将包含子字符串的字符串转换为字典

huangapple
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英文:

convert string which contains sub string to dictionary

问题

  1. 我试图将特定格式的字符串转换为Python字典。
  2. 字符串格式如下，
  3. ```python
  4. st1 = ''key1 key2=value2 key3="key3.1, key3.2=value3.2 , key3.3 = value3.3, key3.4" key4''

我想解析它并将其转换为以下字典，

  1. dict1 {
  2.     key1: None,
  3.     key2: value2,
  4.     key3: {
  5.             key3.1: None,
  6.             key3.2: value3.2,
  7.             key3.3: value3.3,
  8.             key3.4: None
  9.           }
  10.     key4: None,

我尝试使用python的re包和字符串分割函数，但未能实现结果。我有成千上万个相同格式的字符串，我正在尝试自动化处理它。有人能帮忙吗？

  2. <details>
  3. <summary>英文:</summary>
  5. I am tring to convert particular strings which are in particular format to Python dictionary.
  6. String format is like below,

st1 = 'key1 key2=value2 key3="key3.1, key3.2=value3.2 , key3.3 = value3.3, key3.4" key4'

  2. I want to parse it and convert to dictionary as below,

dict1 {
key1: None,
key2: value2,
key3: {
key3.1: None,
key3.2: value3.2,
key3.3: value3.3,
key3.2: None
}
key4: None,

  7. I tried to use python re package and string split function. not able to acheive the result. I have thousands of string in same format, I am trying to automate it. could someone help. 
  9. </details>
  12. # 答案1
  13. **得分**: 0
  15. ```python
  16. 如果您的所有字符串都是一致的，并且只有一层子字典，下面的代码应该能起作用，您可能需要对其进行微调/更改。
  18. import json
  20. st1 = 'key1 key2=item2 key3="key3.1, key3.2=item3.2 , key3.3 = item3.3, key3.4" key4'
  21. st1 = st1.replace(' = ', '=')
  22. st1 = st1.replace(' ,', ',')
  23. new_dict = {}
  24. no_keys=False
  26. while not no_keys:
  27.     st1 = st1.lstrip()
  28.     
  29.     if " " in st1:
  30.         item = st1.split(" ")[0]
  31.     else:
  32.         item = st1
  33.     
  34.     if '=' in item:
  35.         if '="' in item:
  36.             item = item.split('=')[0]
  37.             new_dict[item] = {}        
  38.             
  39.             st1 = st1.replace(f'{item}=', '')
  40.             sub_items = st1.split('"')[1]
  41.             sub_values = sub_items.split(',')
  42.     
  43.             for sub_item in sub_values:
  44.                 if "=" in sub_item:
  45.                     sub_key, sub_value = sub_item.split('=')
  46.                     new_dict[item].update({sub_key.strip():sub_value.strip()})
  47.                 else:
  48.                     new_dict[item].update({sub_item.strip(): None})
  49.                 
  50.                 st1 = st1.replace(f'"{sub_items}"', '')
  51.         else:
  52.             key, value = item.split('=')
  53.             new_dict.update({key:value})
  54.             st1 = st1.replace(f"{item} ", "")
  55.     else:
  56.         new_dict.update({item: None})
  57.         st1 = st1.replace(f"{item}", "")
  58.         
  59.     if st1 == "":
  60.         no_keys=True    
  61.         
  62. print(json.dumps(new_dict, indent=4))
英文:

If all your strings are consistent, and only have 1 layer of sub dict, this code below should do the trick, you may need to make tweaks/changes to it.

  1. import json
  3. st1 = &#39;key1 key2=item2 key3=&quot;key3.1, key3.2=item3.2 , key3.3 = item3.3, key3.4&quot; key4&#39;
  4. st1 = st1.replace(&#39; = &#39;, &#39;=&#39;)
  5. st1 = st1.replace(&#39; ,&#39;, &#39;,&#39;)
  6. new_dict = {}
  7. no_keys=False
  9. while not no_keys:
  10. 	st1 = st1.lstrip()
  11. 	
  12. 	if &quot; &quot; in st1:
  13. 		item = st1.split(&quot; &quot;)[0]
  14. 	else:
  15. 		item = st1
  16. 	
  17. 	if &#39;=&#39; in item:
  18. 		if &#39;=&quot;&#39; in item:
  19. 			item = item.split(&#39;=&#39;)[0]
  20. 			new_dict[item] = {}		
  21. 			
  22. 			st1 = st1.replace(f&#39;{item}=&#39;,&#39;&#39;)
  23. 			sub_items = st1.split(&#39;&quot;&#39;)[1]
  24. 			sub_values = sub_items.split(&#39;,&#39;)
  26. 			for sub_item in sub_values:
  27. 				if &quot;=&quot; in sub_item:
  28. 					sub_key, sub_value = sub_item.split(&#39;=&#39;)
  29. 					new_dict[item].update({sub_key.strip():sub_value.strip()})
  30. 				else:
  31. 					new_dict[item].update({sub_item.strip(): None})
  32. 			
  33. 			st1 = st1.replace(f&#39;&quot;{sub_items}&quot;&#39;, &#39;&#39;)
  34. 		else:
  35. 			key, value = item.split(&#39;=&#39;)
  36. 			new_dict.update({key:value})
  37. 			st1 = st1.replace(f&quot;{item} &quot;,&quot;&quot;)
  38. 	else:
  39. 		new_dict.update({item: None})
  40. 		st1 = st1.replace(f&quot;{item}&quot;,&quot;&quot;)
  41. 		
  42. 	if st1 == &quot;&quot;:
  43. 		no_keys=True	
  44. 	
  45. print(json.dumps(new_dict, indent=4))

