multirember&co
Anton Vredegoor
anton.vredegoor at gmail.com
Thu Apr 19 12:13:13 EDT 2007
Anton Vredegoor wrote:
> Anton Vredegoor wrote:
>> attn.steven.kuo at gmail.com wrote:
>>
>>> Try it with
>>>
>>> def test():
>>> L = 'a', 1, 2, 'a'
>>> it1, it2 = xsplitter(L, lambda x: x == 'a')
>>> print it1.next()
>>> print it2.next()
>>> print it1.next()
>>> print it2.next()
>>>
>>>
>>> The last print statement raises StopIteration...
>>> We, however, expected each iterator to contain
>>> two elements (one yielding 'a' then 'a', and
>>> the other yielding 1 then 2).
>> Ouch! I never understood much about generators anyway.
>
> How about this one?
No that can result in an infinite loop after yet another
print it1.next()
This one however ...
from collections import deque
class sentinel(object):
pass
class myiter(object):
def __init__(self,seq):
self.seq = seq
self.index = -1
def __iter__(self):
return self
def next(self):
self.index +=1
if self.index < len(self.seq):
return self.seq[self.index]
else:
return sentinel
def xsplitter(seq, pred):
Q = deque(),deque()
it = myiter(seq)
def gen(p):
for x in it:
while Q[p]: yield Q[p].popleft()
if x is sentinel: break
if pred(x) == p: yield x
else:
Q[~p].append(x)
for x in gen(p): yield x
return gen(1),gen(0)
def test():
L = 'a', 1, 2, 'a'
it1, it2 = xsplitter(L, lambda x: x == 'a')
print it1.next()
print it2.next()
print it1.next()
print it2.next()
if __name__=='__main__':
test()
A.
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