and at the same time is one of the fastest solutions.

good, bad = [], []
for item in my_list:
    good.append(item) if item in set(goodvals) else bad.append(item)

Tip: Turning goodvals into a set gives us an easy speed boost.

Fastest

For maximum speed, we take the fastest answer and turbocharge it by turning good_list into a set. That alone gives us a 40%+ speed boost, and we end up with a solution that is more than 5.5x as fast as the slowest solution, even while it remains readable.

good_list_set = set(good_list)  # 40%+ faster than a tuple.

good, bad = [], []
for item in my_origin_list:
    if item in good_list_set:
        good.append(item)
    else:
        bad.append(item)

A little shorter

This is a more concise version of the previous answer.

good_list_set = set(good_list)  # 40%+ faster than a tuple.

good, bad = [], []
for item in my_origin_list:
    out = good if item in good_list_set else bad
    out.append(item)

Elegance can be somewhat subjective, but some of the Rube Goldberg style solutions that are cute and ingenious are quite concerning and should not be used in production code in any language, let alone python which is elegant at heart.

Benchmark results:

filter_BJHomer                  80/s       --   -3265%   -5312%   -5900%   -6262%   -7273%   -7363%   -8051%   -8162%   -8244%
zip_Funky                       118/s    4848%       --   -3040%   -3913%   -4450%   -5951%   -6085%   -7106%   -7271%   -7393%
two_lst_tuple_JohnLaRoy         170/s   11332%    4367%       --   -1254%   -2026%   -4182%   -4375%   -5842%   -6079%   -6254%
if_else_DBR                     195/s   14392%    6428%    1434%       --    -882%   -3348%   -3568%   -5246%   -5516%   -5717%
two_lst_compr_Parand            213/s   16750%    8016%    2540%     967%       --   -2705%   -2946%   -4786%   -5083%   -5303%
if_else_1_line_DanSalmo         292/s   26668%   14696%    7189%    5033%    3707%       --    -331%   -2853%   -3260%   -3562%
tuple_if_else                   302/s   27923%   15542%    7778%    5548%    4177%     343%       --   -2609%   -3029%   -3341%
set_1_line                      409/s   41308%   24556%   14053%   11035%    9181%    3993%    3529%       --    -569%    -991%
set_shorter                     434/s   44401%   26640%   15503%   12303%   10337%    4836%    4345%     603%       --    -448%
set_if_else                     454/s   46952%   28358%   16699%   13349%   11290%    5532%    5018%    1100%     469%       --

The full benchmark code for Python 3.7 (modified from FunkySayu):

good_list = ['.jpg','.jpeg','.gif','.bmp','.png']

import random
import string
my_origin_list = []
for i in range(10000):
    fname = ''.join(random.choice(string.ascii_lowercase) for i in range(random.randrange(10)))
    if random.getrandbits(1):
        fext = random.choice(list(good_list))
    else:
        fext = "." + ''.join(random.choice(string.ascii_lowercase) for i in range(3))

    my_origin_list.append((fname + fext, random.randrange(1000), fext))

# Parand
def two_lst_compr_Parand(*_):
    return [e for e in my_origin_list if e[2] in good_list], [e for e in my_origin_list if not e[2] in good_list]

# dbr
def if_else_DBR(*_):
    a, b = list(), list()
    for e in my_origin_list:
        if e[2] in good_list:
            a.append(e)
        else:
            b.append(e)
    return a, b

# John La Rooy
def two_lst_tuple_JohnLaRoy(*_):
    a, b = list(), list()
    for e in my_origin_list:
        (b, a)[e[2] in good_list].append(e)
    return a, b

# # Ants Aasma
# def f4():
#     l1, l2 = tee((e[2] in good_list, e) for e in my_origin_list)
#     return [i for p, i in l1 if p], [i for p, i in l2 if not p]

# My personal way to do
def zip_Funky(*_):
    a, b = zip(*[(e, None) if e[2] in good_list else (None, e) for e in my_origin_list])
    return list(filter(None, a)), list(filter(None, b))

# BJ Homer
def filter_BJHomer(*_):
    return list(filter(lambda e: e[2] in good_list, my_origin_list)), list(filter(lambda e: not e[2] in good_list,                                                                             my_origin_list))

# ChaimG's answer; as a list.
def if_else_1_line_DanSalmo(*_):
    good, bad = [], []
    for e in my_origin_list:
        _ = good.append(e) if e[2] in good_list else bad.append(e)
    return good, bad

# ChaimG's answer; as a set.
def set_1_line(*_):
    good_list_set = set(good_list)
    good, bad = [], []
    for e in my_origin_list:
        _ = good.append(e) if e[2] in good_list_set else bad.append(e)
    return good, bad

# ChaimG set and if else list.
def set_shorter(*_):
    good_list_set = set(good_list)
    good, bad = [], []
    for e in my_origin_list:
        out = good if e[2] in good_list_set else bad
        out.append(e)
    return good, bad

# ChaimG's best answer; if else as a set.
def set_if_else(*_):
    good_list_set = set(good_list)
    good, bad = [], []
    for e in my_origin_list:
        if e[2] in good_list_set:
            good.append(e)
        else:
            bad.append(e)
    return good, bad

# ChaimG's best answer; if else as a set.
def tuple_if_else(*_):
    good_list_tuple = tuple(good_list)
    good, bad = [], []
    for e in my_origin_list:
        if e[2] in good_list_tuple:
            good.append(e)
        else:
            bad.append(e)
    return good, bad

def cmpthese(n=0, functions=None):
    results = {}
    for func_name in functions:
        args = ['%s(range(256))' % func_name, 'from __main__ import %s' % func_name]
        t = Timer(*args)
        results[func_name] = 1 / (t.timeit(number=n) / n) # passes/sec

    functions_sorted = sorted(functions, key=results.__getitem__)
    for f in functions_sorted:
        diff = []
        for func in functions_sorted:
            if func == f:
                diff.append("--")
            else:
                diff.append(f"{results[f]/results[func]*100 - 100:5.0%}")
        diffs = " ".join(f'{x:>8s}' for x in diff)

        print(f"{f:27s} \t{results[f]:,.0f}/s {diffs}")


if __name__=='__main__':
    from timeit import Timer
cmpthese(1000, 'two_lst_compr_Parand if_else_DBR two_lst_tuple_JohnLaRoy zip_Funky filter_BJHomer if_else_1_line_DanSalmo set_1_line set_if_else tuple_if_else set_shorter'.split(" "))

Answer 14

def partition(pred, iterable):
    '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)

Check this

Answer 15

I think a generalization of splitting a an iterable based on N conditions is handy

from collections import OrderedDict
def partition(iterable,*conditions):
    '''Returns a list with the elements that satisfy each of condition.
       Conditions are assumed to be exclusive'''
    d= OrderedDict((i,list())for i in range(len(conditions)))        
    for e in iterable:
        for i,condition in enumerate(conditions):
            if condition(e):
                d[i].append(e)
                break                    
    return d.values()

For instance:

ints,floats,other = partition([2, 3.14, 1, 1.69, [], None],
                              lambda x: isinstance(x, int), 
                              lambda x: isinstance(x, float),
                              lambda x: True)

print " ints: {}\n floats:{}\n other:{}".format(ints,floats,other)

 ints: [2, 1]
 floats:[3.14, 1.69]
 other:[[], None]

If the element may satisfy multiple conditions, remove the break.

Answer 16

bad = []
good = [x for x in mylist if x in goodvals or bad.append(x)]

append returns None, so it works.

Answer 17

Yet another solution to this problem. I needed a solution that is as fast as possible. That means only one iteration over the list and preferably O(1) for adding data to one of the resulting lists. This is very similar to the solution provided by sastanin, except much shorter:

from collections import deque

def split(iterable, function):
    dq_true = deque()
    dq_false = deque()

    # deque - the fastest way to consume an iterator and append items
    deque((
      (dq_true if function(item) else dq_false).append(item) for item in iterable
    ), maxlen=0)

    return dq_true, dq_false

Then, you can use the function in the following way:

lower, higher = split([0,1,2,3,4,5,6,7,8,9], lambda x: x < 5)

selected, other = split([0,1,2,3,4,5,6,7,8,9], lambda x: x in {0,4,9})

If you're not fine with the resulting deque object, you can easily convert it to list, set, whatever you like (for example list(lower)). The conversion is much faster, that construction of the lists directly.

This methods keeps order of the items, as well as any duplicates.

Answer 18

For example, splitting list by even and odd

arr = range(20)
even, odd = reduce(lambda res, next: res[next % 2].append(next) or res, arr, ([], []))

Or in general:

def split(predicate, iterable):
    return reduce(lambda res, e: res[predicate(e)].append(e) or res, iterable, ([], []))

Advantages:

• Shortest posible way
• Predicate applies only once for each element

Disadvantages

• Requires knowledge of functional programing paradigm

Answer 19

Inspired by @gnibbler's great (but terse!) answer, we can apply that approach to map to multiple partitions:

from collections import defaultdict

def splitter(l, mapper):
    """Split an iterable into multiple partitions generated by a callable mapper."""

    results = defaultdict(list)

    for x in l:
        results[mapper(x)] += [x]

    return results

Then splitter can then be used as follows:

>>> l = [1, 2, 3, 4, 2, 3, 4, 5, 6, 4, 3, 2, 3]
>>> split = splitter(l, lambda x: x % 2 == 0)  # partition l into odds and evens
>>> split.items()
>>> [(False, [1, 3, 3, 5, 3, 3]), (True, [2, 4, 2, 4, 6, 4, 2])]

This works for more than two partitions with a more complicated mapping (and on iterators, too):

>>> import math
>>> l = xrange(1, 23)
>>> split = splitter(l, lambda x: int(math.log10(x) * 5))
>>> split.items()
[(0, [1]),
 (1, [2]),
 (2, [3]),
 (3, [4, 5, 6]),
 (4, [7, 8, 9]),
 (5, [10, 11, 12, 13, 14, 15]),
 (6, [16, 17, 18, 19, 20, 21, 22])]

Or using a dictionary to map:

>>> map = {'A': 1, 'X': 2, 'B': 3, 'Y': 1, 'C': 2, 'Z': 3}
>>> l = ['A', 'B', 'C', 'C', 'X', 'Y', 'Z', 'A', 'Z']
>>> split = splitter(l, map.get)
>>> split.items()
(1, ['A', 'Y', 'A']), (2, ['C', 'C', 'X']), (3, ['B', 'Z', 'Z'])]

Answer 20

solution

from itertools import tee

def unpack_args(fn):
    return lambda t: fn(*t)

def separate(fn, lx):
    return map(
        unpack_args(
            lambda i, ly: filter(
                lambda el: bool(i) == fn(el),
                ly)),
        enumerate(tee(lx, 2)))

test

[even, odd] = separate(
    lambda x: bool(x % 2),
    [1, 2, 3, 4, 5])
print(list(even) == [2, 4])
print(list(odd) == [1, 3, 5])

Answer 21

If you don't mind using an external library there two I know that nativly implement this operation:

>>> files = [ ('file1.jpg', 33, '.jpg'), ('file2.avi', 999, '.avi')]
>>> IMAGE_TYPES = ('.jpg','.jpeg','.gif','.bmp','.png')

• iteration_utilities.partition:

  >>> from iteration_utilities import partition
  >>> notimages, images = partition(files, lambda x: x[2].lower() in IMAGE_TYPES)
  >>> notimages
  [('file2.avi', 999, '.avi')]
  >>> images
  [('file1.jpg', 33, '.jpg')]
  

• more_itertools.partition

  >>> from more_itertools import partition
  >>> notimages, images = partition(lambda x: x[2].lower() in IMAGE_TYPES, files)
  >>> list(notimages)  # returns a generator so you need to explicitly convert to list.
  [('file2.avi', 999, '.avi')]
  >>> list(images)
  [('file1.jpg', 33, '.jpg')]
  

Answer 22

Use Boolean logic to assign data to two arrays

>>> images, anims = [[i for i in files if t ^ (i[2].lower() in IMAGE_TYPES) ] for t in (0, 1)]
>>> images
[('file1.jpg', 33, '.jpg')]
>>> anims
[('file2.avi', 999, '.avi')]

Answer 23

For perfomance, try itertools.

