滚动窗口的数据框表示 [英] dataframe representation of a rolling window
问题描述
我想要一个滚动窗口的数据框表示.我希望在一个数据帧中以另一个维度表示该窗口，而不是对滚动窗口执行某些操作.可以是pd.Panel或np.array或pd.DataFrame和pd.MultiIndex.
I want a dataframe representation of of a rolling window. Instead of performing some operation on a rolling window, I want a dataframe where the window is represented in another dimension. This could be as a pd.Panel or np.array or a pd.DataFrame with a pd.MultiIndex.
import pandas as pd
import numpy as np
np.random.seed([3,1415])
df = pd.DataFrame(np.random.rand(10, 3).round(2),
columns=['A', 'B', 'C'],
index=list('abcdefghij'))
print df
A B C
a 0.44 0.41 0.46
b 0.47 0.46 0.02
c 0.85 0.82 0.78
d 0.76 0.93 0.83
e 0.88 0.93 0.72
f 0.12 0.15 0.20
g 0.44 0.10 0.28
h 0.61 0.09 0.84
i 0.74 0.87 0.69
j 0.38 0.23 0.44
预期产量
对于window = 2，我希望结果是
Expected Output
For a window = 2 I'd expect the result to be.
0 1
A B C A B C
a 0.44 0.41 0.46 0.47 0.46 0.02
b 0.47 0.46 0.02 0.85 0.82 0.78
c 0.85 0.82 0.78 0.76 0.93 0.83
d 0.76 0.93 0.83 0.88 0.93 0.72
e 0.88 0.93 0.72 0.12 0.15 0.20
f 0.12 0.15 0.20 0.44 0.10 0.28
g 0.44 0.10 0.28 0.61 0.09 0.84
h 0.61 0.09 0.84 0.74 0.87 0.69
i 0.74 0.87 0.69 0.38 0.23 0.44
我不确定是否以这种方式显示布局，但这是我想要的信息.我正在寻找最有效的方法.
I'm not determined to have the layout presented this way, but this is the information I want. I'm looking for the most efficient way to get at this.
我已经尝试过使用shift的各种方式，但是感觉很笨拙.这就是我用来产生以上输出的内容:
I've experimented with using shift in varying ways but it feels clunky. This is what I use to produce the output above:
print pd.concat([df, df.shift(-1)], axis=1, keys=[0, 1]).dropna()
推荐答案
We could use NumPy to get views into those sliding windows with its esoteric strided tricks. If you are using this new dimension for some reduction like matrix-multiplication, this would be ideal. If for some reason, you want to have a 2D output, we need to use a reshape at the end, which will result in creating a copy though.
因此，实现看起来像这样-
Thus, the implementation would look something like this -
from numpy.lib.stride_tricks import as_strided as strided
def get_sliding_window(df, W, return2D=0):
a = df.values
s0,s1 = a.strides
m,n = a.shape
out = strided(a,shape=(m-W+1,W,n),strides=(s0,s0,s1))
if return2D==1:
return out.reshape(a.shape[0]-W+1,-1)
else:
return out
用于2D/3D输出的样本运行-
Sample run for 2D/3D output -
In [68]: df
Out[68]:
A B
0 0.44 0.41
1 0.46 0.47
2 0.46 0.02
3 0.85 0.82
4 0.78 0.76
In [70]: get_sliding_window(df, 3,return2D=1)
Out[70]:
array([[ 0.44, 0.41, 0.46, 0.47, 0.46, 0.02],
[ 0.46, 0.47, 0.46, 0.02, 0.85, 0.82],
[ 0.46, 0.02, 0.85, 0.82, 0.78, 0.76]])
这是3D视图输出的样子-
Here's how the 3D views output would look like -
In [69]: get_sliding_window(df, 3,return2D=0)
Out[69]:
array([[[ 0.44, 0.41],
[ 0.46, 0.47],
[ 0.46, 0.02]],
[[ 0.46, 0.47],
[ 0.46, 0.02],
[ 0.85, 0.82]],
[[ 0.46, 0.02],
[ 0.85, 0.82],
[ 0.78, 0.76]]])
让我们来看看各种窗口大小的视图3D输出-
Let's time it for views 3D output for various window sizes -
In [331]: df = pd.DataFrame(np.random.rand(1000, 3).round(2))
In [332]: %timeit get_3d_shfted_array(df,2) # @Yakym Pirozhenko's soln
10000 loops, best of 3: 47.9 µs per loop
In [333]: %timeit get_sliding_window(df,2)
10000 loops, best of 3: 39.2 µs per loop
In [334]: %timeit get_3d_shfted_array(df,5) # @Yakym Pirozhenko's soln
10000 loops, best of 3: 89.9 µs per loop
In [335]: %timeit get_sliding_window(df,5)
10000 loops, best of 3: 39.4 µs per loop
In [336]: %timeit get_3d_shfted_array(df,15) # @Yakym Pirozhenko's soln
1000 loops, best of 3: 258 µs per loop
In [337]: %timeit get_sliding_window(df,15)
10000 loops, best of 3: 38.8 µs per loop
让我们确认我们确实在获得观看次数-
Let's verify that we are indeed getting views -
In [338]: np.may_share_memory(get_sliding_window(df,2), df.values)
Out[338]: True
即使在各种窗口大小下，使用get_sliding_window的时间几乎都是恒定的，这表明获取视图而不是复制具有巨大的优势.
The almost constant timings with get_sliding_window even across various window sizes suggest the huge benefit of getting the view instead of copying.
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