Эффективно циклически перебирая данные в пандах

Question

У меня есть некоторые данные о недвижимости, и я хотел бы эффективно рассчитать TimeDelta с момента последней продажи этого имущества.Результат должен быть эффективным, потому что у меня более 2 миллионов строк, поэтому мое решение было слишком медленным.Вот то, что я реализовал до сих пор, но это требует дней, чтобы рассчитать на моем кадре данных.Есть ли более быстрый способ реализовать это?

import pandas as pd
import numpy as np
import datetime #import datetime
pd.set_option('display.max_columns',5)


## Make some dummy data
data_dict = dict(
    ADDRESS=[
        '123 Main Street', '123 Apple Street', '123 Orange Street', '123 Pineapple Street', '123 Pear Street',
        '123 Main Street', '123 Apple Street', '123 Orange Street', '123 Pineapple Street', '123 Pear Street',
        '123 Main Street', '123 Apple Street', '123 Orange Street', '123 Pineapple Street', '123 Pear Street',
    ],

    SALE_DATE=[
        '2002-01-01', '2006-01-01', '2009-01-01', '2011-01-01', '2012-01-01',
        '2013-01-01', '2012-01-01', '2012-01-01', '2012-01-01', '2014-01-01',
        '2016-01-01', '2018-06-01', '2017-01-01', '2017-01-01', '2019-01-01'
    ]
)

# format as a pandas df
sale_data = pd.DataFrame(data_dict)
sale_data['SALE_DATE'] = pd.to_datetime(sale_data['SALE_DATE'])

# instantiate a df that we will append our results to
master_df = pd.DataFrame()

#loop through each address to get the last sale and expected future sale date
for address in enumerate(sale_data.ADDRESS.drop_duplicates()):
    df_slice = sale_data[sale_data.ADDRESS == address[1]].sort_values(by='SALE_DATE')
    df_slice['days_since_last_sale'] = df_slice['SALE_DATE'] - df_slice['SALE_DATE'].shift(1)
    df_slice['days_since_last_sale'] = [x.days if x.days > 0 else np.nan for x in df_slice['days_since_last_sale']]
    df_slice['years_since_last_sale'] = df_slice['days_since_last_sale'] / 365
    days_average = np.mean(df_slice['days_since_last_sale'])
    df_slice['next_sale'] = datetime.datetime.today() + datetime.timedelta(days=days_average)

    master_df = pd.concat([df_slice, master_df],
                            axis=0)

    print(len(master_df))
    print('_________________________________________________________________________________')

print(master_df)

Answer 1

Использование:

#sorting per 2 columns for grouping ADDRESS together and correct diff
sale_data = sale_data.sort_values(by=['ADDRESS','SALE_DATE'])
#get difference per groups, convert timedeltas to days 
sale_data['days_since_last_sale'] = sale_data.groupby('ADDRESS')['SALE_DATE'].diff().dt.days
#divide by scalar 
sale_data['years_since_last_sale'] = sale_data['days_since_last_sale'] / 365
#get mean per groups
days = sale_data.groupby('ADDRESS')['days_since_last_sale'].transform('mean')
#add to datetime timedeltas of days
sale_data['next_sale'] = datetime.datetime.today() + pd.to_timedelta(days, unit='d')

---

print(sale_data)
                 ADDRESS  SALE_DATE  days_since_last_sale  \
1       123 Apple Street 2006-01-01                   NaN   
6       123 Apple Street 2012-01-01                2191.0   
11      123 Apple Street 2018-06-01                2343.0   
0        123 Main Street 2002-01-01                   NaN   
5        123 Main Street 2013-01-01                4018.0   
10       123 Main Street 2016-01-01                1095.0   
2      123 Orange Street 2009-01-01                   NaN   
7      123 Orange Street 2012-01-01                1095.0   
12     123 Orange Street 2017-01-01                1827.0   
4        123 Pear Street 2012-01-01                   NaN   
9        123 Pear Street 2014-01-01                 731.0   
14       123 Pear Street 2019-01-01                1826.0   
3   123 Pineapple Street 2011-01-01                   NaN   
8   123 Pineapple Street 2012-01-01                 365.0   
13  123 Pineapple Street 2017-01-01                1827.0   

    years_since_last_sale                  next_sale  
1                     NaN 2025-09-04 14:37:24.900489  
6                6.002740 2025-09-04 14:37:24.900489  
11               6.419178 2025-09-04 14:37:24.900489  
0                     NaN 2026-06-21 02:37:24.900489  
5               11.008219 2026-06-21 02:37:24.900489  
10               3.000000 2026-06-21 02:37:24.900489  
2                     NaN 2023-06-21 14:37:24.900489  
7                3.000000 2023-06-21 14:37:24.900489  
12               5.005479 2023-06-21 14:37:24.900489  
4                     NaN 2022-12-21 02:37:24.900489  
9                2.002740 2022-12-21 02:37:24.900489  
14               5.002740 2022-12-21 02:37:24.900489  
3                     NaN 2022-06-21 14:37:24.900489  
8                1.000000 2022-06-21 14:37:24.900489  
13               5.005479 2022-06-21 14:37:24.900489  

Answer 2

a groupby + diff() должно работать в целом и быть быстрее, чем цикл:

sale_data.groupby('ADDRESS').SALE_DATE.diff()

Выход:

                ADDRESS  SALE_DATE     delta
0        123 Main Street 2002-01-01       NaT
1       123 Apple Street 2006-01-01       NaT
2      123 Orange Street 2009-01-01       NaT
3   123 Pineapple Street 2011-01-01       NaT
4        123 Pear Street 2012-01-01       NaT
5        123 Main Street 2013-01-01 4018 days
6       123 Apple Street 2012-01-01 2191 days
7      123 Orange Street 2012-01-01 1095 days
8   123 Pineapple Street 2012-01-01  365 days
9        123 Pear Street 2014-01-01  731 days
10       123 Main Street 2016-01-01 1095 days
11      123 Apple Street 2018-06-01 2343 days
12     123 Orange Street 2017-01-01 1827 days
13  123 Pineapple Street 2017-01-01 1827 days
14       123 Pear Street 2019-01-01 1826 days

Answer 3

ИСПОЛЬЗОВАНИЕ Groupby с преобразованием и применением различий, чтобы получить разницу между датами

sale_data['days']= sale_data.groupby(['ADDRESS'],as_index=False)['SALE_DATE'].transform(pd.Series.diff)

              ADDRESS     SALE_DATE        Days
0        123 Main Street  2002-01-01        NaT
1       123 Apple Street  2006-01-01        NaT
2      123 Orange Street  2009-01-01        NaT
3   123 Pineapple Street  2011-01-01        NaT
4        123 Pear Street  2012-01-01        NaT
5        123 Main Street  2013-01-01  4018 days
6       123 Apple Street  2012-01-01  2191 days
7      123 Orange Street  2012-01-01  1095 days
8   123 Pineapple Street  2012-01-01   365 days
9        123 Pear Street  2014-01-01   731 days
10       123 Main Street  2016-01-01  1095 days
11      123 Apple Street  2018-06-01  2343 days
12     123 Orange Street  2017-01-01  1827 days
13  123 Pineapple Street  2017-01-01  1827 days
14       123 Pear Street  2019-01-01  1826 days