英文:
Find Rows And Delete it - Pandas DataFrame
问题
这是示例数据帧：
name stuff floats ints0 Mike a 1.0 11 Joey d 2.2 32 Zendaya c NaN 83 John a 1.0 14 Abruzzi d NaN 3
我有一个'to_delete'列表：
[['Abruzzi', 'd', pd.NA, 3], ['Mike', 'a', 1.0, 1]]
我应该使用哪个pandas方法来根据'to_delete'列表删除数据帧中的数据？
这样，我将获得新的数据帧：
name stuff floats ints1 Joey d 2.2 32 Zendaya c NaN 83 John a 1.0 1
谢谢，
*我是pandas的新手
英文:
Example dataframe:name stuff floats ints0 Mike a 1.0 11 Joey d 2.2 32 Zendaya c NaN 83 John a 1.0 14 Abruzzi d NaN 3
I have 'to_delete' list :
[['Abruzzi', 'd', pd.NA, 3], ['Mike', 'a', 1.0, 1]]
How can i remove data in the dataframe based on the 'to_delete' list?
What pandas method suit this?
So i will get new dataframe like:
name stuff floats ints1 Joey d 2.2 32 Zendaya c NaN 83 John a 1.0 1
Thanks,
*im new to pandas
答案1
得分: 3
以下是您要翻译的内容：
I would use a merge with indicator:
keep = (df.merge(pd.DataFrame(to_delete, columns=df.columns), how='left', indicator=True).query('_merge == "left_only"').index)out = df.loc[keep]print(out)
Output:
name stuff floats ints1 Joey d 2.2 32 Zendaya c <NA> 83 John a 1.0 1
英文:
I would use a merge with indicator:
keep = (df.merge(pd.DataFrame(to_delete, columns=df.columns), how='left', indicator=True).query('_merge == "left_only"').index)out = df.loc[keep]print(out)
Output:
name stuff floats ints1 Joey d 2.2 32 Zendaya c <NA> 83 John a 1.0 1
答案2
得分: 0
你可以使用drop函数来删除Pandas DataFrame中的行和列。
你可以使用以下代码来查找行并删除：
import pandas as pddata = pd.DataFrame({'name': ['Mike', 'Joey', 'Zendaya', 'John', 'Abruzzi'],'stuff': ['a', 'd', 'c', 'a', 'd'],'floats': [1.0, 2.2, pd.NA, 1.0, pd.NA],'ints': [1, 3, 8, 1, 3]})to_remove = [['Abruzzi', 'd', pd.NA, 3], ['Mike', 'a', 1.0, 1]]data = data[~data.isin(to_remove)].dropna(how='all')
英文:
You can use the drop function to delete rows and columns in a Pandas DataFrame.
You can use the following code for your help finding the row and delete.
import pandas as pares = pa.DataFrame({'name': ['Mike', 'Joey', 'Zendaya', 'John', 'Abruzzi'],'stuff': ['a', 'd', 'c', 'a', 'd'],'floats': [1.0, 2.2, pa.NA, 1.0, pa.NA],'ints': [1, 3, 8, 1, 3]})remove = [['Abruzzi', 'd', pa.NA, 3], ['Mike', 'a', 1.0, 1]]res = res[~res.isin(remove)].dropna(how='all')
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