Creating a Pandas rolling-window series of arrays

Your data look like a strided array :

data=np.lib.stride_tricks.as_strided(np.concatenate(([NaN]*2,s))[2:],(5,3),(8,-8))
"""
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]])
"""

Then transform in Series :

pd.Series(map(list,data))
"""" 
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
""""

Here's a vectorized approach using NumPy broadcasting -

n = 3 # window length
idx = np.arange(n)[::-1] + np.arange(len(s))[:,None] - n + 1
out = s.get_values()[idx]
out[idx<0] = np.nan

This gets you the output as a 2D array.

To get a series with each element holding each window as a list -

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

If you wish to have a list of 1D arrays split arrays, you can use np.split on the output, like so -

out_split = np.split(out,out.shape[0],axis=0)

Sample run -

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]])]

Memory-efficiency with strides

For memory efficiency, we can use a strided one - strided_axis0, similar to @B. M.'s solution, but a bit more generic one.

So, to get 2D array of values with NaNs precedding the first element -

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]])

To get 2D array of values with NaNs as fillers coming after the original elements in each row and the order of elements being flipped, as stated in the problem -

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]])

To get a series with each element holding each window as a list, simply wrap the earlier methods with pd.Series(out.tolist()) with out being the 2D array outputs -

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

If you attach the missing nans at the beginning and the end of the series, you use a simple window

def wndw(s,size=3):

    stretched = np.hstack([
        np.array([np.nan]*(size-1)),
        s.values.T,
        np.array([np.nan]*size)
    ])

    for begin in range(len(stretched)-size):
        end = begin+size
        yield stretched[begin:end][::-1]

for arr in wndw(s, 3):
    print arr

Here's one way to do it

In [294]: arr = [s.shift(x).values[::-1][:3] for x in range(len(s))[::-1]]

In [295]: arr
Out[295]:
[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])]

In [296]: pd.Series(arr, index=s.index)
Out[296]:
1    [1.0, nan, nan]
2    [1.1, 1.0, nan]
3    [1.2, 1.1, 1.0]
4    [1.3, 1.2, 1.1]
5    [1.4, 1.3, 1.2]
dtype: object

Tags:

Python

Pandas

Numpy

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