python-3.x - 通过重复获得不同的数字集

Question

我有一个数字列表：

lst = [1, 2, 3, 1,4]


def permutation(lst):
    # If lst is empty then there are no permutations
    if len(lst) == 0:
        return []

    # If there is only one element in lst then, only
    # one permuatation is possible
    if len(lst) == 1:
        return [lst]

        # Find the permutations for lst if there are
    # more than 1 characters

    l = []  # empty list that will store current permutation

    # Iterate the input(lst) and calculate the permutation
    for i in range(len(lst)):
        m = lst[i]

        # Extract lst[i] or m from the list.  remLst is
        # remaining list
        remLst = lst[:i] + lst[i + 1:]

        # Generating all permutations where m is first
        # element
        for p in permutation(remLst):
            l.append([m] + p)
    return l

if __name__ == "__main__":
    lst = [1, 2, 3, 1,4]
    v_out = permutation(lst)
    print(v_out)

我只得到 4 个长度的排列，我想要所有长度的排列，并且只有不同的排列。但在每个排列中，重复是允许的。

标签： python-3.xcombinationsitertools

Answer 1

这应该可以工作...使用permutationsitertools 中的功能并set排除所有内容以防止将重复项添加到整体结果中

In [20]: from itertools import permutations                                     

In [21]: a = [1, 1, 2, 3]                                                       

In [22]: all_results = set()                                                    

In [23]: for i in range(1, len(a)): 
    ...:     all_results.update(set(permutations(a, i))) 
    ...:                                                                        

In [24]: all_results                                                            
Out[24]: 
{(1,),
 (1, 1),
 (1, 1, 2),
 (1, 1, 3),
 (1, 2),
 (1, 2, 1),
 (1, 2, 3),
 (1, 3),
 (1, 3, 1),
 (1, 3, 2),
 (2,),
 (2, 1),
 (2, 1, 1),
 (2, 1, 3),
 (2, 3),
 (2, 3, 1),
 (3,),
 (3, 1),
 (3, 1, 1),
 (3, 1, 2),
 (3, 2),
 (3, 2, 1)}

In [25]: