包括前几天和后几天的熊猫滚动总和

Question

基本上，初学者试图找到包括今天，过去4天和未来几天（在这种情况下，第二天（明天））的总和。

            Test
1995-07-01    1 
1995-07-02    0 
1995-07-03    0 
1995-07-04    1    
1995-07-05    0    
1995-07-06    0    
1995-07-07    0    
1995-07-08    0    
1995-07-09    0    
1995-07-10    0    
1995-07-11    1

获得“今天”和过去4天的总滚动金额df ['test']。rolling（5）.sum（）

1995-07-01    NaN
1995-07-02    NaN
1995-07-03    NaN
1995-07-04    NaN
1995-07-05    2.0
1995-07-06    1.0
1995-07-07    1.0
1995-07-08    1.0
1995-07-09    0.0
1995-07-10    0.0
1995-07-11    1.0

但是将第二天的值包含在该总和中对我来说很困难，我想要的是1995-07-10的输出显示1，因为它需要包含'明天'（因为1995-07-11在输出中是1测试数据）

Answer 1

df['Tomorrow'] = df.shift(-1)
df['Previous'] = df['Test'].rolling(4).sum()
df.sum(axis=1)

Output

1995-07-01    1.0
1995-07-02    0.0
1995-07-03    1.0
1995-07-04    3.0
1995-07-05    1.0
1995-07-06    1.0
1995-07-07    1.0
1995-07-08    0.0
1995-07-09    0.0
1995-07-10    1.0
1995-07-11    2.0

或者即使少于四天，您也希望前三行具有前四行的值：

df['Previous'] = df['Test'].rolling(4, min_periods=1).sum()

Answer 2

我相信您需要的是shift()方法。它可以让您将数据移动几天，然后可以根据需要将其与日期对齐。

See this link用于文档。

df['test'].shift(-1,fill_value=0).rolling(5).sum()

提供：

date
1995-07-01    NaN
1995-07-02    NaN
1995-07-03    NaN
1995-07-04    NaN
1995-07-05    1.0
1995-07-06    1.0
1995-07-07    1.0
1995-07-08    0.0
1995-07-09    0.0
1995-07-10    1.0
1995-07-11    1.0

Answer 3

IIUC，Series.shift

Series.shift

您可以看到最后一天包含

days_before = 4
days_after = 1
df['Test'].rolling(days_before + days_after + 1).sum().shift(-days_after)

1995-07-01    NaN
1995-07-02    NaN
1995-07-03    NaN
1995-07-04    NaN
1995-07-05    2.0
1995-07-06    1.0
1995-07-07    1.0
1995-07-08    1.0
1995-07-09    0.0
1995-07-10    1.0
1995-07-11    NaN
Name: Test, dtype: float64

，因为它没有几天，所以可以用NaN之类的方法填充它>

Answer 4

使用