Question

我需要在不使用itertools的情况下创建一个函数，它将创建一组具有给定任何东西的元组的排列列表。

示例：

perm({1,2,3}, 2)应该返回[(1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)]

这就是我得到的：

def permutacion(conjunto, k):
    a, b = list(), list()
    for i in conjunto:
        if len(b) < k and i not in b:
            b.append(i)

    b = tuple(b)
    a.append(b)
    return a

我知道这并没有做任何事情，它会添加第一个组合，而不是别的。

Answer 1

正如@John在评论中所提到的，itertools.permutations的代码是：

def permutations(iterable, r=None):
    # permutations('ABCD', 2) --> AB AC AD BA BC BD CA CB CD DA DB DC
    # permutations(range(3)) --> 012 021 102 120 201 210
    pool = tuple(iterable)
    n = len(pool)
    r = n if r is None else r
    if r > n:
        return
    indices = range(n)
    cycles = range(n, n-r, -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

哪个适用于您的示例，不使用外部导入或递归调用：

for x in permutations([1,2,3],2):
    print x

(1, 2)
(1, 3)
(2, 1)
(2, 3)
(3, 1)
(3, 2)

Answer 2

我对@Hooked的答案有疑问...

首先，我是一个有关py的完全新手，但是我正在寻找类似上面的代码的东西。我目前正在Repl.it

输入

我的第一个问题是争论

for x in permutations([1,2,3],2):
    print x

返回以下错误

line 26
    print x
          ^
SyntaxError: Missing parentheses in call to 'print'

我这样固定了

for x in permutations([1,2,3],2):
    print (x)

但是现在出现错误：

line 25, in <module>
    for x in permutations([1,2,3],2):
  File "main.py", line 14, in permutations
    cycles[i] -= 1
TypeError: 'range' object does not support item assignment

现在，我不知道该去哪里调试代码。但是，我看到许多人指出itertools的文档中包含代码。我复制了它，并且有效。这是代码：

def permutations(iterable, r=None):
    # permutations('ABCD', 2) --> AB AC AD BA BC BD CA CB CD DA DB DC
    # permutations(range(3)) --> 012 021 102 120 201 210
    pool = tuple(iterable)
    n = len(pool)
    r = n if r is None else r
    if r > n:
        return
    indices = list(range(n))
    cycles = list(range(n, n-r, -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

如何在不使用itertools模块或递归编程的情况下在Python上创建排列？

2 个答案: