Python 2.7版：白天和列值

Question

移的数据帧我有一个名为DF1数据框如下：

DF1：

   a b id 
2010-01-01  2 3 21 
2010-01-01  2 4 22 
2010-01-01  3 5 23 
2010-01-01  4 6 24 
2010-01-02  1 4 21 
2010-01-02  2 5 22 
2010-01-02  3 6 23 
2010-01-02  4 7 24 
2010-01-03  1 8 21 
2010-01-03  2 9 22 
2010-01-03  3 10 23 
2010-01-03  4 11 24 
........................... 

我想转移的A，B和id的值， i行值成为i + 1行值。正如你可以看到df1，同一日期有几行，而id是不同的。我想转移df1，我的意思是2010-01-02的值是基于id的2010-01-03值（我的意思是2010-01-02值为21，是2010-01- 03值为21）。谢谢！

我想要的答案：

   a b id 
2010-01-01  Nan Nan Nan 
2010-01-01  Nan Nan Nan 
2010-01-01  Nan Nan Nan 
2010-01-01  Nan Nan Nan 
2010-01-02  2 3 21 
2010-01-02  2 4 22 
2010-01-02  3 5 23 
2010-01-02  4 6 24 
2010-01-03  1 4 21 
2010-01-03  2 5 22 
2010-01-03  3 6 23 
2010-01-03  4 7 24 
........................... 

Answer 1

如果所有组长度相同（样品4）DatetimeIndex排序：

df2 = df.shift((df.index == df.index[0]).sum()) 
print (df2) 
       a b id 
2010-01-01 NaN NaN NaN 
2010-01-01 NaN NaN NaN 
2010-01-01 NaN NaN NaN 
2010-01-01 NaN NaN NaN 
2010-01-02 2.0 3.0 21.0 
2010-01-02 2.0 4.0 22.0 
2010-01-02 3.0 5.0 23.0 
2010-01-02 4.0 6.0 24.0 
2010-01-03 1.0 4.0 21.0 
2010-01-03 2.0 5.0 22.0 
2010-01-03 3.0 6.0 23.0 
2010-01-03 4.0 7.0 24.0 

但是，如果需要一个日班指数值：

df3 = df.shift(1, freq='D') 
print (df3) 
      a b id 
2010-01-02 2 3 21 
2010-01-02 2 4 22 
2010-01-02 3 5 23 
2010-01-02 4 6 24 
2010-01-03 1 4 21 
2010-01-03 2 5 22 
2010-01-03 3 6 23 
2010-01-03 4 7 24 
2010-01-04 1 8 21 
2010-01-04 2 9 22 
2010-01-04 3 10 23 
2010-01-04 4 11 24 

Answer 2

其中一个方法是用形状的帮助，如果该日期即排序

df.shift(df.loc[df.index[0]].shape[0]) 
# Or len 
df.shift(len(df.loc[df.index[0]])) 

输出：

 
       a b id 
2010-01-01 NaN NaN NaN 
2010-01-01 NaN NaN NaN 
2010-01-01 NaN NaN NaN 
2010-01-01 NaN NaN NaN 
2010-01-02 2.0 3.0 21.0 
2010-01-02 2.0 4.0 22.0 
2010-01-02 3.0 5.0 23.0 
2010-01-02 4.0 6.0 24.0 
2010-01-03 1.0 4.0 21.0 
2010-01-03 2.0 5.0 22.0 
2010-01-03 3.0 6.0 23.0 
2010-01-03 4.0 7.0 24.0