修复Python递归提取嵌套列表数字及索引路径的代码错误

问题原因 #

• 全局indices变量用于存储当前遍历的路径层级，递归进入子列表时追加了当前索引，但递归结束返回父层级时没有移除该索引，导致路径残留了已处理完成的子列表层级，最终生成的索引路径错误。
• 代码中递归调用时函数名错误，写为了Fruits(elem)，实际应该调用自身Numbers(elem)。

修复方案 #

仅需2处改动即可解决问题：

1. 递归调用子列表结束后，新增indices.pop()语句，清除当前层级的路径残留
2. 修正递归调用的函数名

修复后完整代码 #

indices = []
List_numbers = []
Address_numbers = []

def Numbers(T):
    for index, elem in enumerate(T):
        if isinstance(elem, int):
            List_numbers.append(elem)
            Address_numbers.append(indices + [index])

        if isinstance(elem, list):
            indices.append(index)
            Numbers(elem)
            indices.pop() # 递归返回后弹出当前层级索引，重置路径状态

# 测试调用
T = ['apple', [10, 5, ['orange']], 3]
Numbers(T)
print('List of numbers: ' + str(List_numbers))
print('Address of numbers: ' + str(Address_numbers))

输出结果 #

List of numbers: [10, 5, 3]
Address of numbers: [[1, 0], [1, 1], [2]]

优化方案（可选） #

如果需要避免全局变量污染、支持多次独立调用，可以改用路径参数传递的写法，无需维护全局索引变量：

def Numbers(T, current_path=None):
    if current_path is None:
        current_path = []
    list_numbers = []
    address_numbers = []
    for index, elem in enumerate(T):
        if isinstance(elem, int):
            list_numbers.append(elem)
            address_numbers.append(current_path + [index])
        if isinstance(elem, list):
            sub_nums, sub_addrs = Numbers(elem, current_path + [index])
            list_numbers.extend(sub_nums)
            address_numbers.extend(sub_addrs)
    return list_numbers, address_numbers

# 调用测试
T = ['apple', [10, 5, ['orange']], 3]
list_num, addr_num = Numbers(T)
print('List of numbers:', list_num)
print('Address of numbers:', addr_num)

内容的提问来源于stack exchange，提问作者yoyo