首页 > 解决方案 > 如何在熊猫中创建与多列相结合的数据框列

问题描述

我有一些跟踪公司名称随时间变化的数据。但是,我不想将每个名称都更改为一行,而是希望将它们全部连接到一个字段中。

可以使用以下方法构建输入数据:

#Import the modules:
import pandas as pd
import numpy as np

#Create the empty data frame:
df = pd.DataFrame(columns=['dt','old_name','new_name'])

#Populate the data frame:
df.loc[len(df)] = ['01/01/2001', 'AAA', 'BBB']
df.loc[len(df)] = ['02/02/2002', 'BBB', 'CCC']
df.loc[len(df)] = ['03/03/2003', 'CCC', 'DDD']

#View the output:
df

可以使用以下方法创建我希望输出的样子:

#Create the empty data frame:
end_df = pd.DataFrame(columns=['dt','name'])

#Populate:
end_df.loc[len(end_df)] = ['01/01/2001', 'AAA-BBB-CCC-DDD']
end_df.loc[len(end_df)] = ['02/02/2002', 'AAA-BBB-CCC-DDD']
end_df.loc[len(end_df)] = ['03/03/2003', 'AAA-BBB-CCC-DDD']

#View the output:
end_df

编辑:我正在使用 pandas 数据框在 Pyspark2 中运行此代码 - 以防对语法产生任何影响。 此外,我的数据集中有多组名称。我的意思是,有更多的名称更改组与需要连接名称的第一组无关。

样本分组输入:

#Create the empty data frame:
df = pd.DataFrame(columns=['dt','old_name','new_name'])

#Populate the data frame:
df.loc[len(df)] = ['01/01/2001', 'AAA', 'BBB']
df.loc[len(df)] = ['02/02/2002', 'BBB', 'CCC']
df.loc[len(df)] = ['03/03/2003', 'CCC', 'DDD']
df.loc[len(df)] = ['02/01/2001', 'XXX', 'YYY']
df.loc[len(df)] = ['03/02/2002', 'YYY', 'ZZZ']

样本分组输出:

#Create the empty data frame:
end_df = pd.DataFrame(columns=['dt','name'])

#Populate:
end_df.loc[len(end_df)] = ['01/01/2001', 'AAA-BBB-CCC-DDD']
end_df.loc[len(end_df)] = ['02/02/2002', 'AAA-BBB-CCC-DDD']
end_df.loc[len(end_df)] = ['03/03/2003', 'AAA-BBB-CCC-DDD']
end_df.loc[len(end_df)] = ['02/01/2001', 'XXX-YYY-ZZZ']
end_df.loc[len(end_df)] = ['03/02/2002', 'XXX-YYY-ZZZ']

如果您需要任何进一步的说明,请告诉我。

标签: pythonpandaspyspark

解决方案


你需要np.flatten and np.unique

import numpy as np
end_df = pd.DataFrame(columns=['dt','name'])
end_df['dt']=df['dt'].copy()
flat=df[df.columns[1:]].values.flatten()
end_df['name']='-'.join(np.unique(flat))

print(end_df)
    dt          name
0   01/01/2001  AAA-BBB-CCC-DDD
1   02/02/2002  AAA-BBB-CCC-DDD
2   03/03/2003  AAA-BBB-CCC-DDD 

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