Python之函数更改输入列表，虽然我做了一个副本

我正在编写一个小的 python 脚本，它打印出一个行列表，其中行本身是一个字符串列表。这就是函数的作用。它应该将列表作为输入并打印出漂亮的版本，而不实际更改列表。我可以让它打印出正确的输出。但是经过仔细检查，它也会更改原始列表，尽管我在函数中做的第一件事是使用此 grid_copy = grid[:]

问题:脚本将任何 '' 转换为 ' ' 尽管我没有修改列表

[['X', 'X', ''], ['', 'O', 'O'], ['O', 'X', '']]

进入:

[X] [X] [ ] 
 
[ ] [O] [O] 
 
[O] [X] [ ] 

我不知道是什么导致列表被更改，除了我在开始制作副本时，我没有在哪里引用原始列表。

我添加的一些有用的注释使我的代码更容易理解我所做的事情。如果您运行代码，我还提供了一个测试用例。

def show_grid(grid): 
    """ 
    grid: list. A list of lines, where the lines are a list of strings 
    return: None 
        Prints the grid out in the classic 3x3 
    """ 
    grid_copy = grid[:]  # make a copy so we do not want to actually change grid 
    len_grid = len(grid_copy)  # I use this so much, might as well store it 
 
    # add whitespaces to make each single spot the same width 
    len_digits = len(str((len(grid_copy)**2)))  # the number of chars wide the highest number is 
    # modification happens somewhere vvv 
    for line in range(len_grid): 
        for char in range(len_grid): 
            grid_copy[line][char] = str(grid_copy[line][char]).rjust(len_digits)  # proper white spaces 
 
    # modification happens somewhere ^^^ 
 
    # add brackets, THIS IS NOT WHERE THE PROBLEM IS 
    for line in grid_copy: 
        print('[' + '] ['.join(line)+']\n')  # print each symbol in a box 
 
 
# TESTING TESTING 
test_grid = [['X', 'X', ''], ['', 'O', 'O'], ['O', 'X', '']] 
print('test_grid before: {}'.format(test_grid)) 
show_grid(test_grid)  # this should not modify test_grid, but it does 
print('test_grid after: {}'.format(test_grid)) 

输出:

test_grid before: [['X', 'X', ''], ['', 'O', 'O'], ['O', 'X', '']] 
[X] [X] [ ] 
 
[ ] [O] [O] 
 
[O] [X] [ ] 
 
test_grid after: [['X', 'X', ' '], [' ', 'O', 'O'], ['O', 'X', ' ']] 

请您参考如下方法:

当您在列表的列表上编写 grid_copy = grid[:] 时，您只是复制最顶层的列表，而不是列表本身的元素。这称为浅拷贝，与深拷贝相对。你应该写

grid_copy = [x[:] for x in grid] 

或

grid_copy = copy.deepcopy(grid)