具有意外结果的 Python 数字列表排列

Question

我正在尝试获取包含 0-9 的数字列表并返回列表的所有排列。我想出了两个不同的函数，它们在一定程度上返回了预期的结果，但这都不是我的目标。这是一个返回一个周期的正确结果的方法：

x = [0,1,2,3,4,5,6,7,8,9]

def test(x):
  place_holder = 9
  count = 9
  print x
  while count > 1:
    old_x = x[count]
    x[count] = x[count-1]
    x[count-1] = old_x
    count -= 1
    print x
    if count == 1:
      x.sort()
      place_holder -= 1
      count = place_holder

返回：

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
[0, 1, 2, 3, 4, 5, 6, 7, 9, 8]
[0, 1, 2, 3, 4, 5, 6, 9, 7, 8]
[0, 1, 2, 3, 4, 5, 9, 6, 7, 8]
[0, 1, 2, 3, 4, 9, 5, 6, 7, 8]
[0, 1, 2, 3, 9, 4, 5, 6, 7, 8]
[0, 1, 2, 9, 3, 4, 5, 6, 7, 8]
[0, 1, 9, 2, 3, 4, 5, 6, 7, 8]
[0, 9, 1, 2, 3, 4, 5, 6, 7, 8]
[0, 1, 2, 3, 4, 5, 6, 8, 7, 9]
[0, 1, 2, 3, 4, 5, 8, 6, 7, 9]
[0, 1, 2, 3, 4, 8, 5, 6, 7, 9]
[0, 1, 2, 3, 8, 4, 5, 6, 7, 9]
[0, 1, 2, 8, 3, 4, 5, 6, 7, 9]
[0, 1, 8, 2, 3, 4, 5, 6, 7, 9]
[0, 8, 1, 2, 3, 4, 5, 6, 7, 9]
[0, 1, 2, 3, 4, 5, 7, 6, 8, 9]
[0, 1, 2, 3, 4, 7, 5, 6, 8, 9]
[0, 1, 2, 3, 7, 4, 5, 6, 8, 9]
[0, 1, 2, 7, 3, 4, 5, 6, 8, 9]
[0, 1, 7, 2, 3, 4, 5, 6, 8, 9]
[0, 7, 1, 2, 3, 4, 5, 6, 8, 9]
[0, 1, 2, 3, 4, 6, 5, 7, 8, 9]
[0, 1, 2, 3, 6, 4, 5, 7, 8, 9]
[0, 1, 2, 6, 3, 4, 5, 7, 8, 9]
[0, 1, 6, 2, 3, 4, 5, 7, 8, 9]
[0, 6, 1, 2, 3, 4, 5, 7, 8, 9]
[0, 1, 2, 3, 5, 4, 6, 7, 8, 9]
[0, 1, 2, 5, 3, 4, 6, 7, 8, 9]
[0, 1, 5, 2, 3, 4, 6, 7, 8, 9]
[0, 5, 1, 2, 3, 4, 6, 7, 8, 9]
[0, 1, 2, 4, 3, 5, 6, 7, 8, 9]
[0, 1, 4, 2, 3, 5, 6, 7, 8, 9]
[0, 4, 1, 2, 3, 5, 6, 7, 8, 9]
[0, 1, 3, 2, 4, 5, 6, 7, 8, 9]
[0, 3, 1, 2, 4, 5, 6, 7, 8, 9]
[0, 2, 1, 3, 4, 5, 6, 7, 8, 9]

虽然当我在排列中使用另一个列表时，它会给出意想不到的结果：

x = [1,0,2,3,4,5,6,7,8,9]

