我如何使用Numpy或Pandas等效于rollapply(....,by.column = FALSE)的R(xts)?当给定一个数据框时,pandasrolling_apply似乎只能逐列工作,而不是提供向目标函数提供完整(窗口大小)x(数据帧宽度)矩阵的选项。

import pandas as pd
import numpy as np

xx = pd.DataFrame(np.zeros([10, 10]))
pd.rolling_apply(xx, 5, lambda x: np.shape(x)[0])

    0   1   2   3   4   5   6   7   8   9
0 NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN
1 NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN
2 NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN
3 NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN
4   5   5   5   5   5   5   5   5   5   5
5   5   5   5   5   5   5   5   5   5   5
6   5   5   5   5   5   5   5   5   5   5
7   5   5   5   5   5   5   5   5   5   5
8   5   5   5   5   5   5   5   5   5   5
9   5   5   5   5   5   5   5   5   5   5


因此,发生的情况是rolling_apply依次向下移动每一列,并在其中的每一列上向下滑动一个5长度的窗口,而我想要的是每次滑动窗口都是一个5x10的数组,在这种情况下,我会获得单列向量(不是二维数组)结果。

最佳答案

我确实找不到在熊猫中计算“广泛”滚动应用程序的方法
docs,因此我将使用numpy在数组上获取“窗口”视图并应用ufunc
对此。这是一个例子:

In [40]: arr = np.arange(50).reshape(10, 5); arr
Out[40]:
array([[ 0,  1,  2,  3,  4],
       [ 5,  6,  7,  8,  9],
       [10, 11, 12, 13, 14],
       [15, 16, 17, 18, 19],
       [20, 21, 22, 23, 24],
       [25, 26, 27, 28, 29],
       [30, 31, 32, 33, 34],
       [35, 36, 37, 38, 39],
       [40, 41, 42, 43, 44],
       [45, 46, 47, 48, 49]])

In [41]: win_size = 5

In [42]: isize = arr.itemsize; isize
Out[42]: 8


arr.itemsize为8,因为默认dtype为np.int64,因此以下“窗口”视图惯用法需要它:

In [43]: windowed = np.lib.stride_tricks.as_strided(arr,
                                                    shape=(arr.shape[0] - win_size + 1, win_size, arr.shape[1]),
                                                    strides=(arr.shape[1] * isize, arr.shape[1] * isize, isize)); windowed
Out[43]:
array([[[ 0,  1,  2,  3,  4],
        [ 5,  6,  7,  8,  9],
        [10, 11, 12, 13, 14],
        [15, 16, 17, 18, 19],
        [20, 21, 22, 23, 24]],

       [[ 5,  6,  7,  8,  9],
        [10, 11, 12, 13, 14],
        [15, 16, 17, 18, 19],
        [20, 21, 22, 23, 24],
        [25, 26, 27, 28, 29]],

       [[10, 11, 12, 13, 14],
        [15, 16, 17, 18, 19],
        [20, 21, 22, 23, 24],
        [25, 26, 27, 28, 29],
        [30, 31, 32, 33, 34]],

       [[15, 16, 17, 18, 19],
        [20, 21, 22, 23, 24],
        [25, 26, 27, 28, 29],
        [30, 31, 32, 33, 34],
        [35, 36, 37, 38, 39]],

       [[20, 21, 22, 23, 24],
        [25, 26, 27, 28, 29],
        [30, 31, 32, 33, 34],
        [35, 36, 37, 38, 39],
        [40, 41, 42, 43, 44]],

       [[25, 26, 27, 28, 29],
        [30, 31, 32, 33, 34],
        [35, 36, 37, 38, 39],
        [40, 41, 42, 43, 44],
        [45, 46, 47, 48, 49]]])


跨度是沿给定轴的两个相邻元素之间的字节数,
因此strides=(arr.shape[1] * isize, arr.shape[1] * isize, isize)表示跳过5
从windowed [0]到windowed [1]时元素,当
从windowed [0,0]到windowed [0,1]。现在,您可以在
结果数组,例如:

In [44]: windowed.sum(axis=(1,2))
Out[44]: array([300, 425, 550, 675, 800, 925])

07-27 13:29
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