Question

如何使用Numpy或Pandas进行R（xts）等效rollapply（....，by.column = FALSE）？给定数据帧时，pandas rolling_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

所以正在发生的事情是，rolling_apply依次向下移动每一列，并在每一列中应用一个滑动的5长度窗口，而我想要的是每次滑动窗口为5x10阵列，在这种情况下，我会得到一个单列向量（而不是2d数组）的结果。

Answer 1

我确实无法找到一种方法来计算＆＃34;宽＆＃34;在熊猫中滚动应用程序 docs，所以我使用numpy来获得一个＆＃34;窗口＆＃34;查看数组并应用ufunc 它。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，您需要以下＆＃34; window＆＃34;查看成语：

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]]])

Strides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 从窗口[0]转到窗口[1]时跳过5个元素的元素从窗口[0,0]到窗口[0,1]。现在你可以调用任何ufunc了结果数组，例如：

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

带有Pandas或Numpy的n维滑动窗口

1 个答案: