【发布时间】:2018-07-20 00:00:07
【问题描述】:
我正在寻找一种将 numpy 数组分割成重叠块的有效方法。我知道numpy.lib.stride_tricks.as_strided 可能是要走的路,但我似乎无法理解它在适用于任意形状数组的通用函数中的用法。 Here are some examples for specific applications of as_strided.
这是我想要的:
import numpy as np
from numpy.lib.stride_tricks import as_strided
def segment(arr, axis, new_len, step=1, new_axis=None):
""" Segment an array along some axis.
Parameters
----------
arr : array-like
The input array.
axis : int
The axis along which to segment.
new_len : int
The length of each segment.
step : int, default 1
The offset between the start of each segment.
new_axis : int, optional
The position where the newly created axis is to be inserted. By
default, the axis will be added at the end of the array.
Returns
-------
arr_seg : array-like
The segmented array.
"""
# calculate shape after segmenting
new_shape = list(arr.shape)
new_shape[axis] = (new_shape[axis] - new_len + step) // step
if new_axis is None:
new_shape.append(new_len)
else:
new_shape.insert(new_axis, new_len)
# TODO: calculate new strides
strides = magic_command_returning_strides(...)
# get view with new strides
arr_seg = as_strided(arr, new_shape, strides)
return arr_seg.copy()
所以我想指定要被切割成段的轴、段的长度以及它们之间的步长。此外,我想将插入新轴的位置作为参数传递。唯一缺少的是步幅的计算。
我知道这可能无法直接与as_strided 一起工作,即我可能需要实现一个子例程,该子例程返回一个带有step=1 和new_axis 在固定位置的跨步视图,然后与想要的step 然后转置。
这是一段有效的代码,但显然很慢:
def segment_slow(arr, axis, new_len, step=1, new_axis=None):
""" Segment an array along some axis. """
# calculate shape after segmenting
new_shape = list(arr.shape)
new_shape[axis] = (new_shape[axis] - new_len + step) // step
if new_axis is None:
new_shape.append(new_len)
else:
new_shape.insert(new_axis, new_len)
# check if the new axis is inserted before the axis to be segmented
if new_axis is not None and new_axis <= axis:
axis_in_arr_seg = axis + 1
else:
axis_in_arr_seg = axis
# pre-allocate array
arr_seg = np.zeros(new_shape, dtype=arr.dtype)
# setup up indices
idx_old = [slice(None)] * arr.ndim
idx_new = [slice(None)] * len(new_shape)
# get order of transposition for assigning slices to the new array
order = list(range(arr.ndim))
if new_axis is None:
order[-1], order[axis] = order[axis], order[-1]
elif new_axis > axis:
order[new_axis-1], order[axis] = order[axis], order[new_axis-1]
# loop over the axis to be segmented
for n in range(new_shape[axis_in_arr_seg]):
idx_old[axis] = n * step + np.arange(new_len)
idx_new[axis_in_arr_seg] = n
arr_seg[tuple(idx_new)] = np.transpose(arr[idx_old], order)
return arr_seg
这是基本功能的测试:
import numpy.testing as npt
arr = np.array([[0, 1, 2, 3],
[4, 5, 6, 7],
[8, 9, 10, 11]])
arr_seg_1 = segment_slow(arr, axis=1, new_len=3, step=1)
arr_target_1 = np.array([[[0, 1, 2], [1, 2, 3]],
[[4, 5, 6], [5, 6, 7]],
[[8, 9, 10], [9, 10, 11]]])
npt.assert_allclose(arr_target_1, arr_seg_1)
arr_seg_2 = segment_slow(arr, axis=1, new_len=3, step=1, new_axis=1)
arr_target_2 = np.transpose(arr_target_1, (0, 2, 1))
npt.assert_allclose(arr_target_2, arr_seg_2)
arr_seg_3 = segment_slow(arr, axis=0, new_len=2, step=1)
arr_target_3 = np.array([[[0, 4], [1, 5], [2, 6], [3, 7]],
[[4, 8], [5, 9], [6, 10], [7, 11]]])
npt.assert_allclose(arr_target_3, arr_seg_3)
任何帮助将不胜感激!
【问题讨论】:
-
恭喜您提出了非常完善的问题! :)
-
My attempt at a canonical answer 执行所有这些操作,除了将新轴滚动到指定位置。应该很容易在末尾添加
np.rollaxis。 -
感谢@DanielF,我已经成功实现了我需要的功能,并使用了您的函数的包装器。我将根据您的方法发布我的问题的答案。
标签: python arrays performance numpy