关于字符串中有字典格式的怎么能转换成字典

cdzjf

比如这样的字符串
e=[{"id":"11","name":"北京","code":"11"},
                     {"id":"18","name":"河北","code":"18"},
                     {"id":"38","name":"福建","code":"38"},
                     {"id":"75","name":"江西","code":"75"},
                     {"id":"17","name":"山东","code":"17"},
                     {"id":"76","name":"河南","code":"76"},
                     {"id":"71","name":"湖北","code":"71"},
                     {"id":"74","name":"湖南","code":"74"},
                     {"id":"51","name":"广东","code":"51"},
                     {"id":"59","name":"广西","code":"59"},
                     {"id":"13","name":"天津","code":"13"},
                     {"id":"50","name":"海南","code":"50"},
                     {"id":"83","name":"重庆","code":"83"},
                     {"id":"81","name":"四川","code":"81"},
                     {"id":"19","name":"山西","code":"19"},
                     {"id":"85","name":"贵州","code":"85"},
                     {"id":"86","name":"云南","code":"86"},
                     {"id":"79","name":"西藏","code":"79"},
                     {"id":"84","name":"陕西","code":"84"},
                     {"id":"87","name":"甘肃","code":"87"},
                     {"id":"70","name":"青海","code":"70"},
                     {"id":"88","name":"宁夏","code":"88"},
                     {"id":"89","name":"新疆","code":"89"},
                     {"id":"10","name":"内蒙古","code":"10"},
                     {"id":"91","name":"辽宁","code":"91"},
                     {"id":"90","name":"吉林","code":"90"},
                     {"id":"97","name":"黑龙江","code":"97"},
                     {"id":"31","name":"上海","code":"31"},
                     {"id":"34","name":"江苏","code":"34"},
                     {"id":"36","name":"浙江","code":"36"},
                     {"id":"30","name":"安徽","code":"30"}];

最外层是个字符串加列表，列表里面其实是字符串，如果先用正则去掉后大约是
[('{"id":"11","name":"北京","code":"11"},\r\n\t             {"id":"18","name":"河北","code":"18"},\r\n\t       。。。。。。这样的格式，这个怎么能把里面的字典格式都拿出来保存成一个列表加字典格式呢

qq1151985918

1. for i in eval(e):
   
2.     print(i)

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Daniel_Zhang

1. >>> dict1 = "{'a':'b'}"
   
2. >>> type(dict1)
   
3. <class 'str'>
   
4. >>> dict2 = eval(dict1)
   
5. >>> dict2
   
6. {'a': 'b'}
   
7. >>> type(dict2)
   
8. <class 'dict'>
   
9. >>>
   
10. >>> dict2['a']
    
11. 'b'

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z5560636

题主的问题很有意思，学习了：
eval 貌似只能进行一重转换，并且，如果有异常的符号的话，是会报错的
给出测试代码：

1. str = """[('{"id":"11","name":"北京","code":"11"},\r\n\t             {"id":"18","name":"河北","code":"18"}\r\n\t')]"""
   
2. str = str.replace("\r\n\t","")
   
3. str = str.replace(" ","")
   
4. list_a = eval(str)
   
5. 
   
6. for items in list_a:
   
7.     items = eval(items)
   
8.     for item in items:
   
9.         print(type(item))
   
10.         print(item)

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傻眼貓咪

1. class Infomation:
   
2.     def __init__(self, id, code):
   
3.         self.id = id
   
4.         self.code = code
   
5. 
   
6. location = dict() # 创建字典
   
7. 
   
8. e = [
   
9.     {"id":"11","name":"北京","code":"11"},
   
10.     {"id":"18","name":"河北","code":"18"},
    
11.     {"id":"38","name":"福建","code":"38"},
    
12.     {"id":"75","name":"江西","code":"75"},
    
13.     {"id":"17","name":"山东","code":"17"},
    
14.     {"id":"76","name":"河南","code":"76"},
    
15.     {"id":"71","name":"湖北","code":"71"},
    
16.     {"id":"74","name":"湖南","code":"74"},
    
17.     {"id":"51","name":"广东","code":"51"},
    
18.     {"id":"59","name":"广西","code":"59"},
    
19.     {"id":"13","name":"天津","code":"13"},
    
20.     {"id":"50","name":"海南","code":"50"},
    
21.     {"id":"83","name":"重庆","code":"83"},
    
22.     {"id":"81","name":"四川","code":"81"},
    
23.     {"id":"19","name":"山西","code":"19"},
    
24.     {"id":"85","name":"贵州","code":"85"},
    
25.     {"id":"86","name":"云南","code":"86"},
    
26.     {"id":"79","name":"西藏","code":"79"},
    
27.     {"id":"84","name":"陕西","code":"84"},
    
28.     {"id":"87","name":"甘肃","code":"87"},
    
29.     {"id":"70","name":"青海","code":"70"},
    
30.     {"id":"88","name":"宁夏","code":"88"},
    
31.     {"id":"89","name":"新疆","code":"89"},
    
32.     {"id":"10","name":"内蒙古","code":"10"},
    
33.     {"id":"91","name":"辽宁","code":"91"},
    
34.     {"id":"90","name":"吉林","code":"90"},
    
35.     {"id":"97","name":"黑龙江","code":"97"},
    
36.     {"id":"31","name":"上海","code":"31"},
    
37.     {"id":"34","name":"江苏","code":"34"},
    
38.     {"id":"36","name":"浙江","code":"36"},
    
39.     {"id":"30","name":"安徽","code":"30"}]
    
40. 
    
41. for i in e:
    
42.     location[i["name"]] = Infomation(i["id"], i["code"]) # 将资料输入字典
    
43. 
    
44. # 范例
    
45. x = "山东"
    
46. print(x, location[x].id, location[x].code)
    
47. 
    
48. y = "四川"
    
49. print(y, location[y].id, location[y].code)
    
50. 
    
51. z = "上海"
    
52. print(z, location[z].id, location[z].code)

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1. 山东 17 17
   
2. 四川 81 81
   
3. 上海 31 31

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阿奇_o

1. 
   
2. #按你描述，和我的推测，你要处理的应该是 JSON字符串，那就json模块，若进一步处理（如数据分析），那直接 pandas 构建DF
   
3. 
   
4. >>> e='[{"id":"11","name":"北京","code":"11"},{"id":"18","name":"河北","code":"18"}, {"id":"38","name":"福建","code":"38"}]'
   
5. >>> import json
   
6. >>> import pandas as pd
   
7. >>> pd.DataFrame(data=json.loads(e))
   
8.    id name code
   
9. 0  11   北京   11
   
10. 1  18   河北   18
    
11. 2  38   福建   38
    
12. >>>

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cdzjf

qq1151985918 发表于 2021-10-5 21:57

这不是一个简单的列表，最外层的列表里面全是一个一个字符串