The itertools module standardizes a core set of fast, memory efficient tools that are useful by themselves or in combination. Together, they form an “iterator algebra” making it possible to construct specialized tools succinctly and efficiently in pure Python.

See itertools.ifilter or imap.

itertools.ifilter(predicate, iterable)

Make an iterator that filters elements from iterable returning only those for which the predicate is True

Answer 24

If you insist on clever, you could take Winden's solution and just a bit spurious cleverness:

def splay(l, f, d=None):
  d = d or {}
  for x in l: d.setdefault(f(x), []).append(x)
  return d

Answer 25

Sometimes you won't need that other half of the list. For example:

import sys
from itertools import ifilter

trustedPeople = sys.argv[1].split(',')
newName = sys.argv[2]

myFriends = ifilter(lambda x: x.startswith('Shi'), trustedPeople)

print '%s is %smy friend.' % (newName, newName not in myFriends 'not ' or '')

Answer 26

Already quite a few solutions here, but yet another way of doing that would be -

anims = []
images = [f for f in files if (lambda t: True if f[2].lower() in IMAGE_TYPES else anims.append(t) and False)(f)]

Iterates over the list only once, and looks a bit more pythonic and hence readable to me.

>>> files = [ ('file1.jpg', 33L, '.jpg'), ('file2.avi', 999L, '.avi'), ('file1.bmp', 33L, '.bmp')]
>>> IMAGE_TYPES = ('.jpg','.jpeg','.gif','.bmp','.png')
>>> anims = []
>>> images = [f for f in files if (lambda t: True if f[2].lower() in IMAGE_TYPES else anims.append(t) and False)(f)]
>>> print '\n'.join([str(anims), str(images)])
[('file2.avi', 999L, '.avi')]
[('file1.jpg', 33L, '.jpg'), ('file1.bmp', 33L, '.bmp')]
>>>

Answer 27

I'd take a 2-pass approach, separating evaluation of the predicate from filtering the list:

def partition(pred, iterable):
    xs = list(zip(map(pred, iterable), iterable))
    return [x[1] for x in xs if x[0]], [x[1] for x in xs if not x[0]]

What's nice about this, performance-wise (in addition to evaluating pred only once on each member of iterable), is that it moves a lot of logic out of the interpreter and into highly-optimized iteration and mapping code. This can speed up iteration over long iterables, as described in this answer.

Expressivity-wise, it takes advantage of expressive idioms like comprehensions and mapping.

Answer 28

Not sure if this is a good approach but it can be done in this way as well

IMAGE_TYPES = ('.jpg','.jpeg','.gif','.bmp','.png')
files = [ ('file1.jpg', 33L, '.jpg'), ('file2.avi', 999L, '.avi')]
images, anims = reduce(lambda (i, a), f: (i + [f], a) if f[2] in IMAGE_TYPES else (i, a + [f]), files, ([], []))

Answer 29

If the list is made of groups and intermittent separators, you can use:

def split(items, p):
    groups = [[]]
    for i in items:
        if p(i):
            groups.append([])
        groups[-1].append(i)
    return groups

Usage:

split(range(1,11), lambda x: x % 3 == 0)
# gives [[1, 2], [3, 4, 5], [6, 7, 8], [9, 10]]

Answer 30

Yet another answer, short but "evil" (for list-comprehension side effects).

digits = list(range(10))
odd = [x.pop(i) for i, x in enumerate(digits) if x % 2]

>>> odd
[1, 3, 5, 7, 9]

>>> digits
[0, 2, 4, 6, 8]