如何将 groupby object 转换为列表列表和 append 将新列/值转换为列表中的列表

Question

I have the following sample df

import pandas as pd

list_of_customers =[
[202206,'patrick','lemon','fruit','citrus',10,'tesco'],
[202206,'paul','lemon','fruit','citrus',20,'tesco'],
[202206,'frank','lemon','fruit','citrus',10,'tesco'],
[202206,'jim','lemon','fruit','citrus',20,'tesco'], 
[202206,'wendy','watermelon','fruit','',39,'tesco'],
[202206,'greg','watermelon','fruit','',32,'sainsburys'],
[202209,'wilson','carrot','vegetable','',34,'sainsburys'],    
[202209,'maree','carrot','vegetable','',22,'aldi'],
[202209,'greg','','','','','aldi'], 
[202209,'wilmer','sprite','drink','',22,'aldi'],
[202209,'jed','lime','fruit','citrus',40,'tesco'],    
[202209,'michael','lime','fruit','citrus',12,'aldi'],
[202209,'andrew','','','','33','aldi'], 
[202209,'ahmed','lime','fruit','fruit',33,'aldi'] 
]

df = pd.DataFrame(list_of_customers,columns = ['date','customer','item','item_type','fruit_type','cost','store'])

(df)

I then define variable for each category we need to aggregate

fruit_variable = df['item_type'].isin(['fruit'])

vegetable_variable = df['item_type'].isin(['vegetable'])

citrus_variable = df['fruit_type'].isin(['citrus'])

I then want to aggregate each variable and merge them into one dataframe. For each variable I want to have a separate field (variable_number) that has a number assigned to each, so we know what variable rule was used for aggregation. So for fruit_variable the field will be '01', vegetable variable will be '02' and so on. Note we can't assign a new field with each variable and include it in the grouby fields as there are rows that would not be mutually exclusive (ie rows need to aggregate for both the fruit_variable and citrus_variable).

list_agg = df.where(fruit_variable).groupby(['date','store'])[['cost']].sum().reset_index().agg(list),
df.where(vegetable_variable).groupby(['date','store'])[['cost']].sum().reset_index().agg(list),
df.where(citrus_variable).groupby(['date','store'])[['cost']].sum().reset_index().agg(list)

print(list_agg)
type(list_agg)

df_agg = pd.DataFrame(list_Agg, columns = ['date','store','cost'])
print(df_agg)

I am having trouble converting the tuple to a dataframe.

I can convert the groupby object's to lists using.to_records().tolist() but it still leaves me the problem of how to add the new row with the variable number.

Note this is a much smaller subset of the actual problem. I am hoping to get a dataframe looking like below in this example:

Please let me know if any further information is required.

EDIT1:

I'm adding an additional variable to the example so that the groupby includes a value which will return a null cost. Rather than having an incremental variable_number how can we define it so that the variable numbers are predefined (fruit variable is '01', vegetable_variable is '02'citrus_variable is '01a', meat_variable is '03')

fruit_variable = df['item_type'].isin(['fruit'])

vegetable_variable = df['item_type'].isin(['vegetable'])

citrus_variable = df['fruit_type'].isin(['citrus'])

meat_variable = df['item_type'].isin(['poultry'])


list_agg = [df.where(fruit_variable).groupby(['date','store'])[['cost']].sum().reset_index().agg(list),
            df.where(vegetable_variable).groupby(['date','store'])[['cost']].sum().reset_index().agg(list),
            df.where(citrus_variable).groupby(['date','store'])[['cost']].sum().reset_index().agg(list),
            df.where(meat_variable).groupby(['date','store'])[['cost']].sum().reset_index().agg(list)]

out = (pd.concat(list_agg, keys=[f'{v+1:02}' for v in range(len(list_agg))])
         .rename_axis(['variable_number', None])
         .reset_index('variable_number').reset_index(drop=True))

We are then looking for output without having a failure for the meat_variable which we return no values in the sample dataset

Answer 1

IIUC, you can use concat :

list_agg = [df.where(fruit_variable).groupby(['date','store'])[['cost']].sum().reset_index().agg(list),
            df.where(vegetable_variable).groupby(['date','store'])[['cost']].sum().reset_index().agg(list),
            df.where(citrus_variable).groupby(['date','store'])[['cost']].sum().reset_index().agg(list)]

out = (pd.concat(list_agg, keys=[f'{v+1:02}' for v in range(len(list_agg))])
         .rename_axis(['variable_number', None])
         .reset_index('variable_number').reset_index(drop=True))

Output:

>>> out
  variable_number      date       store  cost
0              01  202206.0  sainsburys    32
1              01  202206.0       tesco    99
2              01  202209.0        aldi    45
3              01  202209.0       tesco    40
4              02  202209.0        aldi    22
5              02  202209.0  sainsburys    34
6              03  202206.0       tesco    60
7              03  202209.0        aldi    12
8              03  202209.0       tesco    40

Answer 2

The exact logic is unclear, but you might want to use concat with a list comprehension of groupby.agg :

variables = {'01': df['item_type'].isin(['fruit']),
             '02': df['item_type'].isin(['vegetable']),
             '03': df['fruit_type'].isin(['citrus']),
            }

out = (pd.concat({k: df[m].groupby(['date', 'store'], as_index=False)['cost'].sum()
                  for k, m in variables.items()}, names=['variable_number', None])
         .reset_index('variable_number')
      )

print(out)

Output:

  variable_number    date       store  cost
0              01  202206  sainsburys    32
1              01  202206       tesco    99
2              01  202209        aldi    45
3              01  202209       tesco    40
0              02  202209        aldi    22
1              02  202209  sainsburys    34
0              03  202206       tesco    60
1              03  202209        aldi    12
2              03  202209       tesco    40