python - 使用cython加速功能

Question

我正在尝试加快我的一项功能。

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]

这是我使用 Cython 加速它的尝试:

# 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]

但我看不出速度有任何提升。我还尝试用 memoryview cnp.double_t[:, :,::1] 替换 cnp.ndarray[cnp.double_t, ndim=3] 但它只会减慢速度下来整个代码。我将感谢对我的代码的任何提示或更正。我对 Cython 比较陌生，我可能做错了什么。

编辑:

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]

然后我用我的函数和建议的矢量化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 命名空间代表这两个使用 Cython 编译的函数。

dynamic 命名空间代表普通的 Python 文件。

因此，我得出结论，在这种情况下，numpy 方法更好。

Answer 1

您的 Python 代码仍然可以改进，因为您还没有“在 numpy 中完成 98%”:您仍在迭代坐标数组的行并对每行执行 1-2 次检查。

您可以使用 numpy 的“花式索引”和掩码以矢量化形式完全获取它:

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))

我得到以下计时结果:

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

当然，性能测试要小心，因为它通常取决于所使用的数据。

我对 Cython 没有什么经验，无法帮助你。