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 pandasdocs, so I'd use numpy to get a "windowing" view on the array and apply a ufuncto 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 5elements when going from windowed[0] to windowed[1] and skip 5 elements whengoing from windowed[0, 0] to windowed[0, 1]. Now you can call any ufunc on theresulting array, e.g.:
In [44]: windowed.sum(axis=(1,2))
Out[44]: array([300, 425, 550, 675, 800, 925])
这篇关于与 pandas 或numpy的n维滑动窗口的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!