Python确保基于列值的每一行都有一组数据，如果没有add的话

Question

我正在组织用于标记的AWS资源，并已将数据捕获到CSV文件中。CSV文件的输出示例如下。我试图确保每个资源的\u id都有一个标记\u key的数据集，我需要确保它存在。此数据集是

标记\u键

Application
Client
Environment
Name
Owner
Project
Purpose

我刚接触熊猫，我只把CSV文件读作数据帧

import pandas as pd

file_name = "z.csv"

df = pd.read_csv(file_name, names=['resource_id', 'resource_type', 'tag_key', 'tag_value'])

print (df)

CSV文件

vol-00441b671ca48ba41,volume,Environment,Development
vol-00441b671ca48ba41,volume,Name,Database Files
vol-00441b671ca48ba41,volume,Project,Application Development
vol-00441b671ca48ba41,volume,Purpose,Web Server
i-1234567890abcdef0,instance,Environment,Production
i-1234567890abcdef0,instance,Owner,Fast Company

我希望输出如下

vol-00441b671ca48ba41,volume,Environment,Development
vol-00441b671ca48ba41,volume,Name,Database Files
vol-00441b671ca48ba41,volume,Project,Application Development
vol-00441b671ca48ba41,volume,Purpose,Web Server
vol-00441b671ca48ba41,volume,Client,
vol-00441b671ca48ba41,volume,Owner,
vol-00441b671ca48ba41,volume,Application,
i-1234567890abcdef0,instance,Environment,Production
i-1234567890abcdef0,instance,Owner,Fast Company
i-1234567890abcdef0,instance,Application,
i-1234567890abcdef0,instance,Client,
i-1234567890abcdef0,instance,Name,
i-1234567890abcdef0,instance,Project,
i-1234567890abcdef0,instance,Purpose,

Answer 1

一种方法是使用多索引from_product和renindex：

taglist = ['Application',
           'Client',
           'Environment',
           'Name',
           'Owner',
           'Project',
           'Purpose']

df_out = df.set_index(['resource_id','tag_key'])\
           .reindex(pd.MultiIndex.from_product([df['resource_id'].unique(), taglist],
                                              names=['resource_id','tag_key']))

df_out.assign(resource_type = df_out.groupby('resource_id')['resource_type']\
                                    .ffill().bfill()).reset_index()

输出：

              resource_id      tag_key resource_type                tag_value
0   vol-00441b671ca48ba41  Application        volume                      NaN
1   vol-00441b671ca48ba41       Client        volume                      NaN
2   vol-00441b671ca48ba41  Environment        volume              Development
3   vol-00441b671ca48ba41         Name        volume           Database Files
4   vol-00441b671ca48ba41        Owner        volume                      NaN
5   vol-00441b671ca48ba41      Project        volume  Application Development
6   vol-00441b671ca48ba41      Purpose        volume               Web Server
7     i-1234567890abcdef0  Application      instance                      NaN
8     i-1234567890abcdef0       Client      instance                      NaN
9     i-1234567890abcdef0  Environment      instance               Production
10    i-1234567890abcdef0         Name      instance                      NaN
11    i-1234567890abcdef0        Owner      instance             Fast Company
12    i-1234567890abcdef0      Project      instance                      NaN
13    i-1234567890abcdef0      Purpose      instance                      NaN

Answer 2

举个稍微简单一点的例子。我有数据帧df:

df = pd.DataFrame(data={'a': [1, 1, 2, 2], 'b': [[1, 2], [3, 5], [1, 2], [5]]})

返回

   a       b
0  1  [1, 2]
1  1  [3, 5]
2  2  [1, 2]
3  2     [5]

有要求的b:1、2、3、4和5。你知道吗

然后我们需要找出我们已经拥有了什么。我们这样做：

def flatten(lsts):
    return [j for i in lsts for j in i]

df_new = df.groupby(by=['a'])['b'].apply(flatten)

退货：

a
1    [1, 2, 3, 5]
2       [1, 2, 5]

现在我们需要列出缺少的列并添加这些列：

df_new = df_new.reset_index()
lst_wanted = [1, 2, 3, 4, 5]

for row in df_new.itertuples():
    for j in lst_wanted:
        if j not in row.b:
            df = df.append({'a': row.a, 'b': j}, ignore_index=True)
print(df)

返回：

   a       b
0  1  [1, 2]
1  1  [3, 5]
2  2  [1, 2]
3  2     [5]
4  1       4
5  2       3
6  2       4