我正在努力加快我的一项职能。
def get_scale_local_maximas(cube_coordinates, laplacian_cube):
"""
Check provided cube coordinate for scale space local maximas.
Returns only the points that satisfy the criteria.
A point is considered to be a local maxima if its value is greater
than the value of the point on the next scale level and the point
on the previous scale level. If the tested point is located on the
first scale level or on the last one, then only one inequality should
hold in order for this point to be local scale maxima.
Parameters
----------
cube_coordinates : (n, 3) ndarray
A 2d array with each row representing 3 values, ``(y,x,scale_level)``
where ``(y,x)`` are coordinates of the blob and ``scale_level`` is the
position of a point in scale space.
laplacian_cube : ndarray of floats
Laplacian of Gaussian scale space.
Returns
-------
output : (n, 3) ndarray
cube_coordinates that satisfy the local maximum criteria in
scale space.
Examples
--------
>>> one = np.array([[1, 2, 3], [4, 5, 6]])
>>> two = np.array([[7, 8, 9], [10, 11, 12]])
>>> three = np.array([[0, 0, 0], [0, 0, 0]])
>>> check_coords = np.array([[1, 0, 1], [1, 0, 0], [1, 0, 2]])
>>> lapl_dummy = np.dstack([one, two, three])
>>> get_scale_local_maximas(check_coords, lapl_dummy)
array([[1, 0, 1]])
"""
amount_of_layers = laplacian_cube.shape[2]
amount_of_points = cube_coordinates.shape[0]
# Preallocate index. Fill it with False.
accepted_points_index = np.ones(amount_of_points, dtype=bool)
for point_index, interest_point_coords in enumerate(cube_coordinates):
# Row coordinate
y_coord = interest_point_coords[0]
# Column coordinate
x_coord = interest_point_coords[1]
# Layer number starting from the smallest sigma
point_layer = interest_point_coords[2]
point_response = laplacian_cube[y_coord, x_coord, point_layer]
# Check the point under the current one
if point_layer != 0:
lower_point_response = laplacian_cube[y_coord, x_coord, point_layer-1]
if lower_point_response >= point_response:
accepted_points_index[point_index] = False
continue
# Check the point above the current one
if point_layer != (amount_of_layers-1):
upper_point_response = laplacian_cube[y_coord, x_coord, point_layer+1]
if upper_point_response >= point_response:
accepted_points_index[point_index] = False
continue
# Return only accepted points
return cube_coordinates[accepted_points_index]
这是我尝试使用Cython来加速它：
# cython: cdivision=True
# cython: boundscheck=False
# cython: nonecheck=False
# cython: wraparound=False
import numpy as np
cimport numpy as cnp
def get_scale_local_maximas(cube_coordinates, cnp.ndarray[cnp.double_t, ndim=3] laplacian_cube):
"""
Check provided cube coordinate for scale space local maximas.
Returns only the points that satisfy the criteria.
A point is considered to be a local maxima if its value is greater
than the value of the point on the next scale level and the point
on the previous scale level. If the tested point is located on the
first scale level or on the last one, then only one inequality should
hold in order for this point to be local scale maxima.
Parameters
----------
cube_coordinates : (n, 3) ndarray
A 2d array with each row representing 3 values, ``(y,x,scale_level)``
where ``(y,x)`` are coordinates of the blob and ``scale_level`` is the
position of a point in scale space.
laplacian_cube : ndarray of floats
Laplacian of Gaussian scale space.
Returns
-------
output : (n, 3) ndarray
cube_coordinates that satisfy the local maximum criteria in
scale space.
Examples
--------
>>> one = np.array([[1, 2, 3], [4, 5, 6]])
>>> two = np.array([[7, 8, 9], [10, 11, 12]])
>>> three = np.array([[0, 0, 0], [0, 0, 0]])
>>> check_coords = np.array([[1, 0, 1], [1, 0, 0], [1, 0, 2]])
>>> lapl_dummy = np.dstack([one, two, three])
>>> get_scale_local_maximas(check_coords, lapl_dummy)
array([[1, 0, 1]])
"""
cdef Py_ssize_t y_coord, x_coord, point_layer, point_index
cdef cnp.double_t point_response, lower_point_response, upper_point_response
cdef Py_ssize_t amount_of_layers = laplacian_cube.shape[2]
cdef Py_ssize_t amount_of_points = cube_coordinates.shape[0]
# amount_of_layers = laplacian_cube.shape[2]
# amount_of_points = cube_coordinates.shape[0]
# Preallocate index. Fill it with False.
accepted_points_index = np.ones(amount_of_points, dtype=bool)
for point_index in range(amount_of_points):
interest_point_coords = cube_coordinates[point_index]
# Row coordinate
y_coord = interest_point_coords[0]
# Column coordinate
x_coord = interest_point_coords[1]
# Layer number starting from the smallest sigma
point_layer = interest_point_coords[2]
point_response = laplacian_cube[y_coord, x_coord, point_layer]
# Check the point under the current one
if point_layer != 0:
lower_point_response = laplacian_cube[y_coord, x_coord, point_layer-1]
if lower_point_response >= point_response:
accepted_points_index[point_index] = False
continue
# Check the point above the current one
if point_layer != (amount_of_layers-1):
upper_point_response = laplacian_cube[y_coord, x_coord, point_layer+1]
if upper_point_response >= point_response:
accepted_points_index[point_index] = False
continue
# Return only accepted points
return cube_coordinates[accepted_points_index]
但我看不出速度没有增加。此外，我尝试将cnp.ndarray[cnp.double_t, ndim=3]替换为memoryview cnp.double_t[:, :, ::1]，但它只会降低整个代码的速度。
我将很感激我的代码的任何提示或更正。我对Cython比较新，我可能做错了。
修改
我在Cython中完全重写了我的函数：
def get_scale_local_maximas(cnp.ndarray[cnp.int_t, ndim=2] cube_coordinates, cnp.ndarray[cnp.double_t, ndim=3] laplacian_cube):
"""
Check provided cube coordinate for scale space local maximas.
