我有这样的代码:
good = [x for x in mylist if x in goodvals]
bad = [x for x in mylist if x not in goodvals]
目标是根据mylist的内容是否满足条件,将它们拆分为另外两个列表。
我怎样才能做得更优雅呢?我能避免在mylist上做两个单独的迭代吗?我可以通过这样做来提高性能吗?
我有这样的代码:
good = [x for x in mylist if x in goodvals]
bad = [x for x in mylist if x not in goodvals]
目标是根据mylist的内容是否满足条件,将它们拆分为另外两个列表。
我怎样才能做得更优雅呢?我能避免在mylist上做两个单独的迭代吗?我可以通过这样做来提高性能吗?
当前回答
就我个人而言,我喜欢你引用的版本,假设你已经有了一个好的列表。如果没有,就像这样:
good = filter(lambda x: is_good(x), mylist)
bad = filter(lambda x: not is_good(x), mylist)
当然,这真的非常类似于使用列表理解,就像你最初做的,但用一个函数而不是一个查找:
good = [x for x in mylist if is_good(x)]
bad = [x for x in mylist if not is_good(x)]
总的来说,我发现列表推导式的美学非常令人满意。当然,如果您实际上不需要保留顺序,也不需要重复,那么在集合上使用交集和差分方法也会很好。
其他回答
下面是惰性迭代器方法:
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)
它对每个项计算一次条件,并返回两个生成器,第一个生成条件为真时序列中的值,另一个生成条件为假时序列中的值。
因为它是惰性的,你可以在任何迭代器上使用它,甚至是无限迭代器:
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)))
通常情况下,非惰性列表返回方法会更好:
def split_on_condition(seq, condition):
a, b = [], []
for item in seq:
(a if condition(item) else b).append(item)
return a, b
编辑:对于您更具体的用例,将项目按某些键分割到不同的列表中,这里有一个通用函数:
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
用法:
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']
所有提出的解决方案的问题是,它将扫描和应用过滤功能两次。我会做一个简单的小函数,像这样:
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
这样你就不会重复处理任何东西,也不会重复代码。
使用布尔逻辑将数据分配给两个数组
>>> 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')]
清晰快速
这个列表理解是简单的阅读和快速。这正是上级要求的。
set_good_vals = set(good_vals) # Speed boost.
good = [x for x in my_list if x in set_good_vals]
bad = [x for x in my_list if x not in set_good_vals]
我更喜欢一个列表理解而不是两个,但不像张贴的许多答案(其中一些相当巧妙),它是可读的和清晰的。这也是网页上最快的答案之一。
唯一(稍微)快一点的答案是:
set_good_vals = set(good_vals)
good, bad = [], []
for item in my_list:
_ = good.append(item) if item in set_good_vals else bad.append(item)
...还有它的变体。(见我的另一个答案)。但我觉得第一种方法更优雅,而且几乎一样快。
之前的答案似乎并不能满足我所有的四种强迫症:
尽可能的懒惰, 只对原始Iterable求值一次 每个项只计算谓词一次 提供良好的类型注释(适用于python 3.7)
我的解决方案并不漂亮,我不认为我可以推荐使用它,但它是:
def iter_split_on_predicate(predicate: Callable[[T], bool], iterable: Iterable[T]) -> Tuple[Iterator[T], Iterator[T]]:
deque_predicate_true = deque()
deque_predicate_false = deque()
# define a generator function to consume the input iterable
# the Predicate is evaluated once per item, added to the appropriate deque, and the predicate result it yielded
def shared_generator(definitely_an_iterator):
for item in definitely_an_iterator:
print("Evaluate predicate.")
if predicate(item):
deque_predicate_true.appendleft(item)
yield True
else:
deque_predicate_false.appendleft(item)
yield False
# consume input iterable only once,
# converting to an iterator with the iter() function if necessary. Probably this conversion is unnecessary
shared_gen = shared_generator(
iterable if isinstance(iterable, collections.abc.Iterator) else iter(iterable)
)
# define a generator function for each predicate outcome and queue
def iter_for(predicate_value, hold_queue):
def consume_shared_generator_until_hold_queue_contains_something():
if not hold_queue:
try:
while next(shared_gen) != predicate_value:
pass
except:
pass
consume_shared_generator_until_hold_queue_contains_something()
while hold_queue:
print("Yield where predicate is "+str(predicate_value))
yield hold_queue.pop()
consume_shared_generator_until_hold_queue_contains_something()
# return a tuple of two generators
return iter_for(predicate_value=True, hold_queue=deque_predicate_true), iter_for(predicate_value=False, hold_queue=deque_predicate_false)
用下面的测试,我们从print语句中得到如下输出:
t,f = iter_split_on_predicate(lambda item:item>=10,[1,2,3,10,11,12,4,5,6,13,14,15])
print(list(zip(t,f)))
# Evaluate predicate.
# Evaluate predicate.
# Evaluate predicate.
# Evaluate predicate.
# Yield where predicate is True
# Yield where predicate is False
# Evaluate predicate.
# Yield where predicate is True
# Yield where predicate is False
# Evaluate predicate.
# Yield where predicate is True
# Yield where predicate is False
# Evaluate predicate.
# Evaluate predicate.
# Evaluate predicate.
# Evaluate predicate.
# Yield where predicate is True
# Yield where predicate is False
# Evaluate predicate.
# Yield where predicate is True
# Yield where predicate is False
# Evaluate predicate.
# Yield where predicate is True
# Yield where predicate is False
# [(10, 1), (11, 2), (12, 3), (13, 4), (14, 5), (15, 6)]