答案2

得分: 0

考虑使用解析工具，如 lark。对于你的情况，这是一个简单的例子：

  1. _grammar = r&#39;&#39;&#39;
  2.     ?start: value
  3.     
  4.     ?value: object
  5.            | NON_SEPARATOR_STRING?
  7.     object : &quot;\&quot;&quot; [pair (_SEPARATOR pair)*] &quot;\&quot;&quot;
  8.     pair : NON_SEPARATOR_STRING [_PAIRTOR] value
  10.     
  11.     NON_SEPARATOR_STRING: /[a-zA-z0-9\.]+/
  12.     _SEPARATOR: /[,  ]+/
  13.             | &quot;,&quot;
  14.     _PAIRTOR: &quot; = &quot;
  15.             | &quot;=&quot;
  16. &#39;&#39;&#39;
  18. parser = Lark(_grammar)
  20. st1 = &#39;key1 key2=value2 key3=&quot;key3.1, key3.2=value3.2 , key3.3 = value3.3, key3.4&quot; key4&#39;
  22. tree = parser.parse(f&#39;&quot;{st1}&quot;&#39;)
  23. print(tree.pretty())
  25. &quot;&quot;&quot;
  26. object
  27.   pair
  28.     key1
  29.     value
  30.   pair
  31.     key2
  32.     value2
  33.   pair
  34.     key3
  35.     object
  36.       pair
  37.         key3.1
  38.         value
  39.       pair
  40.         key3.2
  41.         value3.2
  42.       pair
  43.         key3.3
  44.         value3.3
  45.       pair
  46.         key3.4
  47.         value
  48.   pair
  49.     key4
  50.     value
  52. &quot;&quot;&quot;

然后，你可以编写自己的 Transformer 来将这个 tree 转换为你想要的日期类型。

英文:

Consider use parsing tool like lark. A simple example to your case:

  1. _grammar = r&#39;&#39;&#39;
  2.     ?start: value
  3.     
  4.     ?value: object
  5.            | NON_SEPARATOR_STRING?
  7.     object : &quot;\&quot;&quot; [pair (_SEPARATOR pair)*] &quot;\&quot;&quot;
  8.     pair : NON_SEPARATOR_STRING [_PAIRTOR] value
  10.     
  11.     NON_SEPARATOR_STRING: /[a-zA-z0-9\.]+/
  12.     _SEPARATOR: /[,  ]+/
  13.             | &quot;,&quot;
  14.     _PAIRTOR: &quot; = &quot;
  15.             | &quot;=&quot;
  16. &#39;&#39;&#39;
  18. parser = Lark(_grammar)
  20. st1 = &#39;key1 key2=value2 key3=&quot;key3.1, key3.2=value3.2 , key3.3 = value3.3, key3.4&quot; key4&#39;
  22. tree = parser.parse(f&#39;&quot;{st1}&quot;&#39;)
  23. print(tree.pretty())
  25. &quot;&quot;&quot;
  26. object
  27.   pair
  28.     key1
  29.     value
  30.   pair
  31.     key2
  32.     value2
  33.   pair
  34.     key3
  35.     object
  36.       pair
  37.         key3.1
  38.         value
  39.       pair
  40.         key3.2
  41.         value3.2
  42.       pair
  43.         key3.3
  44.         value3.3
  45.       pair
  46.         key3.4
  47.         value
  48.   pair
  49.     key4
  50.     value
  52. &quot;&quot;&quot;

Then you can write your own Transformer to transform this tree to your desired date type.

huangapple
  • 本文由 发表于 2023年2月16日 11:15:29
  • 转载请务必保留本文链接：https://go.coder-hub.com/75467500.html
  • python
  • python-regex
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