Groupby and append lists and strings

Question

I am trying to group-by the values in my \"value_1\" column. But my last column is made up of lists. When I try to group-by using my \"value_1\" column, the column made up o

Answer 1

You could groupby value_1 and aggregate the columns containing strings with the following function:

def str_cat(x):
    return x.str.cat(sep=', ')

And use GroupBy.sum to append the lists in the column list:

df.replace('',None).groupby('value_1').agg({'list':'sum', 'value_2': str_cat,
                                            'value_3': str_cat})

                        list                       value_2  \
value_1                                                              
american  [supermarket, connivence, state]  california, nyc, texas   
canadian             [coffee, sipermarket]          toronto, texas   

                    value_3  
value_1                                 
american  walmart, kmart, dunkinDonuts  
canadian         dunkinDonuts, walmart  

Answer 2

Create dynamically dictionary by all columns with no list and value_1 and for list use lambda function with list comprehension with flatenning:

f1 = lambda x: ', '.join(x.dropna())
#alternative for join only strings
#f1 = lambda x: ', '.join([y for y in x if isinstance(y, str)])
f2 = lambda x: [z for y in x for z in y]
d = dict.fromkeys(df.columns.difference(['value_1','list']), f1)
d['list'] = f2 

df = df.groupby('value_1', as_index=False).agg(d)
print (df)
     value_1                 value_2                value_3  \
0   american  california, nyc, texas         walmart, kmart   
1   canadian                 toronto  dunkinDonuts, walmart   

                               list  
0  [supermarket, connivence, state]  
1             [coffee, supermarket]  

Explanation:

f1 and f2 are lambda functions.

First remove missing values (if exist) and join strings with separator:

f1 = lambda x: ', '.join(x.dropna())

First get only strings values (omit missing values, because NaNs) and join strings with separator:

f1 = lambda x: ', '.join([y for y in x if isinstance(y, str)])

First get all string values with filtering empty strings and join strings with separator:

f1 = lambda x: ', '.join([y for y in x if y != '']) 

Function f2 is for flatten lists, because after aggregation get nested lists like [['a','b'], ['c']]

f2 = lambda x: [z for y in x for z in y]