【问题标题】:How can I convert a dictionary-like string to a dictionary?如何将类似字典的字符串转换为字典?
【发布时间】:2021-09-16 07:32:36
【问题描述】:

我有一个类似字符串的字典,如下所示:

str = "Access AR1:\n\tTargets: \n\t\tManagement Name:csw_1\n\t\tObject Name:csw_obj_1\n\t\tdetails:103\n\t\tManagement Name:csw_123\n\t\tObject Name:csw_obj_134\n\t\tdetails:123\n\tSources: \n\t\tIP:10.20.30.40\n\t\tSubnet Mask:255.255.255.255\nAccess AR2:\n\tTargets: \n\t\tManagement Name:csw_2\n\t\tObject Name:csw_obj_2\n\t\tdetails:110\n\t\tManagement Name:csw_431\n\t\tObject Name:csw_obj_21\n\t\tdetails:134\n\tSources: \n\t\tIP:10.20.10.10\n\t\tSubnet Mask:255.255.255.192"

这以相同的格式进行,如下所示:

Access AR1:
    Targets: 
            Management Name:csw_1
            Object Name:csw_obj_1
            details:103
            Management Name:csw_123
            Object Name:csw_obj_134
            details:123
    Sources: 
            IP:10.20.30.40
            Subnet Mask:255.255.255.255
Access AR2:
    Targets: 
            Management Name:csw_2
            Object Name:csw_obj_2
            details:110
            Management Name:csw_431
            Object Name:csw_obj_21
            details:134
    Sources: 
            IP:10.20.10.10
            Subnet Mask:255.255.255.192

这需要改为:

str = {"Access AR1": { "Targets": [{"Management Name:csw_1", "Object Name":"csw_obj_1", "details":"103"}, {"Management Name:csw_123", "Object Name":"csw_obj_134", "details":"123"}]
                      "Sources": {"IP":"10.20.30.40", "Subnet Mask": "255.255.255.255"}
                    },
      "Access AR2": { "Targets": [{"Management Name:csw_2", "Object Name":"csw_obj_2", "details":"110"}, {"Management Name:csw_431", "Object Name":"csw_obj_21", "details":"133"}]
                      "Sources": {"IP":"10.20.10.10", "Subnet Mask": "255.255.255.192"}
                    }
      }

我尝试了ast.literal_eval, eval,结果出现错误。

【问题讨论】:

  • 您能否分享您打算将字符串转换为字典的代码和错误?
  • @DiesanRomero 我使用了 ast.literal_eval(str),它会在给定的字符串空间中引发语法错误并且缺少引号
  • 要使用 ast.literal_eval 字符串需要在括号内。这个方法的作用是评估一个字符串,就好像它是一个 Python 表达式一样。如果您想将该字符串转换为字典,我可以为您提供一些选项。顺便说一句,只是好奇这个字符串是从哪里来的,为什么要使用这种格式?
  • "Sources":"IP":"10.20.30.40" 这不能是 dict 。更正这一行,然后我可以给你从输入输出的代码
  • @BahaeElHmimdi 它的错误,现在编辑

标签: python json dictionary


【解决方案1】:

您的字符串几乎是 yaml - 将 \t 替换为一个或多个空格并将其加载为 YAML。

第一:

pip install pyyaml

那么这段代码就可以工作了:

import yaml
import pprint

str = "Access AR1:\n\tTargets: \n\t\tManagement Name:csw_1\n\t\tObject Name:csw_obj_1\n\t\tdetails:103\n\tSources: \n\t\tIP:10.20.30.40\n\t\tSubnet Mask:255.255.255.255\nAccess AR2:\n\tTargets: \n\t\tManagement Name:csw_2\n\t\tObject Name:csw_obj_2\n\t\tdetails:110\n\tSources: \n\t\tIP:10.20.10.10\n\t\tSubnet Mask:255.255.255.192"

str1 = str.replace( '\t', '    ' )

res = yaml.load(str1)

pprint.pprint( res )

输出:

{'Access AR1': {'Sources': 'IP:10.20.30.40 Subnet Mask:255.255.255.255',
                'Targets': 'Management Name:csw_1 Object Name:csw_obj_1 '
                           'details:103'},
 'Access AR2': {'Sources': 'IP:10.20.10.10 Subnet Mask:255.255.255.192',
                'Targets': 'Management Name:csw_2 Object Name:csw_obj_2 '
                           'details:110'}}

或者,如果你真的想把它作为一个字符串,那么:

str = repr(res)

更新

刚刚意识到,例如'管理名称:csw_1' 未被检测为键:值。需要一个正则表达式 re.sub() 将它们拆分为单独的行来解决这个问题:

import yaml
import pprint
import re

str = "Access AR1:\n\tTargets: \n\t\tManagement Name:csw_1\n\t\tObject Name:csw_obj_1\n\t\tdetails:103\n\tSources: \n\t\tIP:10.20.30.40\n\t\tSubnet Mask:255.255.255.255\nAccess AR2:\n\tTargets: \n\t\tManagement Name:csw_2\n\t\tObject Name:csw_obj_2\n\t\tdetails:110\n\tSources: \n\t\tIP:10.20.10.10\n\t\tSubnet Mask:255.255.255.192"

# replace \t with four-space indent
str1 = str.replace( '\t', '    ' )

# further tweak to split sub-keys like " Management Name:csw_1" onto separate lines
str1 = re.sub(r"^(\s+)(.*?\S:)(\S.*)", r"\1\2\n\1    \3",str1,flags=re.MULTILINE )
   
res = yaml.load(str1)

pprint.pprint( res )

