计算 itertools.product() 的第 n 个结果

我正在尝试计算 itertools.product() 的第 n 个结果

test = list(product('01', repeat=3))
print(test)

desired_output = test[0]
print(desired_output)

所以而不是得到：

[('0', '0', '0'), ('0', '0', '1'), ('0', '1', '0'), ('0', '1', '1'), ('1', '0', '0'), ('1', '0', '1'), ('1', '1', '0'), ('1', '1', '1')]

我试图得到：

('0', '0', '0')

然而，正如您可能已经猜到的那样，它的扩展性不佳。这就是为什么我试图只计算第 n 个值。

我通读了第 N 个组合 https://stackoverflow.com/questions/1776442/nth-combination但我需要 Product() 提供的重复功能

提前致谢！

The repeat功能可以很容易地被模拟。这是 Ruby 代码的 Python 版本，描述于this http://blog.sethladd.com/2010/03/calculating-nth-term-of-cartesian.html博客文章。

def product_nth(lists, num):
    res = []
    for a in lists:
        res.insert(0, a[num % len(a)])
        num //= len(a)

    return ''.join(res)

将此函数称为

>>> repeats = 50
>>> chars = '01'
>>> product_nth([chars] * repeats, 12345673)
'00000000000000000000000000101111000110000101001001'

这是一些 timeit 测试：

repeat = 50
idx = 112345673

%timeit i = product_nth(['01'] * repeat, idx)
%%timeit
test = product('01', repeat=repeat)
one = islice(test, idx, idx+1)
j = ''.join(next(one))

2.01 s ± 22.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
36.5 µs ± 201 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

print(i == j)
True

另一个答案具有误导性，因为它歪曲了islice。作为示例，请参阅：

def mygen(r):
    i = 0
    while i < r:
        print("Currently at", i)
        yield i
        i += 1

list(islice(mygen(1000), 10, 11))

# Currently at 0
# Currently at 1
# Currently at 2
# Currently at 3
# Currently at 4
# Currently at 5
# Currently at 6
# Currently at 7
# Currently at 8
# Currently at 9
# Currently at 10
# Out[1203]: [10]

islice将逐步通过每一次迭代，丢弃结果直到指定的索引。当对输出进行切片时也会发生同样的事情product— 对于较大的 N，该解决方案效率较低。