如何检测无限序列中的重复数字？我试过Floyd&Brent检测算法,但什么都没有......我有一个生成器,产生0到9(含)的数字,我必须认识到它的一个时期.
示例测试用例:
import itertools
# of course this is a fake one just to offer an example
def source():
return itertools.cycle((1, 0, 1, 4, 8, 2, 1, 3, 3, 1))
>>> gen = source()
>>> period(gen)
(1, 0, 1, 4, 8, 2, 1, 3, 3, 1)
在从元组向量(残差,模数)计算中国剩余定理时,以下代码失败:
c = ((1,5),(3,7),(11,13),(19,23))
def crt(c):
residues, moduli = zip(*c)
N = product(moduli)
complements = (N/ni for ni in moduli)
scaled_residues = (product(pair) for pair in zip(residues,complements))
inverses = (modular_inverse(*pair) for pair in zip(complements,moduli))
si = (product(u) for u in zip(scaled_residues,inverses))
result = sum(si) % N
return result
将结果赋予0(我猜生成的iterables为空).但以下代码完美运行:
def crt(c):
residues, moduli = zip(*c)
N = product(moduli)
complements = list((N/ni for ni in moduli)) # <-- listed
scaled_residues = (product(pair) for pair in zip(residues,complements))
inverses = (modular_inverse(*pair) …Python 3.4.2 (default, Oct 8 2014, 13:44:52)
[GCC 4.9.1 20140903 (prerelease)] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> gen = (x for x in range(10)) ## Need to wrap range into ()'s to create a generator, next(range(10)) is invalid
>>> list(zip(gen, [1,2,3])) ## zip will "eat up" the number 3
[(0, 1), (1, 2), (2, 3)]
>>> next(gen) ## Here i need next to return 3
4
>>>
问题是我在拉链电话后丢失了一个值.如果gen不是纯粹的代码,这将是一个更大的问题.
我不知道是否可以创建一个行为类似的函数,如果zip函数的一个参数只是一个生成器,其余的是"正常"迭代器,其中所有的值都是已知,并存储在内存中.如果是这种情况,您可以最后检查发电机.
基本上我想知道的是,如果python标准库中有任何函数,就像我在这种情况下需要的那样.
当然,在某些情况下,人们可以做类似的事情
xs …