Объединить строки с одним и тем же ключом в одну строку

Question

У меня есть Dataframe, и я хотел бы создать еще один столбец, который объединяет столбцы, имя которых начинается с того же value в Answer и QID.

То есть со следующим Dataframe

    QID     Category    Text    QType   Question:   Answer0     Answer1     Country
0   16  Automotive  Access to car   Single  Do you have access to a car?    I own a car/cars    I own a car/cars  UK
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  UK
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     UK
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    UK
4   16  Automotive  Access to car   Single  Do you have access to a car?    Prefer not to say   Prefer not to say   UK

Я бы хотел получить в результате следующее:

        QID     Category    Text    QType   Question:   Answer0     Answer1     Answer2    Answer3  Country    Answers
    0   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    UK    ['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     Prefer not to say']

Пока что пробовал следующее:

previous_qid = None
i = 0
j = 0
answers = []
new_row = {}
new_df = pd.DataFrame(columns=df.columns)
for _, row in df.iterrows():
    # get QID
    qid = row['QID']
    if qid == previous_qid:
        i+=1
        new_row['Answer'+str(i)]=row['Answer0']
        answers.append(row['Answer0'])
    elif new_row != {}:
        # we moved to a new row
        new_row['QID'] = qid
        new_row['Question'] = row['Question']
        new_row['Answers'] = answers
        # we create a new row in the new_dataframe
        new_df.append(new_row, ignore_index=True)
        # we clean up everything to receive the next row
        answers = []
        i=0
        j+=1
        new_row = {}
        # we add the information of the current row
        new_row['Answer'+str(i)]=row['Answer0']
        answers.append(row['Answer0'])
    previous_qid = qid

Но new_df результаты пустые.

Answer 1

Это логическая группировка по QID получение списка ответов с последующим разделением списка на столбцы

import re
data = """    QID     Category    Text    QType   Question:   Answer0     Answer1     Country
0   16  Automotive  Access to car   Single  Do you have access to a car?    I own a car/cars    I own a car/cars  UK
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  UK
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     UK
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    UK
4   16  Automotive  Access to car   Single  Do you have access to a car?    Prefer not to say   Prefer not to say   UK"""
a = [[t.strip() for t in re.split("  ",l) if t!=""]  for l in [re.sub("([0-9]?[ ])*(.*)", r"\2", l) for l in data.split("\n")]]

df = pd.DataFrame(data=a[1:], columns=a[0])

# lazy - want first of all attributes except QID and Answer columns
agg = {col:"first" for col in list(df.columns) if col!="QID" and "Answer" not in col}
# get a list of all answers in Answer0 for a QID
agg = {**agg, **{"Answer0":lambda s: list(s)}}

# helper function for row call.  not needed but makes more readable
def ans(r, i):
    return "" if i>=len(r["AnswerT"]) else r["AnswerT"][i]

# split list from aggregation back out into columns using assign
# rename Answer0 to AnserT from aggregation so that it can be referred to.  
# AnswerT drop it when don't want it any more
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))

вывод

QID    Category           Text   QType                     Question: Country           Answer0                      Answer1                      Answer2                                Answer3            Answer4 Answer5 Answer6
 16  Automotive  Access to car  Single  Do you have access to a car?      UK  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                

больше динамики c

Это немного расширяет python. Использование **kwargs и functools.partial. На самом деле это все еще стат c, столбцы определены как константа MAXANS

import functools 
MAXANS=8
def ansassign(dfa, row=0):
    return dfa.apply(lambda r: "" if row>=len(r["AnswerT"]) else r["AnswerT"][row], axis=1)
dfgrouped = df.groupby("QID").agg(agg).reset_index().rename(columns={"Answer0":"AnswerT"}).assign(
    **{f"Answer{i}":functools.partial(ansassign, row=i) for i in range(MAXANS)}
).drop("AnswerT", axis=1)