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
如果你不想用两行代码来完成一个语义只需要一次的操作,你可以把上面的一些方法(甚至是你自己的方法)包装在一个函数中:
def part_with_predicate(l, pred):
return [i for i in l if pred(i)], [i for i in l if not pred(i)]
这不是一种惰性计算方法,它确实对列表进行了两次迭代,但是它允许您在一行代码中对列表进行分区。
其他回答
有时候你并不需要清单上的另一半。 例如:
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 '')
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)]
Good = [x for x in mylist if x in goodvals] Bad = [x for x in mylist if x not in goodvals] 我怎样才能做得更优雅呢?
代码已经非常优雅了。
使用集合可能会有轻微的性能改进,但差异是微不足道的。基于集合的方法也会丢弃重复项,并且不会保留元素的顺序。我发现列表理解也更容易阅读。
事实上,我们甚至可以更简单地使用for循环:
good, bad = [], []
for x in mylist:
if x in goodvals:
good.append(f)
else:
bad.append(f)
这种方法可以更容易地添加额外的逻辑。例如,代码很容易被修改为丢弃None值:
good, bad = [], []
for x in mylist:
if x is None:
continue
if x in goodvals:
good.append(f)
else:
bad.append(f)
def partition(pred, seq):
return reduce( lambda (yes, no), x: (yes+[x], no) if pred(x) else (yes, no+[x]), seq, ([], []) )
我转向numpy来解决这个问题,以限制行数,并使其成为一个简单的函数。
我能够得到一个条件满足,将一个列表分为两个,使用np。在哪里分离出一个列表。这适用于数字,但这可以扩展使用字符串和列表,我相信。
在这儿……
from numpy import where as wh, array as arr
midz = lambda a, mid: (a[wh(a > mid)], a[wh((a =< mid))])
p_ = arr([i for i in [75, 50, 403, 453, 0, 25, 428] if i])
high,low = midz(p_, p_.mean())