如何在Python中执行range（）？

Question

我想使用n个不同的变量，比如说var {0}，var {1}，var {2}，...，var {n-1}

我怎样才能做到这一点 ？

for var{0} in range(n)
  for var{1} in [x for x in range(n) if x!=var{0}]:
    for var{2} in [x for x in range(n) if (x!=var{0} and x!=var{1})]:
      ...
        for var{n-1} in [x for x in range(n) if (x!=var{0} and x!=var{1} and ... and x!=var{n-2})]:

谢谢

Answer 1

似乎这只是range(n)排列：

from itertools import permutations
for var in permutations(range(n)):
    # do something with var[0], var[1], ..., var[n-1]
    print var

根据文档

排列以字典顺序排序。 因此，如果对输入的iterable进行排序，则将按排序顺序生成置换元组。

我认为这将为您提供与示例中的方法相同的排序。 对于n=3您将获得：

(0, 1, 2)
(0, 2, 1)
(1, 0, 2)
(1, 2, 0)
(2, 0, 1)
(2, 1, 0)

如果您对它的生成方式感兴趣，那么itertools.permutations的源代码在其注释中具有等效的python代码（实际实现在C中）：

def permutations(iterable, r=None):
    'permutations(range(3), 2) --> (0,1) (0,2) (1,0) (1,2) (2,0) (2,1)'
    pool = tuple(iterable)
    n = len(pool)
    r = n if r is None else r
    indices = range(n)
    cycles = range(n-r+1, n+1)[::-1]
    yield tuple(pool[i] for i in indices[:r])
    while n:
        for i in reversed(range(r)):
            cycles[i] -= 1
            if cycles[i] == 0:
                indices[i:] = indices[i+1:] + indices[i:i+1]
                cycles[i] = n - i
            else:
                j = cycles[i]
                indices[i], indices[-j] = indices[-j], indices[i]
                yield tuple(pool[i] for i in indices[:r])
                break
        else:
            return

Answer 2

在Python中，您可以执行以下操作：

for i in range(n**n):
vars = [(int(i/(n**(n-j))))%n for j in range(1,n+1)]

vars是您想要的var列表