[1, 0, 2, 3, 4, 5, 6, 7, 8, 9]
[1, 0, 2, 3, 4, 5, 6, 7, 9, 8]
[1, 0, 2, 3, 4, 5, 6, 9, 7, 8]
[1, 0, 2, 3, 4, 5, 9, 6, 7, 8]
[1, 0, 2, 3, 4, 9, 5, 6, 7, 8]
[1, 0, 2, 3, 9, 4, 5, 6, 7, 8]
[1, 0, 2, 9, 3, 4, 5, 6, 7, 8]
[1, 0, 9, 2, 3, 4, 5, 6, 7, 8]
[1, 9, 0, 2, 3, 4, 5, 6, 7, 8]
[0, 1, 2, 3, 4, 5, 6, 8, 7, 9]
[0, 1, 2, 3, 4, 5, 8, 6, 7, 9]
[0, 1, 2, 3, 4, 8, 5, 6, 7, 9]
[0, 1, 2, 3, 8, 4, 5, 6, 7, 9]
[0, 1, 2, 8, 3, 4, 5, 6, 7, 9]
[0, 1, 8, 2, 3, 4, 5, 6, 7, 9]
[0, 8, 1, 2, 3, 4, 5, 6, 7, 9]
[0, 1, 2, 3, 4, 5, 7, 6, 8, 9]
[0, 1, 2, 3, 4, 7, 5, 6, 8, 9]
[0, 1, 2, 3, 7, 4, 5, 6, 8, 9]
[0, 1, 2, 7, 3, 4, 5, 6, 8, 9]
[0, 1, 7, 2, 3, 4, 5, 6, 8, 9]
[0, 7, 1, 2, 3, 4, 5, 6, 8, 9]
[0, 1, 2, 3, 4, 6, 5, 7, 8, 9]
[0, 1, 2, 3, 6, 4, 5, 7, 8, 9]
[0, 1, 2, 6, 3, 4, 5, 7, 8, 9]
[0, 1, 6, 2, 3, 4, 5, 7, 8, 9]
[0, 6, 1, 2, 3, 4, 5, 7, 8, 9]
[0, 1, 2, 3, 5, 4, 6, 7, 8, 9]
[0, 1, 2, 5, 3, 4, 6, 7, 8, 9]
[0, 1, 5, 2, 3, 4, 6, 7, 8, 9]
[0, 5, 1, 2, 3, 4, 6, 7, 8, 9]
[0, 1, 2, 4, 3, 5, 6, 7, 8, 9]
[0, 1, 4, 2, 3, 5, 6, 7, 8, 9]
[0, 4, 1, 2, 3, 5, 6, 7, 8, 9]
[0, 1, 3, 2, 4, 5, 6, 7, 8, 9]
[0, 3, 1, 2, 4, 5, 6, 7, 8, 9]
[0, 2, 1, 3, 4, 5, 6, 7, 8, 9]

正确地经过九个循环然后回到 0-9。所以我可以看到这是因为 x.sort() 调用。所以我把这个函数改成这样：

def exp_test(x):
  static = []
  for i in x:
    static.append(i)
  place_holder = 9
  count = 9
  print x
  while count > 1:
    old_x = x[count]
    x[count] = x[count-1]
    x[count-1] = old_x
    count -= 1
    print x
    if count == 1:
      x = static
      place_holder -= 1
      count = place_holder

现在这工作正常，直到七的转变，它去每隔一个数字。我想计数搞混了，但是，我通过并没有看到它？

Answer 1

尝试将最后一个if 语句中的x = static 修改为x = static[:]。问题是您只是将名称x 重新绑定到static 绑定到的同一个列表。您确实想复制 static 所绑定的内容。

Answer 2

最好的解决办法是

from itertools import permutations

但如果一定要自己写，通常的解决方案是递归：

def permutations(seq):
    _len = len(seq)
    if _len:
        if _len==1:
            yield seq
        else:
            for p in permutations(seq[1:]):
                for i in range(_len):
                    yield p[:i] + seq[0:1] + p[i:]

编辑：好吧，欧拉问题需要完全不同的方法......诀窍不是生成所有排列到 1,000,000（这太慢了），而是计算必须是百万分之一的排列。有n！排列 n 项的方法 - 您可以在序列的尾部递归地使用它来计算必须重新排列多少子序列才能达到第 100 万个排列，并从中计算出必须是什么排列。

你需要写一些类似的东西

def nth_arrangement(seq, n):
    # you have to figure this bit out!