用Pandas将suds对象转换为数据帧

Question

我有一份清单如下：

`[(deliveryObject){
   id = "0bf003ee0000000000000000000002a11cb6"
   start = 2019-01-02 09:30:00
   messageId = "68027b94b892396ed29581cde9ad07ff"
   status = "sent"
   type = "normal"
   }, (deliveryObject){
   id = "0bf0BE3ABFFDF8744952893782139E82793B"
   start = 2018-12-29 23:00:00
   messageId = "0bc403eb0000000000000000000000113404"
   status = "sent"
   type = "transactional"
 }, (deliveryObject){
   id = "0bf0702D03CB42D848CBB0B0AF023A87FA65"
   start = 2018-12-29 23:00:00
   messageId = "0bc403eb0000000000000000000000113403"
   status = "sent"
   type = "transactional"
   }
]`

当我调用type()时，Python告诉我它是一个列表。在

当我用pd.DataFrame(df)将其转换为数据帧时，结果如下：

有人能帮我吗？dataframe应该有列名，如“Id”、“Start”、“messageId”等，但它们只是作为每个观察的第一个元素出现，列名显示为0、1、2等

感谢任何帮助，谢谢！在

Answer 1

如果这是针对bronto并且正在使用SOAP和suds实现。那么deliverObject就是一个suds对象。在

你能做到的

from suds.client import Client

list_of_deliveryObjects = [(deliveryObject){
   id = "0bf003ee0000000000000000000002a11cb6"
   start = 2019-01-02 09:30:00
   messageId = "68027b94b892396ed29581cde9ad07ff"
   status = "sent"
   type = "normal"
   }, (deliveryObject){
   id = "0bf0BE3ABFFDF8744952893782139E82793B"
   start = 2018-12-29 23:00:00
   messageId = "0bc403eb0000000000000000000000113404"
   status = "sent"
   type = "transactional"
 }, (deliveryObject){
   id = "0bf0702D03CB42D848CBB0B0AF023A87FA65"
   start = 2018-12-29 23:00:00
   messageId = "0bc403eb0000000000000000000000113403"
   status = "sent"
   type = "transactional"
   }
]


data = [Client.dict(suds_object) for suds_object in list_of_deliveryObjects]
df = pd.DataFrame(data)

Answer 2

好吧，这看起来不漂亮，但很管用。 我把你的列表转换成了一个字符串：

import re
import pandas as pd

x = """[(deliveryObject){
   id = "0bf003ee0000000000000000000002a11cb6"
   start = 2019-01-02 09:30:00
   messageId = "68027b94b892396ed29581cde9ad07ff"
   status = "sent"
   type = "normal"
   }, (deliveryObject){
   id = "0bf0BE3ABFFDF8744952893782139E82793B"
   start = 2018-12-29 23:00:00
   messageId = "0bc403eb0000000000000000000000113404"
   status = "sent"
   type = "transactional"
 }, (deliveryObject){
   id = "0bf0702D03CB42D848CBB0B0AF023A87FA65"
   start = 2018-12-29 23:00:00
   messageId = "0bc403eb0000000000000000000000113403"
   status = "sent"
   type = "transactional"
   }
]"""

然后我用regex列出了一个字典列表：

^{pr2}$

现在你有了一个字典列表。第一排看起来像这样

list_of_dict = eval(a)
df = pd.DataFrame(list_of_dict[0])
print(df.head())

                                     id                start                         messageId status    type
0  0bf003ee0000000000000000000002a11cb6  2019-01-02 09:30:00  68027b94b892396ed29581cde9ad07ff   sent  normal

从字典的列表中添加其余的字典

请随意改进我的正则表达式，我知道它看起来很糟糕。在

Answer 3

我这样做了：

import pandas as pd
lst =[{
   'id':"0bf003ee0000000000000000000002a11cb6",
   'start' : "2019-01-02 09:30:00",
   'messageId': "68027b94b892396ed29581cde9ad07ff",
   'status' : "sent",
   'type' : "normal"
   },{
   'id' :  "0bf0BE3ABFFDF8744952893782139E82793B",
   'start' :  "2018-12-29 23:00:00",
   'messageId' :  "0bc403eb0000000000000000000000113404",
   'status' :  "sent",
   'type' :  "transactional"
 }, {
   'id' :  "0bf0702D03CB42D848CBB0B0AF023A87FA65",
   'start' :  "2018-12-29 23:00:00",
   'messageId' :  "0bc403eb0000000000000000000000113403",
   'status' :  "sent",
   'type' :  "transactional"
   }]
df = pd.DataFrame(lst)
df

得到了这个（见附图）：

^{pr2}$

Result