Question

我编写了一个脚本，用于按字典顺序查找数字列表的所有排列。立即打印时输出正确，但如果我将其附加到列表中则会更改。在我的脚本输出中，首先看到包含单个元素的列表，然后是我追加到该元素的列表，最后在添加子列表后看到完成的列表：

Find all permutations of 1 -> x. x = ? 3
[[1, 2, 3]] 


[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]


 [[1, 3, 2], [2, 3, 1], [2, 3, 1], [3, 2, 1], [3, 2, 1], [3, 2, 1]]

如您所见，初始[1,2,3]列表在最终输出中甚至不是。以下是我的代码。有人能告诉我到底发生了什么事吗？

permutation_range = int(input('Find all permutations of 1 -> x. x = ? '))

def permutation_finder(x):
    permutation = list(range(1, 1+x))
    rev_permutation = permutation[::-1]
    permutation_list = [permutation]
    print(permutation_list, '\n\n')          #bugcheck print
    while permutation != rev_permutation:
        permutation, index_k = step_1(permutation)
        permutation, index_k, index_l = step_2(permutation, index_k)
        permutation, index_k = step_3(permutation, index_k, index_l)
        permutation = step_4(permutation, index_k)
        permutation_list.append(permutation)    
        print(permutation)                  #bugcheck print 
    return permutation_list

def step_1(permutation):
    for val in permutation[:-1][::-1]:
        index_k = permutation.index(val)
        if val < permutation[index_k+1]:
            return permutation, index_k

def step_2(permutation, index_k):
    for val in permutation[index_k+1:][::-1]:
        if val > permutation[index_k]:
            index_l = permutation.index(val)
            return permutation, index_k, index_l

def step_3(permutation, index_k, index_l):
    a_k, a_l = permutation[index_k], permutation[index_l]
    permutation[index_k], permutation[index_l] = a_l, a_k
    return permutation, index_k

def step_4(permutation, index_k):
    front = permutation[:index_k+1]
    back = permutation[index_k+1:]
    back.reverse()
    permutation = front + back
    return permutation

print('\n\n', permutation_finder(permutation_range))

Answer 1

因为列表中的列表实际上是一个引用，而不是在使用append()时转换为值。因此，当您之后编辑列表时，之前添加的列表也会更改。

在追加之前添加[:]来复制列表可以修复它：

def permutation_finder(x):
    permutation = list(range(1, 1+x))
    rev_permutation = permutation[::-1]
    permutation_list = [permutation[:]]
    print(permutation_list, '\n\n')          #bugcheck print
    while permutation != rev_permutation:
        permutation, index_k = step_1(permutation)
        permutation, index_k, index_l = step_2(permutation, index_k)
        permutation, index_k = step_3(permutation, index_k, index_l)
        permutation = step_4(permutation, index_k)
        permutation_list.append(permutation[:])    
        print(permutation)                  #bugcheck print 
    return permutation_list

（注意[:]部分）

改变之后，我可以得到：

$ python permutation.py
Find all permutations of 1 -> x. x = ? 3
[[1, 2, 3]]


[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]


 [[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]

应该是你想要的：）

.append（）不会将列表添加到列表列表中

1 个答案: