这是一个基于pypy库代码的解决方案(感谢agf在评论中的建议)。
状态可以通过.state
属性获得,也可以通过.goto(state)
位置state
序列的索引(从0开始)进行重置。最后有一个演示(恐怕您需要向下滚动)。
这比丢弃值要快得多。
> cat prod.py
class product(object):
def __init__(self, *args, **kw):
if len(kw) > 1:
raise TypeError("product() takes at most 1 argument (%d given)" %
len(kw))
self.repeat = kw.get('repeat', 1)
self.gears = [x for x in args] * self.repeat
self.num_gears = len(self.gears)
self.reset()
def reset(self):
# initialization of indicies to loop over
self.indicies = [(0, len(self.gears[x]))
for x in range(0, self.num_gears)]
self.cont = True
self.state = 0
def goto(self, n):
self.reset()
self.state = n
x = self.num_gears
while n > 0 and x > 0:
x -= 1
n, m = divmod(n, len(self.gears[x]))
self.indicies[x] = (m, self.indicies[x][1])
if n > 0:
self.reset()
raise ValueError("state exceeded")
def roll_gears(self):
# Starting from the end of the gear indicies work to the front
# incrementing the gear until the limit is reached. When the limit
# is reached carry operation to the next gear
self.state += 1
should_carry = True
for n in range(0, self.num_gears):
nth_gear = self.num_gears - n - 1
if should_carry:
count, lim = self.indicies[nth_gear]
count += 1
if count == lim and nth_gear == 0:
self.cont = False
if count == lim:
should_carry = True
count = 0
else:
should_carry = False
self.indicies[nth_gear] = (count, lim)
else:
break
def __iter__(self):
return self
def next(self):
if not self.cont:
raise StopIteration
l = []
for x in range(0, self.num_gears):
index, limit = self.indicies[x]
l.append(self.gears[x][index])
self.roll_gears()
return tuple(l)
p = product('abc', '12')
print list(p)
p.reset()
print list(p)
p.goto(2)
print list(p)
p.goto(4)
print list(p)
> python prod.py
[('a', '1'), ('a', '2'), ('b', '1'), ('b', '2'), ('c', '1'), ('c', '2')]
[('a', '1'), ('a', '2'), ('b', '1'), ('b', '2'), ('c', '1'), ('c', '2')]
[('b', '1'), ('b', '2'), ('c', '1'), ('c', '2')]
[('c', '1'), ('c', '2')]
您应该对其进行更多测试-我可能犯了一个愚蠢的错误-但是这个想法很简单,因此您应该可以解决它:o)您可以自由使用我的更改;不知道原始的pypy许可证是什么。
也不state
是真正的完整状态-它不包括原始参数-只是序列的索引。也许最好把它称为索引,但是代码中已经有索引了。
这是一个简单的版本,具有相同的想法,但是可以通过转换数字序列来工作。因此,您只需将imap
其count(n)
偏移量即可n
。
> cat prod2.py
from itertools import count, imap
def make_product(*values):
def fold((n, l), v):
(n, m) = divmod(n, len(v))
return (n, l + [v[m]])
def product(n):
(n, l) = reduce(fold, values, (n, []))
if n > 0: raise StopIteration
return tuple(l)
return product
print list(imap(make_product(['a','b','c'], [1,2,3]), count()))
print list(imap(make_product(['a','b','c'], [1,2,3]), count(3)))
def product_from(n, *values):
return imap(make_product(*values), count(n))
print list(product_from(4, ['a','b','c'], [1,2,3]))
> python prod2.py
[('a', 1), ('b', 1), ('c', 1), ('a', 2), ('b', 2), ('c', 2), ('a', 3), ('b', 3), ('c', 3)]
[('a', 2), ('b', 2), ('c', 2), ('a', 3), ('b', 3), ('c', 3)]
[('b', 2), ('c', 2), ('a', 3), ('b', 3), ('c', 3)]
(这里的缺点是,如果您要停止并重新启动,则需要跟踪自己已使用了多少个)