这是调整后的字符串:

Access AR1:
    Targets:
        Management Name:
            csw_1
        Object Name:
            csw_obj_1
        details:
            103
    Sources:
        IP:
            10.20.30.40
        Subnet Mask:
            255.255.255.255
Access AR2:
    Targets:
        Management Name:
            csw_2
        Object Name:
            csw_obj_2
        details:
            110
    Sources:
        IP:
            10.20.10.10
        Subnet Mask:
            255.255.255.192

结果:

{'Access AR1': {'Sources': {'IP': '10.20.30.40',
                            'Subnet Mask': '255.255.255.255'},
                'Targets': {'Management Name': 'csw_1',
                            'Object Name': 'csw_obj_1',
                            'details': 103}},
 'Access AR2': {'Sources': {'IP': '10.20.10.10',
                            'Subnet Mask': '255.255.255.192'},
                'Targets': {'Management Name': 'csw_2',
                            'Object Name': 'csw_obj_2',
                            'details': 110}}}

【讨论】:

  • (+1) 我不知道,但我和yaml.safe_load(StringIO(s.replace('\t', '\s\s\s\s'))) 纠缠了很长时间但想不通,str.replace( '\t', ' ' ) 看起来很棒?
  • 这对我来说很有道理,我正在寻找字符串的来源?。
  • 我努力将正则表达式替换字符串为 dict 字符串(添加引号和括号),没有运气
  • 上面的代码和数据适用于我在 Win10 上使用 Python 3.9.5
  • @nonuser yaml 严格依赖缩进。所以检查你的标签数量。
【解决方案2】:

您好,这是您需要的解决方案

strr = "Access AR1:\n\tTargets: \n\t\tManagement Name:csw_1\n\t\tObject Name:csw_obj_1\n\t\tdetails:103\n\tSources: \n\t\tIP:10.20.30.40\n\t\tSubnet Mask:255.255.255.255\nAccess AR2:\n\tTargets: \n\t\tManagement Name:csw_2\n\t\tObject Name:csw_obj_2\n\t\tdetails:110\n\tSources: \n\t\tIP:10.20.10.10\n\t\tSubnet Mask:255.255.255.192"
nvstr=strr.replace("\n\t\t","-").replace("\n\t","+")
#print(nvstr)
nvdd={}
for u in nvstr.split("\n"):
   #print(u) 
   dts=u.split("+") 
   
   nvdd[dts[0]]={}
   
   for el in dts[1:]:
       
     dts1=el.split("-")
     nvdd[dts[0]][dts1[0][:-1]]={}
     for el1 in dts1[1:]:
         
       k,v=el1.split(":")
       nvdd[dts[0]][dts1[0][:-1]][k]=v
       
print(nvdd)    

【讨论】:

  • 感谢k,v=el1.split(":") 一些错误,顺便说一句,上面建议的解决方案已经足够好了
  • 此方法可能有效,但不可扩展。如果输入有超过 3 个级别怎么办?
  • 您想要可扩展的解决方案吗?
  • 是的。可扩展性是软件设计的一个方面,对吧?不过,我不是 OP。
【解决方案3】:

您可以使用递归将输入转换为字典,而无需事先对字符串本身进行任何调整:

import re
s = "Access AR1:\n\tTargets: \n\t\tManagement Name:csw_1\n\t\tObject Name:csw_obj_1\n\t\tdetails:103\n\tSources: \n\t\tIP:10.20.30.40\n\t\tSubnet Mask:255.255.255.255\nAccess AR2:\n\tTargets: \n\t\tManagement Name:csw_2\n\t\tObject Name:csw_obj_2\n\t\tdetails:110\n\tSources: \n\t\tIP:10.20.10.10\n\t\tSubnet Mask:255.255.255.192"
def to_dict(d):
   k, v, r = None, [], {}
   for *b, a in d:
     if not b:
        if k is not None:
           if not v:
              r[j[0]] = (j:=re.split(':\s*', k))[-1]
           else:
              r[re.split(':\s*', k)[0]] = to_dict(v)
        k, v = a, []
     else:
         v.append([*b[1:], a])
   if k is not None:
       if not v:
          r[j[0]] = (j:=re.split(':\s*', k))[-1]
       else:
          r[re.split(':\s*', k)[0]] = to_dict(v)
   return r

import json       
new_s = [re.findall('\t|[^\t]+$', i) for i in s.split('\n')]
print(json.dumps(to_dict(new_s), indent=4))

输出:

{
    "Access AR1": {
        "Targets": {
            "Management Name": "csw_1",
            "Object Name": "csw_obj_1",
            "details": "103"
        },
        "Sources": {
            "IP": "10.20.30.40",
            "Subnet Mask": "255.255.255.255"
        }
    },
    "Access AR2": {
        "Targets": {
            "Management Name": "csw_2",
            "Object Name": "csw_obj_2",
            "details": "110"
        },
        "Sources": {
            "IP": "10.20.10.10",
            "Subnet Mask": "255.255.255.192"
        }
    }
}

【讨论】:

  • 我确信你的代码可以工作,但我个人觉得它难以理解;我很感激一些 cmets 来解释发生了什么。以及更好的变量名;例如d 并不意味着任何有用的东西,abv 也不是
猜你喜欢
  • 2020-10-23
  • 2021-05-16
  • 1970-01-01
  • 2021-07-25
  • 2019-11-09
  • 1970-01-01
  • 1970-01-01
  • 1970-01-01
  • 2021-08-21
相关资源
最近更新 更多