Как соединить линии, которые имеют один и тот же ключ, в одну линию?

Eсть DataFrame, и хотел бы сделать еще один столбец, который объединяет столбцы, чье имя начинается с того же самого значения в Answer и QID.

Вот данные, и их пример:

    QID     Category    Text    QType   Question    Answer0     Answer1
0   16  Automotive  Access to car   Single  Do you have access to a car?    I own a car/cars    I own a car/cars
1   16  Automotive  Access to car   Single  Do you have access to a car?    I lease/ have a company car     I lease/have a company car
2   16  Automotive  Access to car   Single  Do you have access to a car?    I have access to a car/cars     I have access to a car/cars
3   16  Automotive  Access to car   Single  Do you have access to a car?    No, I don’t have access to a car/cars   No, I don't have access to a car
4   16  Automotive  Access to car   Single  Do you have access to a car?    Prefer not to say   Prefer not to say
5   17  Automotive  Make of car/cars    Multiple    If you own/lease a car(s), which brand are they?    Audi    Audi
6   17  Automotive  Make of car/cars    Multiple    If you own/lease a car(s), which brand are they?    Alfa Romeo  Alfa Romeo
7   17  Automotive  Make of car/cars    Multiple    If you own/lease a car(s), which brand are they?    BMW     BMW
8   17  Automotive  Make of car/cars    Multiple    If you own/lease a car(s), which brand are they?    Cadillac    Cadillac
9   17  Automotive  Make of car/cars    Multiple    If you own/lease a car(s), which brand are they?    Chevrolet   Chevrolet
10  17  Automotive  Make of car/cars    Multiple    If you own/lease a car(s), which brand are they?    Chrysler    Chrysler
11  17  Automotive  Make of car/cars    Multiple    If you own/lease a car(s), which brand are they?    Citroen     Citroen
12  17  Automotive  Make of car/cars    Multiple    If you own/lease a car(s), which brand are they?    Daihatsu    Daihatsu
13  17  Automotive  Make of car/cars    Multiple    If you own/lease a car(s), which brand are they?    Fiat    Fiat
14  17  Automotive  Make of car/cars    Multiple    If you own/lease a car(s), which brand are they?    Ford    Ford
15  17  Automotive  Make of car/cars    Multiple    If you own/lease a car(s), which brand are they?    Honda   Honda
16  17  Automotive  Make of car/cars    Multiple    If you own/lease a car(s), which brand are they?    Hyundai     Hyundai
...

Xотел бы получить что-то подобное:

    QID     Category    Text    QType   Question    Answer0     Answer1     Answer3     Answer4     Answer5     Answer6     Answer7     Answer8     Answer9     Answer10    Answer11     Answer12     ...      
4   16  Automotive  Access to car   Single  Do you have access to a car?    I own a car/cars    I lease/ have a company car     I have access to a car/cars     No, I don’t have access to a car/cars   Prefer not to say       
5   17  Automotive  Make of car/cars    Multiple    If you own/lease a car(s), which brand are they?    Audi    Alfa Romeo  BMW     Cadillac    Chevrolet   Chrysler    Citroen     ...

Я могу комбинировать подаренное/статическое число столбцов, чье имя начинается с одного и того же значения в Answer и QID:

df = pd.DataFrame('path/to/file')

# ленивый - в первую очередь нужны атрибуты, кроме столбцов QID и Answer
agg = {col:"first" for col in list(df.columns) if col!="QID" and "Answer" not in col}
# получить список всех ответов в Answer0 для QID
agg = {**agg, **{"Answer0":lambda s: list(s)}}

# вспомогательная функция для вызова ряда. не нужна, но делает более читабельной.
def ans(r, i):
    return "" if i>=len(r["AnswerT"]) else r["AnswerT"][i]

# разделить список от объединения обратно на столбцы с помощью назначения
# переименовать Answer0 в AnserT из агрегирования, чтобы на него можно было ссылаться.  
# AnswerT бросай, когда больше не хочешь.
dfgrouped = df.groupby("QID").agg(agg).reset_index().rename(columns={"Answer0":"AnswerT"}).assign(
    Answer0=lambda dfa: dfa.apply(lambda r: ans(r, 0), axis=1),
    Answer1=lambda dfa: dfa.apply(lambda r: ans(r, 1), axis=1),
    Answer2=lambda dfa: dfa.apply(lambda r: ans(r, 2), axis=1),
    Answer3=lambda dfa: dfa.apply(lambda r: ans(r, 3), axis=1),
    Answer4=lambda dfa: dfa.apply(lambda r: ans(r, 4), axis=1),
    Answer5=lambda dfa: dfa.apply(lambda r: ans(r, 5), axis=1),
    Answer6=lambda dfa: dfa.apply(lambda r: ans(r, 6), axis=1),
).drop("AnswerT", axis=1)

print(dfgrouped.to_string(index=False))

А как можно комбинировать динамическое число столбцов, где эти столбцы имеют имена, которые начинаются с одного и того же значения в Answer и QID?