numpy stride_tricks.as_strided vs滚动窗口的列表理解

In this issue concerning why stride_tricks.as_strided is undocumented, it's mentioned that 故意！这很危险！只是低级管道即可帮助实现broadcast_arrays(). Intentionally! It's dangerous! It was just low-level plumbing to help implement broadcast_arrays(). stride_tricks.as_strided 是否比列表理解或for循环有任何优势?我看了 stride_tricks的源代码 ，但收益不大.Is there any advantage for stride_tricks.as_strided over the list comprehension or a for loop? I had a look at the source code of stride_tricks but gained little.推荐答案来自 此帖子 ，我们可以使用 strided_app 基本上将滑动视图获取到数组中，并且还可以指定hopsize/stepsize.然后，我们只需沿第二个轴使用 np.std 即可获得最终输出，就像这样-From this post, we can use strided_app to basically get sliding views into the array and it also allows us to specify the hopsize/stepsize. Then, we simply use np.std along the second axis for the final output, like so -np.std(strided_app(x, framesize, hopsize), axis=1)运行示例以进行验证-In [162]: x = np.random.randint(0,9,(11))In [163]: framesize = 5In [164]: hopsize = 3In [165]: np.array([np.std(x[i:i+framesize]) \ for i in range(0, len(x)-framesize+1, hopsize)])Out[165]: array([ 1.62480768, 2.05912603, 1.78885438])In [166]: np.std(strided_app(x, framesize, hopsize), axis=1)Out[166]: array([ 1.62480768, 2.05912603, 1.78885438])作为输入数组的视图，这些跨步操作必须非常有效.让我们找出答案吧！Being a view into the input array, these strided operations must be really efficient. Let's find it out! 运行时测试 循环方法-def loopy_app(x, framesize, hopsize): return [np.std(x[i:i+framesize]) \ for i in range(0, len(x)-framesize+1, hopsize)]时间-In [185]: x = np.random.randint(0,9,(1001))In [186]: framesize = 5In [187]: hopsize = 3In [188]: %timeit loopy_app(x, framesize, hopsize)10 loops, best of 3: 17.8 ms per loopIn [189]: %timeit np.std(strided_app(x, framesize, hopsize), axis=1)10000 loops, best of 3: 111 µs per loop因此，要通过 stride 来回答效率问题，时间安排应该有助于证明这一点！So, to answer the question on efficiency with strides, the timings should help prove a point there! 这篇关于numpy stride_tricks.as_strided vs滚动窗口的列表理解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！