python - 如何反转滚动总和？

Question

我有一个在分组数据框上计算的滚动总和，但它的加法是错误的，它是 future 的总和，而我需要过去的总和。

我在这里做错了什么？

我导入数据并按维度和日期排序(我已经尝试删除日期排序)

df = pd.read_csv('Input.csv', parse_dates=True)
df.sort_values(['Dimension','Date'])
print(df)

然后我创建一个新列，它是一个按滚动窗口分组的多索引

new_column = df.groupby('Dimension').Value1.apply(lambda x: 
x.rolling(window=3).sum())

然后我将索引重置为与原始索引相同

df['Sum_Value1'] = new_column.reset_index(level=0, drop=True)
print(df)

我也试过在计算前反转索引，但也失败了。

输入

Dimension,Date,Value1,Value2
1,4/30/2002,10,20
1,1/31/2002,10,20
1,10/31/2001,10,20
1,7/31/2001,10,20
1,4/30/2001,10,20
1,1/31/2001,10,20
1,10/31/2000,10,20
2,4/30/2002,10,20
2,1/31/2002,10,20
2,10/31/2001,10,20
2,7/31/2001,10,20
2,4/30/2001,10,20
2,1/31/2001,10,20
2,10/31/2000,10,20
3,4/30/2002,10,20
3,1/31/2002,10,20
3,10/31/2001,10,20
3,7/31/2001,10,20
3,1/31/2001,10,20
3,10/31/2000,10,20

输出:

    Dimension        Date  Value1  Value2  Sum_Value1
0           1   4/30/2002      10      20         NaN
1           1   1/31/2002      10      20         NaN
2           1  10/31/2001      10      20        30.0
3           1   7/31/2001      10      20        30.0
4           1   4/30/2001      10      20        30.0
5           1   1/31/2001      10      20        30.0
6           1  10/31/2000      10      20        30.0
7           2   4/30/2002      10      20         NaN
8           2   1/31/2002      10      20         NaN
9           2  10/31/2001      10      20        30.0
10          2   7/31/2001      10      20        30.0
11          2   4/30/2001      10      20        30.0
12          2   1/31/2001      10      20        30.0
13          2  10/31/2000      10      20        30.0

目标输出:

    Dimension        Date  Value1  Value2  Sum_Value1
0           1   4/30/2002      10      20        30.0
1           1   1/31/2002      10      20        30.0
2           1  10/31/2001      10      20        30.0
3           1   7/31/2001      10      20        30.0
4           1   4/30/2001      10      20        30.0
5           1   1/31/2001      10      20         NaN
6           1  10/31/2000      10      20         NaN
7           2   4/30/2002      10      20        30.0
8           2   1/31/2002      10      20        30.0
9           2  10/31/2001      10      20        30.0
10          2   7/31/2001      10      20        30.0
11          2   4/30/2001      10      20        30.0
12          2   1/31/2001      10      20         Nan
13          2  10/31/2000      10      20         NaN

Answer 1

你需要一个反向求和，因此在求和之前反转你的序列:

lambda x: x[::-1].rolling(window=3).sum()