Python：从“图形”（数据库）数据递归创建字典

Question

一个在 MySQL 数据库中映射的层次结构（我通过 Peewee 访问）。我正在尝试遍历数据以将其重新组合成嵌套字典（用于最终转换为 XML）。

以下函数将我的数据向下传递到父节点并打印出我想要在我的 dict 中结构化的数据：

def build_dict(current):
    query = (ParamLevel
            .select()
            # If we are looking for parents, we are matching on child
            .join(ParamLevelParamLevels, JOIN.LEFT_OUTER, on = (ParamLevelParamLevels.parent == ParamLevel.id))
            .where(ParamLevelParamLevels.child == current)
            )

    # If we have a parent node, recurse further
    if query.exists():
        parent = query.get()
        build_dict(parent)
        print('Current ParamLevel "%s" parent: "%s"' % ( current.name, parent.name ))
    else:
        print('Found root node: %s' % current.name)

这样做时，它会打印出来：

Found root node: polycomConfig
Current ParamLevel "device" parent: "polycomConfig"
Current ParamLevel "device.dhcp" parent: "device"
Current ParamLevel "device.dhcp.bootSrvOptType" parent: "device.dhcp"

我正在寻找有关如何生成以下数据结构的输入：

{polycomConfig : { device : { device.dhcp : { device.dhcp.bootSrvOptType: {} } } } }

我确信这相当简单，但我对实现递归函数感到生疏。

谢谢！

Answer 1

使用while 循环而不是递归来执行此操作，并且只需构建嵌套的字典即可。在这种情况下，递归实际上没有任何好处。

def build_dict(current):
    print(f'Found root node {current.name}')
    temp_dict = {}
    query = (ParamLevel
            .select()
            # If we are looking for parents, we are matching on child
            .join(ParamLevelParamLevels, JOIN.LEFT_OUTER, on = (ParamLevelParamLevels.parent == ParamLevel.id))
            .where(ParamLevelParamLevels.child == current)
            )
    while query.exists():
        result = query.get()
        temp_dict = {result.name: temp_dict}
        query = (ParamLevel
            .select()
            # If we are looking for parents, we are matching on child
            .join(ParamLevelParamLevels, JOIN.LEFT_OUTER, on = (ParamLevelParamLevels.parent == ParamLevel.id))
            .where(ParamLevelParamLevels.child == result)
            )

没有办法对其进行测试，我无法为您提供输出，也无法检查任何错误，但您应该了解它的要点。你想要的结果应该在temp_dict。

我用这个测试过：

d = {}
for i in range(5):
    d = {i: d}
print(d)

输出：

{4: {3: {2: {1: {0: {}}}}}}

Answer 2

如果没有一个好的例子很难尝试，但我相信这应该可行：

def build_dict(current, root={}):
    query = (
        ParamLevel.select()
        # If we are looking for parents, we are matching on child
        .join(
            ParamLevelParamLevels,
            JOIN.LEFT_OUTER,
            on=(ParamLevelParamLevels.parent == ParamLevel.id),
        ).where(ParamLevelParamLevels.child == current)
    )

    # If we have a parent node, recurse further
    if query.exists():
        parent = query.get()
        root, parent_dict = build_dict(parent, root)
        parent_dict[parent.name] = {current.name: {}}
        return root, parent_dict[parent.name]
    else:
        root[current.name] = {}
        return root, root

answer, _ = build_dict(<starting_node>)