Question

我有一个函数f，我想在滑动窗口中高效计算。

def efficient_f(x):
   # do stuff
   wSize=50
   return another_f(rolling_window_using_strides(x, wSize), -1)

我在SO上看到使用步幅来做这件事特别有效：来自numpy.lib.stride_tricks导入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)

然后我尝试将它应用于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.

它的工作非常好，确实带来了显着的性能提升。但是，我仍然收到以下错误：

ValueError: Shape of passed values is (1, 179951), indices imply (1, 180000).

这是因为我使用wSize = 50，产生

rolling_window_using_strides(df['foo'].values,50).shape
(1L, 179951L, 50L)

有没有办法通过边界处的零/ np.nan填充来获得

(1L, 180000, 50L)

因此与原始载体的大小相同

Answer 1

这是使用np.lib.stride_tricks.as_strided -

解决问题的一种方法

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))

示例运行 -

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]])

padding numpy滚动窗口操作使用步幅

1 个答案: