使用跨度填充numpy滚动窗口操作 [英] padding numpy rolling window operations using strides
问题描述
我有一个函数f,我想在滑动窗口中有效地进行计算.
I have a function f that I would like to efficiently compute in a sliding window.
def efficient_f(x):
# do stuff
wSize=50
return another_f(rolling_window_using_strides(x, wSize), -1)
我在SO上看到使用跨步执行此操作特别有效: 从numpy.lib.stride_tricks导入as_strided
I have seen on SO that is particularly efficient to do that using strides: from numpy.lib.stride_tricks import as_strided
def rolling_window_using_strides(a, window):
shape = a.shape[:-1] + (a.shape[-1] - window + 1, window)
strides = a.strides + (a.strides[-1],)
print np.lib.stride_tricks.as_strided(a, shape=shape, strides=strides).shape
return np.lib.stride_tricks.as_strided(a, shape=shape, strides=strides)
然后我尝试将其应用于df:
Then I try to apply it on a df:
df=pd.DataFrame(data=np.random.rand(180000,1),columns=['foo'])
df['bar']=df[['foo']].apply(efficient_f,raw=True)
# note the double [[, otherwise pd.Series.apply
# (not accepting raw, and axis kwargs) will be called instead of pd.DataFrame.
它运行得非常好,并且确实带来了显着的性能提升. 但是,我仍然收到以下错误:
It is working very nicely, and it indeed led to significant performance gains. However, I still get the following error:
ValueError: Shape of passed values is (1, 179951), indices imply (1, 180000).
这是因为我正在使用wSize = 50,它会产生
This is because I am using wSize=50, which yields
rolling_window_using_strides(df['foo'].values,50).shape
(1L, 179951L, 50L)
有没有一种方法可以通过零/np.nan边界填充来获得
Is there a way by zero/np.nan padding at the borders to get
(1L, 180000, 50L)
因此具有与原始矢量相同的大小
hence same size as the original vector
推荐答案
这是使用 np.lib.stride_tricks.as_strided
-
def strided_axis0(a, fillval, L): # a is 1D array
a_ext = np.concatenate(( np.full(L-1,fillval) ,a))
n = a_ext.strides[0]
strided = np.lib.stride_tricks.as_strided
return strided(a_ext, shape=(a.shape[0],L), strides=(n,n))
样品运行-
In [95]: np.random.seed(0)
In [96]: a = np.random.rand(8,1)
In [97]: a
Out[97]:
array([[ 0.55],
[ 0.72],
[ 0.6 ],
[ 0.54],
[ 0.42],
[ 0.65],
[ 0.44],
[ 0.89]])
In [98]: strided_axis0(a[:,0], fillval=np.nan, L=3)
Out[98]:
array([[ nan, nan, 0.55],
[ nan, 0.55, 0.72],
[ 0.55, 0.72, 0.6 ],
[ 0.72, 0.6 , 0.54],
[ 0.6 , 0.54, 0.42],
[ 0.54, 0.42, 0.65],
[ 0.42, 0.65, 0.44],
[ 0.65, 0.44, 0.89]])
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