我有一些房地产数据，我想有效地计算自该房产上次出售日期以来的TimeDelta。结果必须高效，因为我有超过200万行，所以我的解决方案太慢了。到目前为止，这是我已经实现的内容，但这需要几天的时间才能在我的数据框中进行计算。有没有更快的方法来实现这一点？
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)
答案 0 :(得分：3)
使用：
#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
答案 1 :(得分：1)
groupby + diff()应该可以正常工作，并且比循环快：
sale_data.groupby('ADDRESS').SALE_DATE.diff()
输出：
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
答案 2 :(得分：0)
使用Groupby进行转换并应用diff来获取日期之间的差异
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