Returns only the points that satisfy the criteria.
A point is considered to be a local maxima if its value is greater
than the value of the point on the next scale level and the point
on the previous scale level. If the tested point is located on the
first scale level or on the last one, then only one inequality should
hold in order for this point to be local scale maxima.
Parameters
----------
cube_coordinates : (n, 3) ndarray
A 2d array with each row representing 3 values, ``(y,x,scale_level)``
where ``(y,x)`` are coordinates of the blob and ``scale_level`` is the
position of a point in scale space.
laplacian_cube : ndarray of floats
Laplacian of Gaussian scale space.
Returns
-------
output : (n, 3) ndarray
cube_coordinates that satisfy the local maximum criteria in
scale space.
Examples
--------
>>> one = np.array([[1, 2, 3], [4, 5, 6]])
>>> two = np.array([[7, 8, 9], [10, 11, 12]])
>>> three = np.array([[0, 0, 0], [0, 0, 0]])
>>> check_coords = np.array([[1, 0, 1], [1, 0, 0], [1, 0, 2]])
>>> lapl_dummy = np.dstack([one, two, three])
>>> get_scale_local_maximas(check_coords, lapl_dummy)
array([[1, 0, 1]])
"""
cdef Py_ssize_t y_coord, x_coord, point_layer, point_index
cdef cnp.double_t point_response, lower_point_response, upper_point_response
cdef Py_ssize_t amount_of_layers = laplacian_cube.shape[2]
cdef Py_ssize_t amount_of_points = cube_coordinates.shape[0]
# Preallocate index. Fill it with False.
accepted_points_index = np.ones(amount_of_points, dtype=bool)
for point_index in range(amount_of_points):
interest_point_coords = cube_coordinates[point_index]
# Row coordinate
y_coord = interest_point_coords[0]
# Column coordinate
x_coord = interest_point_coords[1]
# Layer number starting from the smallest sigma
point_layer = interest_point_coords[2]
point_response = laplacian_cube[y_coord, x_coord, point_layer]
# Check the point under the current one
if point_layer != 0:
lower_point_response = laplacian_cube[y_coord, x_coord, point_layer-1]
if lower_point_response >= point_response:
accepted_points_index[point_index] = False
continue
# Check the point above the current one
if point_layer != (amount_of_layers-1):
upper_point_response = laplacian_cube[y_coord, x_coord, point_layer+1]
if upper_point_response >= point_response:
accepted_points_index[point_index] = False
continue
# Return only accepted points
return cube_coordinates[accepted_points_index]
之后我用我的函数和建议的函数做了一些基准测试：
%timeit compiled.get_scale_local_maximas_np(coords, lapl_dummy)
%timeit compiled.get_scale_local_maximas(coords, lapl_dummy)
%timeit dynamic.get_scale_local_maximas_np(coords, lapl_dummy)
%timeit dynamic.get_scale_local_maximas(coords, lapl_dummy)
10000 loops, best of 3: 101 µs per loop
1000 loops, best of 3: 328 µs per loop
10000 loops, best of 3: 103 µs per loop
1000 loops, best of 3: 1.6 ms per loop
compiled命名空间表示使用Cython编译的这两个函数。
dynamic命名空间代表通常的Python文件。
所以，我得出结论，在这种情况下，numpy方法更好。
答案 0 :(得分：4)
你的Python代码仍然可以改进，因为你并没有＆＃34;已经做了98％的numpy＆＃34;：你仍在迭代坐标数组的行并执行1-2每行检查。
你可以使用numpy＆＃34;花式索引＆＃34;和掩码以矢量化形式完全：
def get_scale_local_maximas_full_np(coords, cube):
x, y, z = [ coords[:, ind] for ind in range(3) ]
point_responses = cube[x, y, z]
lowers = point_responses.copy()
uppers = point_responses.copy()
not_layer_0 = z > 0
lower_responses = cube[x[not_layer_0], y[not_layer_0], z[not_layer_0]-1]
lowers[not_layer_0] = lower_responses
not_max_layer = z < (cube.shape[2] - 1)
upper_responses = cube[x[not_max_layer], y[not_max_layer], z[not_max_layer]+1]
uppers[not_max_layer] = upper_responses
lo_check = np.ones(z.shape, dtype=np.bool)
lo_check[not_layer_0] = (point_responses > lowers)[not_layer_0]
hi_check = np.ones(z.shape, dtype=np.bool)
hi_check[not_max_layer] = (point_responses > uppers)[not_max_layer]
return coords[lo_check & hi_check]
我已经生成了一组更大的数据来测试性能：
lapl_dummy = np.random.rand(100,100,100)
coords = np.random.random_integers(0,99, size=(1000,3))
我得到以下时间结果：
In [146]: %timeit get_scale_local_maximas_full_np(coords, lapl_dummy)
10000 loops, best of 3: 175 µs per loop
In [147]: %timeit get_scale_local_maximas(coords, lapl_dummy)
100 loops, best of 3: 2.24 ms per loop
但当然，要小心性能测试，因为它通常取决于使用的数据。
我对Cython没什么经验，无法帮助你。