熊猫组，然后根据日期范围+/- x-天计算和求和

Question

我希望在将数据帧分组到特定列之后，获得列的+/- 7天期间的值的计数和值。

示例数据(编辑以反映我的真实数据集)：

group  |        date          |   amount
-------------------------------------------
A      |  2017-12-26 04:20:20 |    50000.0
A      |  2018-01-17 00:54:15 |    60000.0
A      |  2018-01-27 06:10:12 |   150000.0
A      |  2018-02-01 01:15:06 |   100000.0
A      |  2018-02-11 05:05:34 |   150000.0
A      |  2018-03-01 11:20:04 |   150000.0
A      |  2018-03-16 12:14:01 |   150000.0
A      |  2018-03-23 05:15:07 |   150000.0
A      |  2018-04-02 10:40:35 |   150000.0

按group分组，然后基于date-7 < date < date+7进行和

我想要的结果：

group  |        date          |   amount    |  grouped_sum
-----------------------------------------------------------
A      |  2017-12-26 04:00:00 |    50000.0  |    50000.0
A      |  2018-01-17 00:00:00 |    60000.0  |    60000.0
A      |  2018-01-27 06:00:00 |   150000.0  |   250000.0
A      |  2018-02-01 01:00:00 |   100000.0  |   250000.0
A      |  2018-02-11 05:05:00 |   150000.0  |   150000.0
A      |  2018-03-01 11:00:04 |   150000.0  |   150000.0
A      |  2018-03-16 12:00:01 |   150000.0  |   150000.0
A      |  2018-03-23 05:00:07 |   100000.0  |   100000.0
A      |  2018-04-02 10:00:00 |   100000.0  |   100000.0

实现数据集的快速片段：

group = 9 * ['A']
date = pd.to_datetime(['2017-12-26 04:20:20', '2018-01-17 00:54:15', 
                       '2018-01-27 06:10:12', '2018-02-01 01:15:06', 
                       '2018-02-11 05:05:34', '2018-03-01 11:20:04', 
                       '2018-03-16 12:14:01', '2018-03-23 05:15:07', 
                       '2018-04-02 10:40:35'])
amount = [50000.0, 60000.0, 150000.0, 100000.0, 150000.0, 
          150000.0, 150000.0, 150000.0, 150000.0]
df = pd.DataFrame({'group':group, 'date':date, 'amount':amount})

一点解释：

• 第二行为40，因为它将2018-01-14和2018-01-15期间A的数据相加。
• 第四行是30，因为它在2018-01-03 +未来7天内将B的数据相加。
• 第六行是30，因为它在2018-01-03 + prev 7天内将B的数据相加。

我不知道如何在一段时间内计算和。如果我这样做的话，我也许能做到：

1.为每一行创建显示日期-7和date+7的另一列。

group  |    date     |  amount  |    date-7    |    date+7 
-------------------------------------------------------------
A      |  2017-12-26 |  50000.0 |  2017-12-19  |  2018-01-02
A      |  2018-01-17 |  60000.0 |  2018-01-10  |  2018-01-24

2.计算日期范围之间的金额：df[df.group == 'A' & df.date > df.date-7 & df.date < df.date+7].amount.sum()

3.但这种方法相当繁琐。

编辑(2018-09-01)：基于@jezrael答案找到以下方法，该方法对我有效，但只适用于单个组：

t = pd.Timedelta(7, unit='d')
def g(row):
    res = df[(df.created > row.created - t) & (df.created < row.created + t)].amount.sum()
    return res

df['new'] = df.apply(g, axis=1)

Answer 1

下面是每个行和每个组的问题需要循环：

t = pd.Timedelta(7, unit='d')

def f(x):
    return x.apply(lambda y: x.loc[x['date'].between(y['date'] - t, 
                                                     y['date'] + t,
                                                     inclusive=False),'amount'].sum() ,axis=1)

df['new'] = df.groupby('group', group_keys=False).apply(f)
print (df)
  group       date  amount   new
0     A 2018-01-01      10  10.0
1     A 2018-01-14      20  40.0
2     A 2018-01-15      20  40.0
3     B 2018-02-03      10  30.0
4     B 2018-02-04      10  30.0
5     B 2018-02-05      10  30.0

感谢@jpp的改进：

def f(x, t):
    return x.apply(lambda y: x.loc[x['date'].between(y['date'] - t, 
                                                     y['date'] + t,
                                                     inclusive=False),'amount'].sum(),axis=1)

df['new'] = df.groupby('group', group_keys=False).apply(f, pd.Timedelta(7, unit='d'))

验证解决方案：

t = pd.Timedelta(7, unit='d')


df = df[df['group'] == 'A']

def test(y):
    a = df.loc[df['date'].between(y['date'] - t,  y['date'] + t,inclusive=False)]
    print (a)
    print (a['amount'])
    return a['amount'].sum()

  group       date  amount
0     A 2018-01-01      10
0    10
Name: amount, dtype: int64
  group       date  amount
1     A 2018-01-14      20
2     A 2018-01-15      20
1    20
2    20
Name: amount, dtype: int64
  group       date  amount
1     A 2018-01-14      20
2     A 2018-01-15      20
1    20
2    20
Name: amount, dtype: int64

df['new'] = df.apply(test,axis=1)
print (df)
  group       date  amount  new
0     A 2018-01-01      10   10
1     A 2018-01-14      20   40
2     A 2018-01-15      20   40

Answer 2

添加列和一周中的第一天：

df['week_start'] = df['date'].dt.to_period('W').apply(lambda x: x.start_time)

结果：

  group       date  amount week_start
0     A 2018-01-01      10 2017-12-26
1     A 2018-01-14      20 2018-01-09
2     A 2018-01-15      20 2018-01-09
3     B 2018-02-03      10 2018-01-30
4     B 2018-02-04      10 2018-01-30
5     B 2018-02-05      10 2018-01-30

按新列分组并查找每周总金额：

grouped_sum = df.groupby('week_start')['amount'].sum().reset_index()

结果：

  week_start  amount
0 2017-12-26      10
1 2018-01-09      40
2 2018-01-30      30

在week_start上合并数据

pd.merge(df.drop('amount', axis=1), grouped_sum, on='week_start').drop('week_start', axis=1)

结果：

  group       date  amount
0     A 2018-01-01      10
1     A 2018-01-14      40
2     A 2018-01-15      40
3     B 2018-02-03      30
4     B 2018-02-04      30
5     B 2018-02-05      30