过滤映射值

Question

抱歉，新来的。

我在用熊猫来看地板。映射数据集中的列。我将根据映射的值进行筛选，只返回符合我条件的行。

我的数据如下所示:列: UUID，UUID_c，approvalTimestamp，approvalTimestamp列看起来如下(并且是一个对象数据类型)：

[('US', 'IB'), ('EU', 'IA'), ('CA', 'IIC'), ('CH', 'III'), ('UK', 'IA'), ('AU', 'IB'), ('TW', 'III'), ('TK', 'IV')]

我想过滤美国的价值，无论是"IA“或"IB”。

这将返回地图中"US“的所有实例：

df2 = df[df['Rating'].str.contains("US")]

这将返回一个空的dataframe：

df2 = df[df['Rating'].str.contains("IA")]

如何返回分配给美国的值为"IA“或"IB”的实例？

dataframe看起来像：

UUID  |  UUID_c  |   Rating   |  approvalTimeStamp|
---------------------------------------------------
037a9db2-c91f-4e93-a36e-3b6e7adb885f   |   ['8b2c409b-6c01-0100-2d32-670000010368','1fdfa790-a001-0100-5efe-b90000060013'] | [('US', 'IB'), ('EU', 'IA'), ('CA', 'IIC'), ('CH', 'III'), ('UK', 'IA'), ('AU', 'IB'), ('TW', 'III'), ('TK', 'IV')]   |   2022-01-06T19:10:46.304734Z
037a9db2-c91f-4e93-a36e-3b6e7adb885f   |   ['8b2c409b-6c01-0100-2d32-670000010368','691aa282-e1ec-4904-b6c3-18a20ba3cda2'] | [('US', 'IIC'), ('EU', 'IA'), ('CA', 'IIC'), ('CH', 'III'), ('UK', 'IA'), ('AU', 'IB'), ('TW', 'III'), ('TK', 'IV')]   |   2022-01-06T19:10:46.304734Z
037a9db2-c91f-4e93-a36e-3b6e7adb885f   |   ['8b2c409b-6c01-0100-2d32-670000010368','eb8d409b-6c01-0100-0f90-bd0000410011'] | [('US', 'IA'), ('EU', 'IA'), ('CA', 'IIC'), ('CH', 'III'), ('UK', 'IA'), ('AU', 'IB'), ('TW', 'III'), ('TK', 'IV')]   |   2022-01-06T19:10:46.304734Z

想要返回这个：(装修出美国，IIC行)

 UUID  |  UUID_c  |   Rating   |  approvalTimeStamp|
    ---------------------------------------------------
    037a9db2-c91f-4e93-a36e-3b6e7adb885f   |   ['8b2c409b-6c01-0100-2d32-670000010368','1fdfa790-a001-0100-5efe-b90000060013'] | [('US', 'IB'), ('EU', 'IA'), ('CA', 'IIC'), ('CH', 'III'), ('UK', 'IA'), ('AU', 'IB'), ('TW', 'III'), ('TK', 'IV')]   |   2022-01-06T19:10:46.304734Z
    037a9db2-c91f-4e93-a36e-3b6e7adb885f   |   ['8b2c409b-6c01-0100-2d32-670000010368','eb8d409b-6c01-0100-0f90-bd0000410011'] | [('US', 'IA'), ('EU', 'IA'), ('CA', 'IIC'), ('CH', 'III'), ('UK', 'IA'), ('AU', 'IB'), ('TW', 'III'), ('TK', 'IV')]   |   2022-01-06T19:10:46.304734Z

Answer 1

这可能有助于：

df = pd.DataFrame({'id': ['037a9db2-c91f-4e93-a36e-3b6e7adb885f','037a9db2-c91f-4e93-a36e-3b6e7adb885f','037a9db2-c91f-4e93-a36e-3b6e7adb885f'],
                   'Rating' : [[('US', 'IB'), ('EU', 'IA'), ('CA', 'IIC'), ('CH', 'III'), ('UK', 'IA'), ('AU', 'IB'), ('TW', 'III'), ('TK', 'IV')],[('US', 'IIC'), ('EU', 'IA'), ('CA', 'IIC'), ('CH', 'III'), ('UK', 'IA'), ('AU', 'IB'), ('TW', 'III'), ('TK', 'IV')],[('US', 'IA'), ('EU', 'IA'), ('CA', 'IIC'), ('CH', 'III'), ('UK', 'IA'), ('AU', 'IB'), ('TW', 'III'), ('TK', 'IV')]]})
df

    id                                      Rating
0   037a9db2-c91f-4e93-a36e-3b6e7adb885f    [(US, IB), (EU, IA), (CA, IIC), (CH, III), (UK...
1   037a9db2-c91f-4e93-a36e-3b6e7adb885f    [(US, IIC), (EU, IA), (CA, IIC), (CH, III), (U...
2   037a9db2-c91f-4e93-a36e-3b6e7adb885f    [(US, IA), (EU, IA), (CA, IIC), (CH, III), (UK...

Lambda功能：

lst = df.apply(lambda row : True if ('US','IIC') not in row['Rating'] else False, axis= 1)
df[lst]

结果

    id                                      Rating
0   037a9db2-c91f-4e93-a36e-3b6e7adb885f    [(US, IB), (EU, IA), (CA, IIC), (CH, III), (UK...
2   037a9db2-c91f-4e93-a36e-3b6e7adb885f    [(US, IA), (EU, IA), (CA, IIC), (CH, III), (UK...