递归搜索父子组合并在python和XML中构建树

Question

我试图遍历这个充满父 - >子关系的XML数据,并且需要一种构建树的方法.任何帮助将非常感激.此外,在这种情况下,为父 - >子关系设置属性或节点是否更好？

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<nodes>
    <node name="Car" child="Engine"/>
    <node name="Car" child="Wheel"/>
    <node name="Engine" child="Piston"/>
    <node name="Engine" child="Carb"/>
    <node name="Carb" child="Bolt"/>
    <node name="Spare Wheel"/>
    <node name="Bolt" child="Thread"/>
    <node name="Carb" child="Foat"/>
    <node name="Truck" child="Engine"/>
    <node name="Engine" child="Bolt"/>
    <node name="Wheel" child="Hubcap"/>
</nodes>

在Python脚本上,这就是我所拥有的.我的大脑是炒的,我无法理解逻辑？请帮忙

import xml.etree.ElementTree as ET
tree = ET.parse('rec.xml')
root = tree.getroot()
def find_node(data,search):
    #str = root.find('.//node[@child="1.2.1"]')
    for node in data.findall('.//node'):
        if node.attrib['name']==search:
            print('Child-->', node)

for nodes in root.findall('node'):
    parent = nodes.attrib.get('name')
    child = nodes.attrib.get('child')
    print (parent,'-->', child)
    find_node(root,child)

预期的可能输出是这样的(实际上不关心排序顺序,只要所有节点项都在树中的某处表示.

Car --> Engine --> Piston
Car --> Engine --> Carb --> Float
Car --> Engine --> Carb --> Bolt --> Thread
Car --> Wheel --> Hubcaps
Truck --> Engine --> Piston
Truck --> Engine --> Carb --> Bolt --> Thread
Truck --> Loading Bin
Spare Wheel -->

Answer 1

我用图表做了很长时间,但这应该非常接近它不是最优化的方法:

x = """<?xml version="1.0"?>
<nodes>
    <node name="Car" child="Engine"></node>
    <node name="Engine" child="Piston"></node>
    <node name="Engine" child="Carb"></node>
    <node name="Car" child="Wheel"></node>
    <node name="Wheel" child="Hubcaps"></node>
    <node name="Truck" child="Engine"></node>
    <node name="Truck" child="Loading Bin"></node>
    <nested>
        <node name="Spare Wheel" child="Engine"></node>
    </nested>
    <node name="Spare Wheel" child=""></node>

</nodes>"""

from lxml import etree

xml = etree.fromstring(x)
graph = {}
nodes = set()
for x in xml.xpath("//node"):
    par, child = x.xpath(".//@name")[0], x.xpath(".//@child")[0]
    graph.setdefault(par, set())
    graph[par].add(child)
    nodes.update([child, par])


def find_all_paths(graph, start, end, path=None):
    if path is None:
        path = []
    path = path + [start]
    if start == end:
        yield path
    for node in graph.get(start, []):
        if node not in path:
            for new_path in find_all_paths(graph, node, end, path):
                yield new_path


for n in graph:
    for e in nodes:
        if n != e:
            for path in find_all_paths(graph, n, e):
                if path:
                    print("--> ".join(path))

哪个更新的输入会给你:

Engine--> Carb
Engine--> Piston
Car--> Engine
Car--> Wheel
Car--> Wheel--> Hubcaps
Car--> Engine--> Carb
Car--> Engine--> Piston
Spare Wheel--> Engine
Spare Wheel--> 
Spare Wheel--> Engine--> Carb
Spare Wheel--> Engine--> Piston
Wheel--> Hubcaps
Truck--> Engine
Truck--> Engine--> Carb
Truck--> Engine--> Piston
Truck--> Loading Bin

Answer 2

这是一个纯粹的 XSLT 解决方案—— 有效地使用键（相当于哈希表）并且只有 23 行 —— 迄今为止最短的解决方案。

这在计算上也是最简单的——将嵌套级别 1 与嵌套级别 4 - 5 进行比较......

该解决方案是尾递归的，这意味着任何好的 XSLT 处理器都会通过迭代对其进行优化，从而避免堆栈溢出的可能性，因为最大调用堆栈深度保持不变 (1)。

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output method="text"/>

 <xsl:key name="kNodeByChild" match="node" use="@child"/>
 <xsl:key name="kNodeByName" match="node" use="@name"/>

  <xsl:template match="/*">
    <xsl:apply-templates select="node[not(key('kNodeByChild', @name))]"/>
  </xsl:template>

  <xsl:template match="node[not(key('kNodeByName', @child))]">
    <xsl:param name="pParentPath"/>
    <xsl:value-of select="concat($pParentPath, @name, ' ---> ', @child, '&#xA;')"/>
  </xsl:template>

  <xsl:template match="node">
    <xsl:param name="pParentPath"/>

    <xsl:apply-templates select="key('kNodeByName', @child)">
      <xsl:with-param name="pParentPath" select="concat($pParentPath, @name, ' ---> ')"/>
    </xsl:apply-templates>
  </xsl:template>
</xsl:stylesheet>

当此转换应用于提供的 XML 文档时：

<nodes>
    <node name="Car" child="Engine"/>
    <node name="Car" child="Wheel"/>
    <node name="Engine" child="Piston"/>
    <node name="Engine" child="Carb"/>
    <node name="Carb" child="Bolt"/>
    <node name="Spare Wheel"/>
    <node name="Bolt" child="Thread"/>
    <node name="Carb" child="Foat"/>
    <node name="Truck" child="Engine"/>
    <node name="Engine" child="Bolt"/>
    <node name="Wheel" child="Hubcap"/>
</nodes>

产生了想要的正确结果：

Car ---> Engine ---> Piston
Car ---> Engine ---> Carb ---> Bolt ---> Thread
Car ---> Engine ---> Carb ---> Foat
Car ---> Engine ---> Bolt ---> Thread
Car ---> Wheel ---> Hubcap
Spare Wheel ---> 
Truck ---> Engine ---> Piston
Truck ---> Engine ---> Carb ---> Bolt ---> Thread
Truck ---> Engine ---> Carb ---> Foat
Truck ---> Engine ---> Bolt ---> Thread

Answer 3

记录.xml：

<?xml version="1.0"?>
<nodes>
    <node name="Car" child="Engine"></node>
    <node name="Engine" child="Piston"></node>
    <node name="Engine" child="Carb"></node>
    <node name="Car" child="Wheel"></node>
    <node name="Wheel" child="Hubcaps"></node>
    <node name="Truck" child="Engine"></node>
    <node name="Truck" child="Loading Bin"></node>
    <node name="Piston" child="Loa"></node>
    <node name="Piston" child="Loaqq"></node>
    <node name="Piston" child="Loaww"></node>
    <node name="Loaww" child="Loawwqqqqq"></node>
    <node name="Spare Wheel" child=""></node>
</nodes>

解析.py:-

import xml.etree.ElementTree as ET
tree = ET.parse('rec.xml')
root = tree.getroot()
data = {}
child_list = []
def recursive_print(string,x):
    if x in data.keys():
     for x_child in data[x]:
        if x_child in data.keys():
          recursive_print(string+'-------->'+x_child,x_child)
        else:
         print string+'-------->'+x_child
    else:
       print string

for nodes in root.findall('node'):
    parent = nodes.attrib.get('name')
    child = nodes.attrib.get('child')
    child_list.append(child)
    if parent not in data.keys():
        data[parent] = []
    data[parent].append(child)
for key in data.keys():
    if key not in child_list:
      for x in data[key]:
        string = key+'------->'+x
        recursive_print(string,x)

输出：-

Spare Wheel------->
Car------->Engine-------->Piston-------->Loa
Car------->Engine-------->Piston-------->Loaqq
Car------->Engine-------->Piston-------->Loaww-------->Loawwqqqqq
Car------->Engine-------->Carb
Car------->Wheel-------->Hubcaps
Truck------->Engine-------->Piston-------->Loa
Truck------->Engine-------->Piston-------->Loaqq
Truck------->Engine-------->Piston-------->Loaww-------->Loawwqqqqq
Truck------->Engine-------->Carb
Truck------->Loading Bin