Question

我需要一个递归函数，它可以替换给定列表和长度的所有可能的排列：

DataBind()

我找到了下面的代码，但它只给出了组合;它不重复元素，如[222] [322]等。

>>> rec_offspring(3,[2,3])
[[2, 2, 2], [3, 2, 2], [2, 3, 2], [3, 3, 2], [2, 2, 3], [3, 2, 3], [2, 3, 3], [3, 3, 3]]

Answer 1

这称为笛卡尔积。您可以使用itertools module：

>>> from itertools import product
>>> list(product([2,3], repeat=3))
[(2, 2, 2), (2, 2, 3), (2, 3, 2), (2, 3, 3), (3, 2, 2), (3, 2, 3), (3, 3, 2), (3, 3, 3)]

Answer 2

您有三个基本问题：

（1）在您的递归调用中，您切断了所选元素。相反，将它们全部传递到下一个级别。取代

for j in choose_sets(mylist[k+1:],length-1):

与

for j in choose_sets(mylist,length-1):

（2）您限制了您考虑的元素数量，使元素处于非递减顺序。删除行

    if mylen - k + 1> length:

（3）最后，您已将返回列表限制为输入列表的长度。这意味着只有2个元素（例如[2,3]），您不能返回3个元素的选择列表（例如[2,3,2]）。删除有问题的代码：

if length > mylen:
    return []

结果程序看起来像你想要的那样，返回：

>>> print choose_sets([2,3], 3)
[[2, 2, 2], [2, 2, 3], [2, 3, 2], [2, 3, 3], [3, 2, 2], [3, 2, 3], [3, 3, 2], [3, 3, 3]]
>>> print choose_sets([1,2,3,4,5], 3)
[[1, 1, 1], [1, 1, 2], [1, 1, 3], [1, 1, 4], [1, 1, 5], [1, 2, 1], [1, 2, 2], [1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 3, 1], [1, 3, 2], [1, 3, 3], [1, 3, 4], [1, 3, 5], [1, 4, 1], [1, 4, 2], [1, 4, 3], [1, 4, 4], [1, 4, 5], [1, 5, 1], [1, 5, 2], [1, 5, 3], [1, 5, 4], [1, 5, 5], [2, 1, 1], [2, 1, 2], [2, 1, 3], [2, 1, 4], [2, 1, 5], [2, 2, 1], [2, 2, 2], [2, 2, 3], [2, 2, 4], [2, 2, 5], [2, 3, 1], [2, 3, 2], [2, 3, 3], [2, 3, 4], [2, 3, 5], [2, 4, 1], [2, 4, 2], [2, 4, 3], [2, 4, 4], [2, 4, 5], [2, 5, 1], [2, 5, 2], [2, 5, 3], [2, 5, 4], [2, 5, 5], [3, 1, 1], [3, 1, 2], [3, 1, 3], [3, 1, 4], [3, 1, 5], [3, 2, 1], [3, 2, 2], [3, 2, 3], [3, 2, 4], [3, 2, 5], [3, 3, 1], [3, 3, 2], [3, 3, 3], [3, 3, 4], [3, 3, 5], [3, 4, 1], [3, 4, 2], [3, 4, 3], [3, 4, 4], [3, 4, 5], [3, 5, 1], [3, 5, 2], [3, 5, 3], [3, 5, 4], [3, 5, 5], [4, 1, 1], [4, 1, 2], [4, 1, 3], [4, 1, 4], [4, 1, 5], [4, 2, 1], [4, 2, 2], [4, 2, 3], [4, 2, 4], [4, 2, 5], [4, 3, 1], [4, 3, 2], [4, 3, 3], [4, 3, 4], [4, 3, 5], [4, 4, 1], [4, 4, 2], [4, 4, 3], [4, 4, 4], [4, 4, 5], [4, 5, 1], [4, 5, 2], [4, 5, 3], [4, 5, 4], [4, 5, 5], [5, 1, 1], [5, 1, 2], [5, 1, 3], [5, 1, 4], [5, 1, 5], [5, 2, 1], [5, 2, 2], [5, 2, 3], [5, 2, 4], [5, 2, 5], [5, 3, 1], [5, 3, 2], [5, 3, 3], [5, 3, 4], [5, 3, 5], [5, 4, 1], [5, 4, 2], [5, 4, 3], [5, 4, 4], [5, 4, 5], [5, 5, 1], [5, 5, 2], [5, 5, 3], [5, 5, 4], [5, 5, 5]]

从n大小列表中的k大小的python组合（递归）

2 个答案: