Here's a vectorized approach using - NumPy broadcasting
idx = np.arange(n)[::-1] + np.arange(len(s))[:,None] - n + 1
out = s.get_values()[idx]
out[idx<0] = np.nan
This gives you output in the form of a 2D array.
If you need the final conclusion as a series -
In [40]: pd.Series(out.tolist())
Out[40]:
0 [1.0, nan, nan]
1 [1.1, 1.0, nan]
2 [1.2, 1.1, 1.0]
3 [1.3, 1.2, 1.1]
4 [1.4, 1.3, 1.2]
dtype: object
, np.split , , :
out_split = np.split(out,out.shape[0],axis=0)
-
In [100]: s
Out[100]:
1 1.0
2 1.1
3 1.2
4 1.3
5 1.4
dtype: float64
In [101]: n = 3
In [102]: idx = np.arange(n)[::-1] + np.arange(len(s))[:,None] - n + 1
...: out = s.get_values()[idx]
...: out[idx<0] = np.nan
...:
In [103]: out
Out[103]:
array([[ 1. , nan, nan],
[ 1.1, 1. , nan],
[ 1.2, 1.1, 1. ],
[ 1.3, 1.2, 1.1],
[ 1.4, 1.3, 1.2]])
In [104]: np.split(out,out.shape[0],axis=0)
Out[104]:
[array([[ 1., nan, nan]]),
array([[ 1.1, 1. , nan]]),
array([[ 1.2, 1.1, 1. ]]),
array([[ 1.3, 1.2, 1.1]]),
array([[ 1.4, 1.3, 1.2]])]
strides
- strided_axis0, @BM solution, .
, NaN -
In [35]: strided_axis0(s.values, fillval=np.nan, L=3)
Out[35]:
array([[nan, nan, 1. ],
[nan, 1. , 1.1],
[1. , 1.1, 1.2],
[1.1, 1.2, 1.3],
[1.2, 1.3, 1.4]])
NaN , , , -
In [36]: strided_axis0(s.values, fillval=np.nan, L=3)[:,::-1]
Out[36]:
array([[1. , nan, nan],
[1.1, 1. , nan],
[1.2, 1.1, 1. ],
[1.3, 1.2, 1.1],
[1.4, 1.3, 1.2]])
, , pd.Series(out.tolist()) out 2D .
In [38]: pd.Series(strided_axis0(s.values, fillval=np.nan, L=3)[:,::-1].tolist())
Out[38]:
0 [1.0, nan, nan]
1 [1.1, 1.0, nan]
2 [1.2, 1.1, 1.0]
3 [1.3, 1.2, 1.1]
4 [1.4, 1.3, 1.2]
dtype: object