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python - Pandas 分组然后根据日期范围 +/- x-days 计数和求和

问题描述

在将数据框分组到某个列之后，我想在 +/- 7 天的时间内获得一个列的值的计数和总和

示例数据（经过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

group然后根据date-7< date<进行分组date+7

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：

我不知道如何在一段日期范围内求和。如果我这样做，我也许可以做到：

1.创建另一列，每行显示 date-7 和 date+7

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.但是这种方法相当繁琐。

编辑（2018-09-01）：根据@jezrael 答案在下面找到此方法，该方法适用于我，但仅适用于单个组：

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)

标签： pythonpandasdataframe

解决方案

这是每行和每组的问题需要循环：

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

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