如何在Python中创建排列,不使用itertools模块或递归编程?

1 投票
2 回答
8867 浏览
提问于 2025-04-18 04:25

我需要创建一个函数,不使用itertools,这个函数可以生成一个包含给定元素的元组的排列列表。

举个例子:

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

这是我写的代码:

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

我知道这段代码没有实现什么,它只会添加第一个组合,其他的都不会有。

2 个回答

1

我对@Hooked的回答有点问题...

首先,我对Python完全是个新手,但我在找类似上面那段代码的东西。我现在正在Repl.it上输入这段代码。

我遇到的第一个问题是参数

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

这导致了以下错误

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

我这样修复了它

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

但现在又出现了这个错误:

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

到目前为止,我完全不知道该去哪里调试代码。不过,我看到很多人提到过itertools,说它的文档里有相关代码。我复制了那个,结果可以运行。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
3

正如@John在评论中提到的,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 = 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

这个代码可以在你的例子中使用,不需要引入其他库或者使用递归调用:

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

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

撰写回答