如何获得大熊猫数据框，其中列是来自另一列数据框的后续n元素？

Question

A very simple example just for understanding.

I have the following pandas dataframe:

import pandas as pd
df = pd.DataFrame({'A':pd.Series([1, 2, 13, 14, 25, 26, 37, 38])})
df 
        A
    0   1
    1   2
    2  13
    3  14
    4  25
    5  26
    6  37
    8  38

Set n = 3

First example

How to get a new dataframe df1 (in an efficient way), like the following:

   D1  D2  D3     T
0   1   2  13    14
1   2  13  14    25
2  13  14  25    26
3  14  25  26    37
4  25  26  37    38

Hint: think at the first n-columns as the data (Dx) and the last columns as the target (T). In the 1st example the target (eg 25) depends on the preceding n-elements (2, 13, 14).

Second example

What if the target is some element ahead (eg+3)?

   D1  D2  D3     T
0   1   2  13    26
1   2  13  14    37
2  13  14  25    38

Thank you for your help,
Gilberto

PS If you think that the title can be improved, please suggest me how to modify it.

Update

Thanks to @Divakar and this post the rolling function can be defined as:

import numpy as np
def rolling(a, window):
    shape = (a.size - window + 1, window)
    strides = (a.itemsize, a.itemsize)
    return np.lib.stride_tricks.as_strided(a, shape=shape, strides=strides)

a = np.arange(1000000000)
b = rolling(a, 4)

In less than 1 second!

Answer 1

Let's see how we can solve it with NumPy tools. So, let's imagine you have the column data as a NumPy array, let's call it a . For such sliding windowed operations, we have a very efficient tool in NumPy as strides , as they are views into the input array without actually making copies.

Let's directly use the methods with the sample data and start with case #1 -

In [29]: a  # Input data
Out[29]: array([ 1,  2, 13, 14, 25, 26, 37, 38])

In [30]: m = a.strides[0] # Get strides

In [31]: n = 3 # parameter

In [32]: nrows = a.size - n # Get number of rows in o/p

In [33]: a2D = np.lib.stride_tricks.as_strided(a,shape=(nrows,n+1),strides=(m,m))

In [34]: a2D
Out[34]: 
array([[ 1,  2, 13, 14],
       [ 2, 13, 14, 25],
       [13, 14, 25, 26],
       [14, 25, 26, 37],
       [25, 26, 37, 38]])

In [35]: np.may_share_memory(a,a2D) 
Out[35]: True    # a2D is a view into a

Case #2 would be similar with an additional parameter for the Target column -

In [36]: n2 = 3 # Additional param

In [37]: nrows = a.size - n - n2 + 1

In [38]: part1 = np.lib.stride_tricks.as_strided(a,shape=(nrows,n),strides=(m,m))

In [39]: part1 # These are D1, D2, D3, etc.
Out[39]: 
array([[ 1,  2, 13],
       [ 2, 13, 14],
       [13, 14, 25]])

In [43]: part2 = a[n+n2-1:] # This is target col

In [44]: part2
Out[44]: array([26, 37, 38])

Answer 2

I found another method: view_as_windows

import numpy as np
from skimage.util.shape import view_as_windows
window_shape = (4, )

aa = np.arange(1000000000) # 1 billion!
bb = view_as_windows(aa, window_shape)
bb

array([[        0,         1,         2,         3],
       [        1,         2,         3,         4],
       [        2,         3,         4,         5],
       ..., 
       [999999994, 999999995, 999999996, 999999997],
       [999999995, 999999996, 999999997, 999999998],
       [999999996, 999999997, 999999998, 999999999]])

Around 1 second.

What do you think?