numpy.stride_tricks返回垃圾值 [英] numpy.stride_tricks returns junk values

问题描述

我正在尝试使用numpy.strided_tricks重塑一个numpy数组.这是我要遵循的指南: https://stackoverflow.com/a/2487551/4909087

I'm trying to reshape a numpy array using numpy.strided_tricks. This is the guide I'm following: https://stackoverflow.com/a/2487551/4909087

我的用例非常相似，不同之处在于我需要跨度为3.

My use case is very similar, with the difference being that I need strides of 3.

给出此数组:

a = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9])

我想要得到:

array([[1, 2, 3],
       [2, 3, 4],
       [3, 4, 5],
       [4, 5, 6],
       [5, 6, 7],
       [6, 7, 8],
       [7, 8, 9]])

这是我尝试过的:

import numpy as np

as_strided = np.lib.stride_tricks.as_strided
a = np.arange(1, 10)

as_strided(a, (len(a) - 2, 3), (3, 3))

array([[                 1,      2199023255552,             131072],
       [     2199023255552,             131072, 216172782113783808],
       [            131072, 216172782113783808,        12884901888],
       [216172782113783808,        12884901888,                768],
       [       12884901888,                768,   1125899906842624],
       [               768,   1125899906842624,           67108864],
       [  1125899906842624,           67108864,                  4]])

我很确定我将示例跟在了T上，但显然没有.我要去哪里错了?

I was pretty sure I'd followed the example to a T, but evidently not. Where am I going wrong?

推荐答案

我不知道为什么您认为需要跨度为3.您需要跨度跨度以字节为单位的a元素与下一个元素之间的距离.可以使用a.strides:

I have no idea why you think you need strides of 3. You need strides the distance in bytes between one element of a and the next, which you can get using a.strides:

as_strided(a, (len(a) - 2, 3), a.strides*2)

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