Anton Vredegoor wrote:
> [EMAIL PROTECTED] 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?
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 mygen(seq):
for x in seq:
if x is sentinel: #way past bedtime
raise StopIteration
yield x
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 pred(x) == p: yield x
else:
Q[~p].append(x)
for x in gen(p): yield x
return map(mygen,[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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