与 pandas 或numpy的n维滑动窗口 [英] n-dimensional sliding window with Pandas or Numpy
问题描述
我该怎么办rollapply（....，by.column = FALSE）的R（XTS）等效，使用numpy的还是熊猫？当给定一个数据帧，熊猫rolling_apply似乎只列工作列，而不是提供选项来提供一个完整的（窗口大小）×（数据帧的宽度）矩阵为目标的功能。
How do I do the R(xts) equivalent of rollapply(...., by.column=FALSE), using Numpy or Pandas? When given a dataframe, pandas rolling_apply seems only to work column by column instead of providing the option to provide a full (window-size) x (data-frame-width) matrix to the target function.
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，发生了什么事是rolling_apply正在下降，反过来每一列和应用滑动5长窗下的这些各一个，而我希望的是滑动窗口每次是一个5×10阵列，而在这种情况下， ，我会得到一个列向量（而不是二维数组）的结果。
So what's happening is rolling_apply is going down each column in turn and applying a sliding 5-length window down each one of these, whereas what I want is for the sliding windows to be a 5x10 array each time, and in this case, I would get a single column vector (not 2d array) result.
推荐答案
我确实无法找到一种方法来计算宽，在大熊猫滚动的应用
文档，所以我会使用numpy的，以获得阵列上的窗口化的观点，并应用ufunc
给它。这里有一个例子：
I indeed cannot find a way to compute "wide" rolling application in pandas docs, so I'd use numpy to get a "windowing" view on the array and apply a ufunc to it. Here's an example:
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 是8，因为默认情况下DTYPE是 np.int64 ，你需要它下面的窗口鉴于成语：
arr.itemsize is 8 because default dtype is np.int64, you need it for the following "window" view idiom:
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]]])
健是沿给定轴的两个相邻元件之间的字节数，
因此， =迈进（arr.shape [1] * isize，arr.shape [1] * isize，isize）表示跳过5
从元素窗口去当[0]为窗口[1]，并跳过时，5种元素
从窗口去[0,0]为窗口[0,1]。现在你可以呼吁任何ufunc
结果数组，例如：
Strides are number of bytes between two neighbour elements along given axis,
thus strides=(arr.shape[1] * isize, arr.shape[1] * isize, isize) means skip 5
elements when going from windowed[0] to windowed[1] and skip 5 elements when
going from windowed[0, 0] to windowed[0, 1]. Now you can call any ufunc on the
resulting array, e.g.:
In [44]: windowed.sum(axis=(1,2))
Out[44]: array([300, 425, 550, 675, 800, 925])
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