将一列字典列表转换为一个列列表，以便从列表中每个字典下的键“name”派生值

Question

The input column has a variable number of dictionary lists, it is not fixed.

INPUT column:

Facilities
[{'name': 'Work from home', 'icon': 'WFH.svg'}]
[{'name': 'Gymnasium', 'icon': 'Gym.svg'}, {'name': 'Cafeteria', 'icon': 'Cafeteria.svg'}, {'name': 'Work from home', 'icon': 'WFH.svg'}]
[{'name': 'Free food', 'icon': 'FreeFood.svg'}, {'name': 'Team outings', 'icon': 'TeamOuting.svg'}, {'name': 'Education assistance', 'icon': 'Education.svg'}]
[{'name': 'Soft skill training', 'icon': 'SoftSkillsTraining.svg'}, {'name': 'Job training', 'icon': 'JobTraining.svg'}]
[{'name': 'Free transport', 'icon': 'Transportation.svg'}, {'name': 'Work from home', 'icon': 'WFH.svg'}, {'name': 'Team outings', 'icon': 'TeamOuting.svg'}, {'name': 'Soft skill training', 'icon': 'SoftSkillsTraining.svg'}]

This above input should be filtered so that the column will only have a single list with all the values of keys "name" collected from different dictionaries in the list.

Desired Output column:

Facilities
['Work from home']
['Gymnasium', 'Cafeteria', 'Work from home']
['Free food','Team outings','Education assistance']
['Soft skill training','Job training']
['Free transport', 'Work from home','Team outings','Soft skill training']

Answer 1

Suppose you have this DataFrame:

df = pd.DataFrame({'Facilities':[
[{'name': 'Work from home', 'icon': 'WFH.svg'}],
[{'name': 'Gymnasium', 'icon': 'Gym.svg'}, {'name': 'Cafeteria', 'icon': 'Cafeteria.svg'}, {'name': 'Work from home', 'icon': 'WFH.svg'}],
[{'name': 'Free food', 'icon': 'FreeFood.svg'}, {'name': 'Team outings', 'icon': 'TeamOuting.svg'}, {'name': 'Education assistance', 'icon': 'Education.svg'}],
[{'name': 'Soft skill training', 'icon': 'SoftSkillsTraining.svg'}, {'name': 'Job training', 'icon': 'JobTraining.svg'}],
[{'name': 'Free transport', 'icon': 'Transportation.svg'}, {'name': 'Work from home', 'icon': 'WFH.svg'}, {'name': 'Team outings', 'icon': 'TeamOuting.svg'}, {'name': 'Soft skill training', 'icon': 'SoftSkillsTraining.svg'}],
    ]})

print(df)

                                          Facilities
0    [{'name': 'Work from home', 'icon': 'WFH.svg'}]
1  [{'name': 'Gymnasium', 'icon': 'Gym.svg'}, {'n...
2  [{'name': 'Free food', 'icon': 'FreeFood.svg'}...
3  [{'name': 'Soft skill training', 'icon': 'Soft...
4  [{'name': 'Free transport', 'icon': 'Transport...

---

Then:

df['Facilities'] = df['Facilities'].apply(lambda x: [d['name'] for d in x])
print(df)

Prints:

                                          Facilities
0                                   [Work from home]
1             [Gymnasium, Cafeteria, Work from home]
2    [Free food, Team outings, Education assistance]
3                [Soft skill training, Job training]
4  [Free transport, Work from home, Team outings,...

Answer 2

You can extract it with two list comprehensions:

facility_names = [[facility["name"] for facility in facility_list] for facility_list in facilities]

Assuming that your input data is:

facilities=[
[{'name': 'Work from home', 'icon': 'WFH.svg'}],
[{'name': 'Gymnasium', 'icon': 'Gym.svg'}, {'name': 'Cafeteria', 'icon': 'Cafeteria.svg'}, {'name': 'Work from home', 'icon': 'WFH.svg'}],
[{'name': 'Free food', 'icon': 'FreeFood.svg'}, {'name': 'Team outings', 'icon': 'TeamOuting.svg'}, {'name': 'Education assistance', 'icon': 'Education.svg'}],
[{'name': 'Soft skill training', 'icon': 'SoftSkillsTraining.svg'}, {'name': 'Job training', 'icon': 'JobTraining.svg'}],
[{'name': 'Free transport', 'icon': 'Transportation.svg'}, {'name': 'Work from home', 'icon': 'WFH.svg'}, {'name': 'Team outings', 'icon': 'TeamOuting.svg'}, {'name': 'Soft skill training', 'icon': 'SoftSkillsTraining.svg'}]
]