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])