Python中如何通过辅助函数获取列表待修改值的地址并完成赋值

方案1：路径元组+通用读写函数 #

Python不支持直接对函数返回值赋值，你可以让定位函数返回目标元素所属的容器、以及完整索引路径的元组，配合两个通用读写函数实现类似指针的操作：

# 通用读函数
def get_ref(ref):
    container, *indices = ref
    for idx in indices:
        container = container[idx]
    return container

# 通用写函数
def set_ref(ref, new_val):
    container, *indices = ref
    for idx in indices[:-1]:
        container = container[idx]
    container[indices[-1]] = new_val

各场景使用示例

一维列表修改：

helper = lambda x: (x, 0)
a = [1, 2, 3]
set_ref(helper(a), 0)
# a 变为 [0, 2, 3]

二维列表修改：

helper_1 = lambda x: (x, 0, 0)
a = [[1, 2, 3]]
set_ref(helper_1(a), 0)
# a 变为 [[0, 2, 3]]

90度旋转实现（完全消除冗余索引写法）：

def rotate_90(image, direction):
    rows, columns = len(image), len(image[0])
    rotated = [[0]*rows for _ in range(columns)]
    
    # 定位函数直接返回路径元组
    if direction == 'R':
        rotated_place = lambda row, col: (rotated, col, -1 - row)
    else:
        rotated_place = lambda row, col: (rotated, -1 - col, row)
        
    for row in range(rows):
        for col in range(columns):
            set_ref(rotated_place(row, col), image[row][col])
            
    return rotated

方案2：封装引用类，更贴近指针语法 #

如果你希望写法更接近C风格指针语法，可以封装一个简单的引用类，通过属性读写实现赋值：

class Ref:
    def __init__(self, container, *indices):
        self.container = container
        self.indices = indices
    
    @property
    def val(self):
        res = self.container
        for idx in self.indices:
            res = res[idx]
        return res
    
    @val.setter
    def val(self, new_val):
        res = self.container
        for idx in self.indices[:-1]:
            res = res[idx]
        res[self.indices[-1]] = new_val

各场景使用示例

一维列表修改：

helper = lambda x: Ref(x, 0)
a = [1, 2, 3]
helper(a).val = 0

遍历修改一维列表：

lst = [1, 2, 3]
for i in range(len(lst)):
    ref = Ref(lst, i)
    ref.val *= 2
# lst 变为 [2, 4, 6]

90度旋转实现：

def rotate_90(image, direction):
    rows, columns = len(image), len(image[0])
    rotated = [[0]*rows for _ in range(columns)]
    
    if direction == 'R':
        rotated_place = lambda row, col: Ref(rotated, col, -1 - row)
    else:
        rotated_place = lambda row, col: Ref(rotated, -1 - col, row)
        
    for row in range(rows):
        for col in range(columns):
            rotated_place(row, col).val = image[row][col]
            
    return rotated

以上两种方案都是纯Python原生实现，不需要引入额外依赖，完全符合Python的语法规则，同时可以避免多层索引写法的冗余问题。

内容的提问来源于stack exchange，提问作者gil aharoni