Это должно достичь того, что вы хотите:
import pandas as pd
import operator
from collections import defaultdict
d = { "user":["user1","user1","user1","user1","user1","user2","user2","user2","user2","user2","user2"],
"visit_type":["search","search","booking","search","booking","search","booking","search","booking","booking","booking"],
"hotel_code":[1,2,1,8,8,6,6,4,4,6,4]}
df = pd.DataFrame(data=d)
#Setting default value
df['most_booked']='NaN'
for user in df.user.unique():
#Ignoring searches, only considering bookings
df_bookings = df.loc[(df["visit_type"] == "booking") & (df['user'] == user)]
last_booked = ""
booking_counts = defaultdict(int)
for i, entry in df_bookings.iterrows():
#Skipping first booking
if last_booked != "":
highest = max(booking_counts.values())
#Prefers last booked if it equals max
if booking_counts[last_booked] == highest:
max_booked = last_booked
#Otherwise chooses max
else:
max_booked = max(booking_counts.items(), key=operator.itemgetter(1))[0]
df.loc[i, 'most_booked'] = max_booked
#Update number of bookings in dictionary
current_booking = entry["hotel_code"]
booking_counts[current_booking] += 1
last_booked = current_booking
print(df)
hotel_code user visit_type most_booked
0 1 user1 search NaN
1 2 user1 search NaN
2 1 user1 booking NaN
3 8 user1 search NaN
4 8 user1 booking 1
5 6 user2 search NaN
6 6 user2 booking NaN
7 4 user2 search NaN
8 4 user2 booking 6
9 6 user2 booking 4
10 4 user2 booking 6