从另一个带有熊猫代码列的数据框中获取描述答案

【问题标题】：get description from another data frame with code column in pandas从另一个带有熊猫代码列的数据框中获取描述
【发布时间】：2021-06-08 07:45:55
【问题描述】：

输入：

df1 = pd.DataFrame([[101, 'DC1', 'QA,DEMO'],
               [101, 'EM5', 'QA,DEMO'],
               [102, 'RA6', 'PA,QA,GF'],
               [103, 'DC1', 'AB,LK'],
               [103, 'RA6', 'OO'],
               [103, 'PA4', 'AB,OO']
              ]
              , columns=['Call_id', 'Agent_id', 'Task_code'])

df2 = pd.DataFrame([['QA', 'Enter phone number'],
               ['DEMO', 'ASK EMAIL ID'],
               ['PA', 'make notes'],
               ['GF', 'take call back'],
               ['AB', 'Apply bell mark'],
               ['LK', 'call Low markets'],
               ['OO','Out of order']
              ]
              , columns=['Task_code', 'Task_Desc'])

输出：

df = pd.DataFrame([[101, 'DC1', 'QA,DEMO','Enter phone number and ASK EMAIL ID'],
               [101, 'EM5', 'QA,DEMO','Enter phone number and ASK EMAIL ID'],
               [102, 'RA6', 'PA,QA,GF','make notes and Enter phone number and take call back'],
               [103, 'DC1', 'AB,LK','Apply bell mark and call Low markets'],
               [103, 'RA6', 'OO','Out of order'],
               [103, 'PA4', 'AB,OO','Apply bell mark and Out of order']
              ]
              , columns=['Call_id', 'Agent_id', 'Task_code','Task_desc'])

我想在 Python 3.6 中为每个代理 id 和 call_id 将 TASK_desc 与 df1 上的“和”合并

【问题讨论】：

标签： python python-3.x pandas

【解决方案1】：

task_mapper = dict(zip(df2["Task_code"],df2["Task_Desc"]))

def get_description(values):

    values = values.strip().split(",")
    values = list(filter(None,values))
    descriptions = [task_mapper.get(value,"") for value in values]
    return " and ".join(descriptions)

df1["Task_desc"] = df1["Task_code"].apply(get_description)

print(df1) # your output

【讨论】：

【解决方案2】：

您可以创建字典以通过, 映射拆分值，然后通过and 分隔符连接值：

d = df2.set_index('Task_code')['Task_Desc']

f = lambda x:  ' and '.join(d.get(y) for y in x.split(',') if y in d)

df1['Task_Desc'] = df1['Task_code'].apply(f)
print (df1)
   Call_id Agent_id Task_code  \
0      101      DC1   QA,DEMO   
1      101      EM5   QA,DEMO   
2      102      RA6  PA,QA,GF   
3      103      DC1     AB,LK   
4      103      RA6        OO   
5      103      PA4     AB,OO   

                                           Task_Desc  
0                Enter phone number and ASK EMAIL ID  
1                Enter phone number and ASK EMAIL ID  
2  make notes and Enter phone number and take cal...  
3               Apply bell mark and call Low markets  
4                                       Out of order  
5                   Apply bell mark and Out of order

或者使用我们的 pandas 函数来分解列，使用不同的分隔符合并和聚合 join：

f1 = lambda x: ' and '.join(x.unique())
f2 = lambda x: ','.join(x.unique())
df = (df1.assign(Task_code = df1['Task_code'].str.split(','))
         .explode('Task_code')
         .merge(df2, on='Task_code')
         .groupby(['Call_id','Agent_id'])
         .agg({'Task_code':f2, 'Task_Desc':f1})
         .reset_index())
print (df)
   Call_id Agent_id Task_code  \
0      101      DC1   QA,DEMO   
1      101      EM5   QA,DEMO   
2      102      RA6  QA,PA,GF   
3      103      DC1     AB,LK   
4      103      PA4     AB,OO   
5      103      RA6        OO   

                                           Task_Desc  
0                Enter phone number and ASK EMAIL ID  
1                Enter phone number and ASK EMAIL ID  
2  Enter phone number and make notes and take cal...  
3               Apply bell mark and call Low markets  
4                   Apply bell mark and Out of order  
5                                       Out of order

【讨论】：

【解决方案3】：

这是我的方式：拆分成列表

df1['desc'] = df1.Task_code.str.split(',')

为查找制作字典

dictionary = dict(zip(df2.Task_code,df2.Task_Desc))
def descdic(l):
    return ' and '.join([dictionary[i] for i in l])

应用于数据

df1.desc = df1.desc.apply(lambda x: descdic(x))

【讨论】：