老哥们, 请教一个问题, 是关于排列组合的, 前几天和老哥们讨论了一下, 能是能解决, 但感觉不太方便 看一个简单的例子应该就明白了.

<school type="primary|middle|university">     <class number="1"/> </school> 

to

<school type="primary">     <class number="1"/> </school>  <school type="middle">     <class number="1"/> </school>  <school type="university">     <class number="1"/> </school> 

目前是我这样做的, 用了一个模版 template

<school type="{school_type}">     <class number="{class_number}"/> </school> 

代码和上面的学校例子有点不同，因为我的完整需求是这样的

<!-- base --> <cross>     <rule1 hint="rule 1" name="long" val1="30|60" val2="90|120" condition="1">         <necessary_rule val1="60" val2="120" condition="1"/>     </rule1> </cross>  <!-- generator --> <!-- 1 --> <cross>     <rule1 hint="rule 1" name="long" val1="30" val2="90" condition="1">         <necessary_rule val1="60" val2="120" condition="1"/>     </rule1> </cross>  <!-- 2 --> <cross>     <rule1 hint="rule 1" name="long" val1="30" val2="120" condition="1">         <necessary_rule val1="60" val2="120" condition="1"/>     </rule1> </cross>  <!-- 3 --> <cross> <rule1 hint="rule 1" name="long" val1="60" val2="90" condition="1">     <necessary_rule val1="60" val2="120" condition="1"/> </rule1> </cross>  <!-- 4 --> <cross> <rule1 hint="rule 1" name="long" val1="60" val2="120" condition="1">     <necessary_rule val1="60" val2="120" condition="1"/> </rule1> </cross>  

代码

import xml.etree.ElementTree as ET import itertools  from typing import Dict, Optional   def producer(rule, root_tag: Optional[str] = ''):     rule_tag = rule.tag     rule_temp = {}     rule_attrib = rule.attrib     for k, v in rule_attrib.items():         k = f'{root_tag}{rule_tag}_{k}'         if 'val' in v:             v = rule_attrib.get(v)         rule_temp[k] = v.split('|')     return rule_temp   def params_parser():     tree = ET.parse('rule.xml')     rule_params = {}     root = tree.getroot()     for strategy in root:         for rule in strategy:             rule_params.update(**producer(rule))             for child_rule in rule:                 rule_params.update(**producer(child_rule, root_tag=f'{rule.tag}_'))      return rule_params   def rule_generator(rule_params: Dict):     with open('rule_template.xml', encoding='utf-8') as f:         template = f.read()      for values in itertools.product(*rule_params.values()):         params = dict(zip(rule_params.keys(), values))         data = template.format(**params)         print(data)   def main():     rule_params = params_parser()     rule_generator(rule_params)   if __name__ == '__main__':     main() 

我的思路是取出所有 tag 的 attrib，然后替换到模版. 上面的做法能解决我的问题，但这样每次都要弄一个模版我觉得很麻烦，有没有更简洁的办法