Pandas Rolling Window To Return An Array
Here is a sample code. df = pd.DataFrame(np.random.randn(10, 2), columns=list('AB')) df['C'] = df.B.rolling(window=3) Output: A B
Solution 1:
Since pandas 1.1 rolling objects are iterable, so you can just use the list constructor:
df['C'] = list(df.B.rolling(window=3))
OR if you want to have your windows as lists instead of Series's do:
df['C'] = [window.to_list() forwindowin df.B.rolling(window=3)]
This is short and you are able to use all the handy parameters of the rolling function.
Solution 2:
You could use np.stride_tricks:
import numpy as np
as_strided = np.lib.stride_tricks.as_strided
df
A B
0 -0.272824 -1.606357
1 -0.350643 0.000510
2 0.247222 1.627117
3 -1.601180 0.550903
4 0.803039 -1.231291
5 -0.536713 -0.313384
6 -0.840931 -0.675352
7 -0.930186 -0.189356
8 0.151349 0.522533
9 -0.046146 0.507406
win = 3 # window size# https://stackoverflow.com/a/47483615/4909087
v = as_strided(df.B, (len(df) - (win - 1), win), (df.B.values.strides * 2))
v
array([[ -1.60635669e+00, 5.10129842e-04, 1.62711678e+00],
[ 5.10129842e-04, 1.62711678e+00, 5.50902812e-01],
[ 1.62711678e+00, 5.50902812e-01, -1.23129111e+00],
[ 5.50902812e-01, -1.23129111e+00, -3.13383794e-01],
[ -1.23129111e+00, -3.13383794e-01, -6.75352179e-01],
[ -3.13383794e-01, -6.75352179e-01, -1.89356194e-01],
[ -6.75352179e-01, -1.89356194e-01, 5.22532550e-01],
[ -1.89356194e-01, 5.22532550e-01, 5.07405549e-01]])
df['C'] = pd.Series(v.tolist(), index=df.index[win - 1:])
df
A B C
0 -0.272824 -1.606357 NaN
1 -0.350643 0.000510 NaN
2 0.247222 1.627117 [-1.606356691642917, 0.0005101298424200881, 1....
3 -1.601180 0.550903 [0.0005101298424200881, 1.6271167809032248, 0....
4 0.803039 -1.231291 [1.6271167809032248, 0.5509028122535129, -1.23...
5 -0.536713 -0.313384 [0.5509028122535129, -1.2312911105674484, -0.3...
6 -0.840931 -0.675352 [-1.2312911105674484, -0.3133837943758246, -0....
7 -0.930186 -0.189356 [-0.3133837943758246, -0.6753521794378446, -0....
8 0.151349 0.522533 [-0.6753521794378446, -0.18935619377656243, 0....
9 -0.046146 0.507406 [-0.18935619377656243, 0.52253255045267, 0.507...
Solution 3:
Let's using this pandas approach with a rolling apply trick:
df = pd.DataFrame(np.random.randn(10, 2), columns=list('AB'))
list_of_values = []
df.B.rolling(3).apply(lambda x: list_of_values.append(x.values) or 0, raw=False)
df.loc[2:,'C'] = pd.Series(list_of_values).values
dfOutput:
AB C
01.6100850.354823 NaN
1 -0.241446 -0.304952 NaN
20.524812 -0.240972[0.35482336179318674, -0.30495156795594963, -0.24097191924555197]30.7673540.281625[-0.30495156795594963, -0.24097191924555197, 0.2816249674055174]4 -0.349844 -0.533781[-0.24097191924555197, 0.2816249674055174, -0.5337811449574766]5 -0.1741890.133795[0.2816249674055174, -0.5337811449574766, 0.13379518286397707]62.799437 -0.978349[-0.5337811449574766, 0.13379518286397707, -0.9783488211443795]70.2501290.289782[0.13379518286397707, -0.9783488211443795, 0.2897823417165459]8 -0.385259 -0.286399[-0.9783488211443795, 0.2897823417165459, -0.28639931887491943]9 -0.755363 -1.010891[0.2897823417165459, -0.28639931887491943, -1.0108913605575793]Solution 4:
Perhaps zipping would also help in your case i.e
defget_list(x,m) : returnlist(zip(*(x[i:] for i inrange(m))))
# get_list(df['B'],3) would return
[(-1.606357, 0.0005099999999999999, 1.627117),
(0.0005099999999999999, 1.627117, 0.5509029999999999),
(1.627117, 0.5509029999999999, -1.231291),
(0.5509029999999999, -1.231291, -0.313384),
(-1.231291, -0.313384, -0.6753520000000001),
(-0.313384, -0.6753520000000001, -0.189356),
(-0.6753520000000001, -0.189356, 0.522533),
(-0.189356, 0.522533, 0.507406)]
df['C'] = pd.Series(get_list(df['B'],3), index=df.index[3 - 1:])
# Little help form @coldspeedprint(df)
A B C
0 -0.272824 -1.606357 NaN
1 -0.3506430.000510 NaN
20.2472221.627117 (-1.606357, 0.0005099999999999999, 1.627117)
3 -1.6011800.550903 (0.0005099999999999999, 1.627117, 0.5509029999...
40.803039 -1.231291 (1.627117, 0.5509029999999999, -1.231291)
5 -0.536713 -0.313384 (0.5509029999999999, -1.231291, -0.313384)
6 -0.840931 -0.675352 (-1.231291, -0.313384, -0.6753520000000001)
7 -0.930186 -0.189356 (-0.313384, -0.6753520000000001, -0.189356)
80.1513490.522533 (-0.6753520000000001, -0.189356, 0.522533)
9 -0.0461460.507406 (-0.189356, 0.522533, 0.507406)
Solution 5:
In newer numpy versions there is a sliding_window_view().
It provides identical to as_strided() arrays, but with more transparent syntax.
import pandas as pd
from numpy.lib.stride_tricks import sliding_window_view
x = pd.Series([1, 2, 3, 4, 5, 6, 7, 8, 9])
sliding_window_view(x, 3)
>>>
array([[1, 2, 3],
[2, 3, 4],
[3, 4, 5],
[4, 5, 6],
[5, 6, 7],
[6, 7, 8],
[7, 8, 9]])
But be aware that pandas rolling will add few nans (window_size - 1) at the start because it uses padding. You can check it like this:
x.rolling(3).sum()
>>>
0 NaN
1 NaN
2 6.0
3 9.0
4 12.0
5 15.0
6 18.0
7 21.0
8 24.0
dtype: float64
sliding_window_view(x, 3).sum(axis=1)
>>>
array([ 6, 9, 12, 15, 18, 21, 24])
So real corresponding array should be:
c = np.array([[nan, nan, 1.],
[nan, 1., 2.],
[ 1., 2., 3.],
[ 2., 3., 4.],
[ 3., 4., 5.],
[ 4., 5., 6.],
[ 5., 6., 7.],
[ 6., 7., 8.],
[ 7., 8., 9.]])
c.sum(axis=1)
>>>
array([nan, nan, 6., 9., 12., 15., 18., 21., 24.])
Post a Comment for "Pandas Rolling Window To Return An Array"