基于停止递归级别修剪python字典

Question

假设我有以下 Python 字典：

d = {1: {2: {3: {}}, 6: {7: {}}, 8: {}, 9: {}}, 10: {11: {}}}

我可以使用以下函数递归遍历对象中的某个级别：

def resurse_stop(d, stop=None, curr=0):
    if stop and curr == stop:
        return
    for k, v in d.items():
        print(f"{k} (level={curr})")
        if v:
            resurse_stop(v, stop, curr + 1)

resurse_stop(d, 1)
# 1 (level=0)
# 10 (level=0)

resurse_stop(d, 2)
# 1 (level=0)
# 2 (level=1)
# 6 (level=1)
# 8 (level=1)
# 9 (level=1)
# 10 (level=0)
# 11 (level=1)

如何基于停止级别创建一个新对象，它类似于上面的resurse_stop函数，但它不是打印数据，而是构建一个对象。 让我们调用函数resurse_stop_obj ，下面是它应该返回的一些示例：

resurse_stop(d, 1)
# d = {1: {}, 10: {}}

resurse_stop(d, 2)
# d = {1: {2: {}, 6: {}, 8: {}, 9: {}}, 10: {11: {}}}

Answer 1

将您的print表达式包装成字典理解：

d = {1: {2: {3: {}}, 6: {7: {}}, 8: {}, 9: {}}, 10: {11: {}}}

def resurse_stop(d, stop=None, curr=0):
    if stop and curr == stop:
        return {}

    return {k: {} if not v else resurse_stop(v, stop, curr + 1)
               for k, v in d.items() }

print(resurse_stop(d, 2))

输出：

{1: {2: {}, 6: {}, 8: {}, 9: {}}, 10: {11: {}}}