Pandas滚动窗口返回一个数组

Question

这是一个示例代码.

df = pd.DataFrame(np.random.randn(10, 2), columns=list('AB'))
df['C'] = df.B.rolling(window=3)

输出:

           A         B                                       C
0 -0.108897  1.877987  Rolling [window=3,center=False,axis=0]
1 -1.276055 -0.424382  Rolling [window=3,center=False,axis=0]
2  1.578561 -1.094649  Rolling [window=3,center=False,axis=0]
3 -0.443294  1.683261  Rolling [window=3,center=False,axis=0]
4  0.674124  0.281077  Rolling [window=3,center=False,axis=0]
5  0.587773  0.697557  Rolling [window=3,center=False,axis=0]
6 -0.258038 -1.230902  Rolling [window=3,center=False,axis=0]
7 -0.443269  0.647107  Rolling [window=3,center=False,axis=0]
8  0.347187  0.753585  Rolling [window=3,center=False,axis=0]
9 -0.369179  0.975155  Rolling [window=3,center=False,axis=0]

我希望我的'C'列是一个像[0.1231,-1.132,0.8766]这样的数组.我尝试使用滚动应用但是徒劳无功.

预期产出:

       A         B                 C
0 -0.108897  1.877987  []
1 -1.276055 -0.424382  []
2  1.578561 -1.094649  [-1.094649, -0.424382, 1.877987]
3 -0.443294  1.683261  [1.683261, -1.094649, -0.424382]
4  0.674124  0.281077  [0.281077, 1.683261, -1.094649]
5  0.587773  0.697557  [0.697557, 0.281077, 1.683261]
6 -0.258038 -1.230902  [-1.230902, 0.697557, 0.281077]
7 -0.443269  0.647107  [0.647107, -1.230902, 0.697557]
8  0.347187  0.753585  [0.753585, 0.647107, -1.230902]
9 -0.369179  0.975155  [0.975155, 0.753585, 0.647107]

Answer 1

让我们通过滚动应用技巧使用这种 Pandas 方法：

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
df

输出：

          A         B                                                                  C
0  1.610085  0.354823                                                                NaN
1 -0.241446 -0.304952                                                                NaN
2  0.524812 -0.240972  [0.35482336179318674, -0.30495156795594963, -0.24097191924555197]
3  0.767354  0.281625   [-0.30495156795594963, -0.24097191924555197, 0.2816249674055174]
4 -0.349844 -0.533781    [-0.24097191924555197, 0.2816249674055174, -0.5337811449574766]
5 -0.174189  0.133795     [0.2816249674055174, -0.5337811449574766, 0.13379518286397707]
6  2.799437 -0.978349    [-0.5337811449574766, 0.13379518286397707, -0.9783488211443795]
7  0.250129  0.289782     [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]

Answer 2

由于熊猫1.1滚动对象是可迭代的，所以你可以这样做：

df['C'] = list(df.B.rolling(window=3))

或者，如果您想要列表，您可以执行以下操作：

df['C'] = [window.to_list() for window in df.B.rolling(window=3)]

这很简短，您可以使用该rolling函数的所有方便参数。

Answer 3

你可以使用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...

Answer 4

在较新的 numpy 版本中，有一个sliding_window_view().

它提供与as_strided()数组相同的功能，但具有更透明的语法。

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]])

但请注意，pandas 滚动会在开始时添加一些 nan (window_size - 1)，因为它使用填充。你可以这样检查：

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])

所以真正对应的数组应该是：

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.])