Sliding standard deviation on a 1D NumPy array

Question

Suppose that you have an array and want to create another array, which's values are equal to standard deviation of first array's 10 elements successively. With the help of for loop, it can be written easily like below code. What I want to do is avoid using for loop for faster execution time. Any suggestions?

Code
a = np.arange(20)
b = np.empty(11)
for i in range(11):
    b[i] = np.std(a[i:i+10])

Answer 1

您可以创建一个滑动窗口的 2D 数组，np.lib.stride_tricks.as_strided它是给定1D数组的视图，因此不会占用更多内存。然后，只需np.std沿第二个轴（轴 = 1）以矢量化方式使用最终结果，就像这样 -

W = 10 # Window size
nrows = a.size - W + 1
n = a.strides[0]
a2D = np.lib.stride_tricks.as_strided(a,shape=(nrows,W),strides=(n,n))
out = np.std(a2D, axis=1)

运行时测试

函数定义 -

def original_app(a, W):
    b = np.empty(a.size-W+1)
    for i in range(b.size):
        b[i] = np.std(a[i:i+W])
    return b
    
def vectorized_app(a, W):
    nrows = a.size - W + 1
    n = a.strides[0]
    a2D = np.lib.stride_tricks.as_strided(a,shape=(nrows,W),strides=(n,n))
    return np.std(a2D,1)

时间和验证 -

In [460]: # Inputs
     ...: a = np.arange(10000)
     ...: W = 10
     ...: 

In [461]: np.allclose(original_app(a, W), vectorized_app(a, W))
Out[461]: True

In [462]: %timeit original_app(a, W)
1 loops, best of 3: 522 ms per loop

In [463]: %timeit vectorized_app(a, W)
1000 loops, best of 3: 1.33 ms per loop

所以，在400x那里加速！

为了完整起见，这是等效的熊猫版本-

import pandas as pd

def pdroll(a, W): # a is 1D ndarray and W is window-size
    return pd.Series(a).rolling(W).std(ddof=0).values[W-1:]