Question

我使用Python字典（product_dict）来表示产品及其所有子部分的层次结构。 dict的键是唯一ID（UUID），值是Class对象，包含有关这些部分的所有信息，包括：

part.name        # A string, containing the actual name of a component
part.idCode      # UUID of component
part.parent      # UUID of parent component
part.children    # List of UUIDs of child components
part.tier        # An integer that specifies its tier/level within the hierarchy

现在为了以有序的方式输出数据，我希望按层次和字母顺序对部分进行排序。对于使用树结构的分层排序，我发现这个问题的答案非常适合打印：Sorting data hierarchically。对于这个使用我的数据结构的例子，我做了一些小的修改：

class Node:
    def __init__(self, article):
        self.article = article
        self.children = []
        self.parent = None
        self.name = None 

    def printer(self, level=0):
        print ('{}{}'.format('\t' * level, self.name))
        for child in self.children:
            child.printer(level + 1)


class Tree:
    def __init__(self):
        self.nodes = {}

    def push(self, article, parent, name):                          
        if parent not in self.nodes:
            self.nodes[parent] = Node(parent)
        if article not in self.nodes:                              
            self.nodes[article] = Node(article)
        if parent == article:
            return
        self.nodes[article].name = name
        self.nodes[article].parent = self.nodes[parent]             
        self.nodes[parent].children.append(self.nodes[article])    

    @property
    def roots(self):
        return (x for x in self.nodes.values() if not x.parent)


t = Tree()

for idCode, part in product_dict.iteritems(): 
    t.push(idCode, part.parent, part.name)
for node in t.roots:
    node.printer()

考虑到我的产品是飞机的示例，输出现在看起来如下（实际订单各不相同）：

Aircraft
    Systems
        Subsystem 2
        Subsystem 1
            Subsubsystem 1.1
    Engines
    Airframe
        Section 2
        Section 1
        Section 4
        Section 3

但是，由于我在这个阶段对Python的理解有限，我很难在这个例程中添加字母排序（基于part.name字符串）。我理解树是如何构建的，但我没有掌握打印例程，因此无法判断在何处添加字母排序例程。

根据给定的例子，我想要的输出应该是：

Aircraft
    Airframe
        Section 1
        Section 2
        Section 3
        Section 4
    Engines     
    Systems
        Subsystem 1
            Subsubsystem 1.1
        Subsystem 2

非常感谢任何帮助。我不遵守上面给出的分层排序方法，所以我对完全不同的方法持开放态度。

Answer 1

您似乎唯一需要做的就是在打印时按名称对孩子进行排序，即

 for child in sorted(self.children, key = lambda x: x.name):
        child.printer(level + 1)

Answer 2

给出一个层次字典：

{
    UUID("71f0967f-bfd3-43d5-883e-f3c971de141f"): {
        "id": UUID("71f0967f-bfd3-43d5-883e-f3c971de141f"),
        "name": "D10:0:0",
        "code": "r0:0:0",
        "level": 0,
        "parent_id": None,
    },
    UUID("f3dcaf21-66c9-48c5-9aae-f9ac84fc7d45"): {
        "id": UUID("f3dcaf21-66c9-48c5-9aae-f9ac84fc7d45"),
        "name": "D11:0:0",
        "code": "r1:0:0",
        "level": 1,
        "parent_id": UUID("71f0967f-bfd3-43d5-883e-f3c971de141f"),
    },
    UUID("4c39d1e8-ba09-455c-8f76-866200531c51"): {
        "id": UUID("4c39d1e8-ba09-455c-8f76-866200531c51"),
        "name": "D12:0:0",
        "code": "r2:0:0",
        "level": 2,
        "parent_id": UUID("f3dcaf21-66c9-48c5-9aae-f9ac84fc7d45"),
    },
    UUID("8c4e89f2-7564-417f-88ba-1cbb3f9ac23f"): {
        "id": UUID("8c4e89f2-7564-417f-88ba-1cbb3f9ac23f"),
        "name": "D12:0:1",
        "code": "r2:0:1",
        "level": 2,
        "parent_id": UUID("f3dcaf21-66c9-48c5-9aae-f9ac84fc7d45"),
    }, ...
}

您可以按以下顺序按键保留层次结构：

TD1 = typing.Dict[str, typing.Union[UUID, str, int, None]]
TD1Dict = typing.Dict[UUID, TD1]
def _sort_dict_hierarchy(dicts: TD1Dict, parent_id: typing.Optional[UUID] = None) -> TD1Dict:
    roots = [(k, v) for k, v in dicts.items() if v['parent_id'] == parent_id]
    sorted_dicts: TD1Dict = {}
    for k, v in sorted(roots, key=lambda r: r[1]['name']):
        sorted_dicts[k] = v
        sorted_dicts.update(_sort_dict_hierarchy(dicts, v['id']))
    return sorted_dicts

测试：


def create_dict_hierarchy(parent_id=None, level=0, pos=0, max_level=4):
    """Create a hierachy of dicts as we expect to be returned from oauth2."""
    dicts = {}
    for i in range(random.randrange(2, 4)):
        uid = uuid.uuid4()
        dicts[uid] = {
            'id': uid,
            'name': f'D1{level}:{pos}:{i}',
            'code': f'r{level}:{pos}:{i}',
            'level': level,
            'parent_id': parent_id,
        }
        if level < max_level:
            dicts.update(create_dict_hierarchy(uid, level+1, i, max_level))
    return dicts


def test_sort_dict_hierarchy():
    """Sort dicts by nane in hirarchy."""
    dicts = create_dict_hierarchy(max_level=3)
    dicts_items = list(dicts.items())
    random.shuffle(dicts_items)
    shuffled_dicts = {k: v for k, v in dicts_items}
    assert list(dicts.keys()) != list(shuffled_dicts.keys())

    sorted_dicts = _sort_dict_hierarchy(shuffled_dicts)

    assert list(dicts.keys()) == list(sorted_dicts.keys())

按层次结构和字母顺序对字典进行排序（Python）

2 个答案: