我正在尝试根据存储在单独表格中的日期值计算Pandas中基于时间的聚合。
第一个表table_a的顶部如下所示：
COMPANY_ID DATE MEASURE
1 2010-01-01 00:00:00 10
1 2010-01-02 00:00:00 10
1 2010-01-03 00:00:00 10
1 2010-01-04 00:00:00 10
1 2010-01-05 00:00:00 10
以下是创建表格的代码：
table_a = pd.concat(\
[pd.DataFrame({'DATE': pd.date_range("01/01/2010", "12/31/2010", freq="D"),\
'COMPANY_ID': 1 , 'MEASURE': 10}),\
pd.DataFrame({'DATE': pd.date_range("01/01/2010", "12/31/2010", freq="D"),\
'COMPANY_ID': 2 , 'MEASURE': 10})])
第二个表table_b看起来像这样：
COMPANY END_DATE
1 2010-03-01 00:00:00
1 2010-06-02 00:00:00
2 2010-03-01 00:00:00
2 2010-06-02 00:00:00
并且创建它的代码是：
table_b = pd.DataFrame({'END_DATE':pd.to_datetime(['03/01/2010','06/02/2010','03/01/2010','06/02/2010']),\
'COMPANY':(1,1,2,2)})
我希望能够在table_b的END_DATE之前的每30天内获得每个COMPANY_ID的度量列的总和。
这是（我认为）SQL等价物：
select
b.COMPANY_ID,
b.DATE
sum(a.MEASURE) AS MEASURE_TO_END_DATE
from table_a a, table_b b
where a.COMPANY = b.COMPANY and
a.DATE < b.DATE and
a.DATE > b.DATE - 30
group by b.COMPANY;
感谢您的帮助
答案 0 :(得分：45)
好吧，我可以想到几个方法。 （1）通过合并company然后在合并后的30天窗口上过滤来基本上炸毁数据帧。这应该很快，但可能会占用大量内存。 （2）将30天窗口中的合并和过滤移动到groupby中。这导致每个组合并，因此它会更慢，但它应该使用更少的内存
选项＃1
假设您的数据如下所示（我扩展了您的示例数据）：
print df
company date measure
0 0 2010-01-01 10
1 0 2010-01-15 10
2 0 2010-02-01 10
3 0 2010-02-15 10
4 0 2010-03-01 10
5 0 2010-03-15 10
6 0 2010-04-01 10
7 1 2010-03-01 5
8 1 2010-03-15 5
9 1 2010-04-01 5
10 1 2010-04-15 5
11 1 2010-05-01 5
12 1 2010-05-15 5
print windows
company end_date
0 0 2010-02-01
1 0 2010-03-15
2 1 2010-04-01
3 1 2010-05-15
为30天窗口创建开始日期：
windows['beg_date'] = (windows['end_date'].values.astype('datetime64[D]') -
np.timedelta64(30,'D'))
print windows
company end_date beg_date
0 0 2010-02-01 2010-01-02
1 0 2010-03-15 2010-02-13
2 1 2010-04-01 2010-03-02
3 1 2010-05-15 2010-04-15
现在进行合并，然后根据date是否在beg_date和end_date范围内进行选择：
df = df.merge(windows,on='company',how='left')
df = df[(df.date >= df.beg_date) & (df.date <= df.end_date)]
print df
company date measure end_date beg_date
2 0 2010-01-15 10 2010-02-01 2010-01-02
4 0 2010-02-01 10 2010-02-01 2010-01-02
7 0 2010-02-15 10 2010-03-15 2010-02-13
9 0 2010-03-01 10 2010-03-15 2010-02-13
11 0 2010-03-15 10 2010-03-15 2010-02-13
16 1 2010-03-15 5 2010-04-01 2010-03-02
18 1 2010-04-01 5 2010-04-01 2010-03-02
21 1 2010-04-15 5 2010-05-15 2010-04-15
23 1 2010-05-01 5 2010-05-15 2010-04-15
25 1 2010-05-15 5 2010-05-15 2010-04-15
您可以通过company和end_date分组来计算30天的窗口总和：
print df.groupby(['company','end_date']).sum()
measure
company end_date
0 2010-02-01 20
2010-03-15 30
1 2010-04-01 10
2010-05-15 15
选项＃2 将所有合并移动到groupby中。这应该在内存上更好，但我认为要慢得多：
windows['beg_date'] = (windows['end_date'].values.astype('datetime64[D]') -
np.timedelta64(30,'D'))
def cond_merge(g,windows):
g = g.merge(windows,on='company',how='left')
g = g[(g.date >= g.beg_date) & (g.date <= g.end_date)]
return g.groupby('end_date')['measure'].sum()
print df.groupby('company').apply(cond_merge,windows)
company end_date
0 2010-02-01 20
2010-03-15 30
1 2010-04-01 10
2010-05-15 15
另一个选项现在，如果您的窗口永远不会重叠（例如示例数据中），您可以执行以下操作，作为替代方案，不会炸毁数据帧，但速度非常快：< / p>
windows['date'] = windows['end_date']
df = df.merge(windows,on=['company','date'],how='outer')
print df
company date measure end_date
0 0 2010-01-01 10 NaT
1 0 2010-01-15 10 NaT
2 0 2010-02-01 10 2010-02-01
3 0 2010-02-15 10 NaT
4 0 2010-03-01 10 NaT
5 0 2010-03-15 10 2010-03-15
6 0 2010-04-01 10 NaT
7 1 2010-03-01 5 NaT
8 1 2010-03-15 5 NaT
9 1 2010-04-01 5 2010-04-01
10 1 2010-04-15 5 NaT
11 1 2010-05-01 5 NaT
12 1 2010-05-15 5 2010-05-15
此合并实际上将您的窗口结束日期插入到数据框中，然后回填结束日期（按组）将为您提供一个结构，以便轻松创建求和窗口：
df['end_date'] = df.groupby('company')['end_date'].apply(lambda x: x.bfill())
print df
company date measure end_date
0 0 2010-01-01 10 2010-02-01
1 0 2010-01-15 10 2010-02-01
2 0 2010-02-01 10 2010-02-01
3 0 2010-02-15 10 2010-03-15
4 0 2010-03-01 10 2010-03-15
5 0 2010-03-15 10 2010-03-15
6 0 2010-04-01 10 NaT
7 1 2010-03-01 5 2010-04-01
8 1 2010-03-15 5 2010-04-01
9 1 2010-04-01 5 2010-04-01
10 1 2010-04-15 5 2010-05-15
11 1 2010-05-01 5 2010-05-15
12 1 2010-05-15 5 2010-05-15
df = df[df.end_date.notnull()]
df['beg_date'] = (df['end_date'].values.astype('datetime64[D]') -
np.timedelta64(30,'D'))
print df
company date measure end_date beg_date
0 0 2010-01-01 10 2010-02-01 2010-01-02
1 0 2010-01-15 10 2010-02-01 2010-01-02
2 0 2010-02-01 10 2010-02-01 2010-01-02
3 0 2010-02-15 10 2010-03-15 2010-02-13
4 0 2010-03-01 10 2010-03-15 2010-02-13
5 0 2010-03-15 10 2010-03-15 2010-02-13
7 1 2010-03-01 5 2010-04-01 2010-03-02
8 1 2010-03-15 5 2010-04-01 2010-03-02
9 1 2010-04-01 5 2010-04-01 2010-03-02
10 1 2010-04-15 5 2010-05-15 2010-04-15
11 1 2010-05-01 5 2010-05-15 2010-04-15
12 1 2010-05-15 5 2010-05-15 2010-04-15
df = df[(df.date >= df.beg_date) & (df.date <= df.end_date)]
print df.groupby(['company','end_date']).sum()
measure
company end_date
0 2010-02-01 20
2010-03-15 30
1 2010-04-01 10
2010-05-15 15
另一种方法是将您的第一个数据帧重新采样为每日数据，然后使用30天的窗口计算rolling_sums;并选择你感兴趣的最后日期。这也可能是内存密集型。
答案 1 :(得分：4)
有一种非常简单，实用（或也许是唯一直接的方法）的条件熊猫联接。由于没有直接的方法可以在熊猫中进行条件连接，因此您将需要一个附加的库，即pandasql
使用命令pandasql从pip安装库pip install pandasql。该库使您可以使用SQL查询来操纵熊猫数据框。
import pandas as pd
from pandasql import sqldf
df = pd.read_excel(r'play_data.xlsx')
df
id Name Amount
0 A001 A 100
1 A002 B 110
2 A003 C 120
3 A005 D 150
现在让我们进行条件连接以比较ID的数量
# Make your pysqldf object:
pysqldf = lambda q: sqldf(q, globals())
# Write your query in SQL syntax, here you can use df as a normal SQL table
cond_join= '''
select
df_left.*,
df_right.*
from df as df_left
join df as df_right
on
df_left.[Amount] > (df_right.[Amount]+10)
'''
# Now, get your queries results as dataframe using the sqldf object that you created
pysqldf(cond_join)
id Name Amount id Name Amount
0 A003 C 120 A001 A 100
1 A005 D 150 A001 A 100
2 A005 D 150 A002 B 110
3 A005 D 150 A003 C 120