使用pandas从数据框中的字典列表中提取值

发布于 2025-01-24 09:22:21 字数 1834 浏览 0 评论 0原文

我有这个带有4列的数据框。我想在一个单独的列中提取ResourceName（IE ID）。我尝试了各种方法和循环，但无法将其分开。

数据集：

用户名	事件名称	资源
xyz-dev_env_post_function	[	“ I-05FBB7A”}
stopInstances	{“ resourceType”： “	aws :: ec2 :: ec2 ::实例”，“ resourcename” ：：ec2 ::实例“，“ resourceName”：“ I-08BD2475”}，{“ resourceType”：“ aws :: ec2 :: ec2 ::实例”，“ resourcename”：“ i-0fd69dc1”} AWS :: ec2 ::实例“，“ resourceName”：“ I-0174DD38AEA”}]

我想要一个将具有来自Resource列的ID的列ID，LL看起来像这样：

用户名	事件名称	xyz	-
dev_env_env_env_env_env_post_function_function	stopinstances	[{{{{ {{{{{{{{{{{{{{{ “ ResourceType”：“ AWS :: ec2 ::实例”，“ ResourceName”：“ I-05FBB7A”}]	I-05FBB7A“
XYZ-DEV_ENV_ENV_ENV_POST_FUNCTICT_FUNCTICTANCES	[	{“ resourceType）[{“ resourcetype” ResourceName“：” I-08BD2475“}，{“ resourceType”：“ aws :: ec2 :: Instance”，“ resourcename”：“ i-0fd69dc1”}，{“ resourceType” ，“ ResourceName”：“ I-0174DD38AEA”}]	I-08BD2475，I-0FD69DC1，I-0174

这是Data.head（2）.TO_DICT（）：

{'date'：'：'：'：'：'： {0：'28 -02-2022'，1：'28 -02-2022'}， “事件名称”： {0：'stopinstances'，1：'startInstances'}， '资源'： { 0：'[{“ resourceType”：“ aws :: ec2 :: instance”，“ resourcename”：“ i-05fbb7a”}]， 1：'[{“ resourceType”：“ aws :: ec2 :: instance”，“ resourceName”：“ i-08BD2475”}，{“ resourceType”：“ aws :: ec2 :: ec2 ::实例”，“ resourceName”，“ resourceName”：“：” i-0fd69dc1“}，{“ resourceType”：“ aws :: ec2 ::实例”，“ resourcename”：“ I-0174DD38AEA”}]'}， '用户名'：{0：'xyz-dev_env_post_function'，1： 'xyz-dev_env_post_function'}}

谢谢，

原文

I have this dataframe with 4 columns. I want to extract resourceName (i.e IDs ) in one separate column. I tried various methods and loops but unable to seperate it.

Dataset:

Username	Event name	Resources
XYZ-DEV_ENV_POST_function	StopInstances	[{"resourceType":"AWS::EC2::Instance","resourceName":"i-05fbb7a"}]
XYZ-DEV_ENV_POST_function	StartInstances	[{"resourceType":"AWS::EC2::Instance","resourceName":"i-08bd2475"},{"resourceType":"AWS::EC2::Instance","resourceName":"i-0fd69dc1"},{"resourceType":"AWS::EC2::Instance","resourceName":"i-0174dd38aea"}]

I want one more column IDs which will have IDS from Resource column and ll look like this:

Username	Event name	Resources	IDS
XYZ-DEV_ENV_POST_function	StopInstances	[{"resourceType":"AWS::EC2::Instance","resourceName":"i-05fbb7a"}]	i-05fbb7a"
XYZ-DEV_ENV_POST_function	StartInstances	[{"resourceType":"AWS::EC2::Instance","resourceName":"i-08bd2475"},{"resourceType":"AWS::EC2::Instance","resourceName":"i-0fd69dc1"},{"resourceType":"AWS::EC2::Instance","resourceName":"i-0174dd38aea"}]	i-08bd2475 , i-0fd69dc1 , i-0174

Here is output of data.head(2).to_dict():

{'Date':
{0: '28-02-2022', 1: '28-02-2022'},
'Event name':
{0: 'StopInstances', 1: 'StartInstances'},
'Resources':
{
0: '[{"resourceType":"AWS::EC2::Instance","resourceName":"i-05fbb7a"}]',
1: '[{"resourceType":"AWS::EC2::Instance","resourceName":"i-08bd2475"},{"resourceType":"AWS::EC2::Instance","resourceName":"i-0fd69dc1"},{"resourceType":"AWS::EC2::Instance","resourceName":"i-0174dd38aea"}]'},
'User name': {0: 'XYZ-DEV_ENV_POST_function', 1:
'XYZ-DEV_ENV_POST_function'}}

Thanks and Regards

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嗳卜坏 2025-01-31 09:22:21

df['ID'] = df['Resources'].apply(lambda x: ','.join([i['resourceName'] for i in eval(x)]))

         Date  ...                                   ID
0  28-02-2022  ...                            i-05fbb7a
1  28-02-2022  ...  i-08bd2475,i-0fd69dc1,i-0174dd38aea

df['ID'] = df['Resources'].apply(lambda x: ','.join([i['resourceName'] for i in eval(x)]))

         Date  ...                                   ID
0  28-02-2022  ...                            i-05fbb7a
1  28-02-2022  ...  i-08bd2475,i-0fd69dc1,i-0174dd38aea

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