Question

我有一些数据可以跟踪公司名称随时间的变化。但是，我不想将每个名称都放在一行上，而是希望将它们全部串联在一个字段中。

输入数据可以使用：

#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输出的外观：

#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-以防对语法造成任何影响。而且，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第一个名称需要连接的组无关。

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']

让我知道是否需要进一步说明。

Answer 1

您需要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

Answer 2

创建了两个dicts：old_new_dict从旧名称遍历到新名称和old_new_dict_rev从新名称遍历到旧名称：

old_new_dict = {k:v for k,v in zip(df.old_name,df.new_name)}          
old_new_dict_rev = {v:k for k,v in zip(df.old_name,df.new_name)}

函数find_tree，在两个方向上遍历并将它们结合在一起以创建名称的完整路径。

def find_tree(name):
    left_list = []
    right_list = []
    name_l, name_r = name, name

    while(name_l in old_new_dict_rev):
        left_list.append(old_new_dict_rev[name_l])
        name_l = old_new_dict_rev[name_l]
    left_list.reverse()

    while(name_r in old_new_dict):
        right_list.append(old_new_dict[name_r])
        name_r = old_new_dict[name_r]

    return "-".join(left_list + [name] + right_list)

将完整路径添加为数据帧name中的df列：

df['name'] = df['old_name'].apply(lambda x: find_tree(x))
end_df = df.drop(['old_name','new_name'], axis = 1)

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
#3  02/01/2001      XXX-YYY-ZZZ
#4  03/02/2002      XXX-YYY-ZZZ

如何在熊猫中创建与多个列组合的数据框列

2 个答案: