Pandas- 连接两列字符串列表

Question

I've got an.. interesting data frame that comes from a database. The data frame has two columns, which are lists of strings. I need to concat the values in these two lists, to create a new column of lists. For example:

data = [ 
    {'id': 1, 'l1': ['Luke', 'Han'], 'l2': ['Skywalker', 'Solo']}, 
    {'id': 2, 'l1': ['Darth', 'Kylo'], 'l2': ['Vader', 'Ren']},
    {'id': 3, 'l1': [], 'l2': []}
]                                                                                                                                                                
df = pd.DataFrame(data)   

Notice the third row has no values. You can also assume that l1 and l2 are of the same length. And I need to concat the values in l1 and l2 (with a space between), eg:

result = [
    {'id': 1, 'name': ['Luke Skywalker', 'Han Solo']},
    {'id': 2, 'name': ['Darth Vader', 'Kylo Ren']},
    {'id': 3, 'name': []}
]
result_df = pd.DataFrame(result)

Answer 1

You you use dict comprehension and ' '.join in combination with zip to iterate over your dataset, for example, this:

import pandas as pd


data = [ 
    {'id': 1, 'l1': ['Luke', 'Han'], 'l2': ['Skywalker', 'Solo']}, 
    {'id': 2, 'l1': ['Darth', 'Kylo'], 'l2': ['Vader', 'Ren']},
    {'id': 3, 'l1': [], 'l2': []}
]                                                                                                                                                                
df = pd.DataFrame(data) 

result = [
    {
        'id': row['id'], 
        'name': [' '.join(l1_l2) for l1_l2 in zip(row['l1'], row['l2'])]
    } for row in data
]

print(pd.DataFrame(result))
>>>
   id                        name
0   1  [Luke Skywalker, Han Solo]
1   2     [Darth Vader, Kylo Ren]
2   3                          []

Answer 2

This should get you to where you want: assuming you only have two columns (if you have more just add one of those ' '+df.iloc[j,3 &or 4 &or...][i])

Voila =[]
for j in range(len(df)):
    Voila.append([df.iloc[j,1][i]+ ' '+df.iloc[j,2][i] for i in range(len(df. 
                                                                loc[j,'l1']))])
df['Voila'] = Voila