刚性转换 - Python - 加速
Rigid transformation - Python - speedup
我有以下关于计算刚性变换的更快方法的问题(是的,我知道我可以简单地使用库,但需要自己编写代码)。
我需要为给定图像中的每个 x,y 计算 x' 和 y'。我的主要瓶颈是所有坐标的点积(之后的插值不是问题)。目前我实现了三个选项:
列表理解
result = np.array([[np.dot(matrix, np.array([x, y, 1])) for x in xs] for y in ys])
简单双for
循环
result = np.zeros((Y, X, 3))
for x in xs:
for y in ys:
result[y, x, :] = np.dot(matrix, np.array([x, y, 1]))
np.ndenumerate
result = np.zeros((Y, X, 3))
for (y, x), value in np.ndenumerate(image):
result[y, x, :] = np.dot(matrix, np.array([x, y, 1]))
512x512 图像中最快的方法是列表理解(比 np.ndenumerate
快 1.5 倍,比 double for 循环快 1.1 倍)。
有什么方法可以更快地做到这一点?
您可以使用 np.indices
和 np.rollaxis
生成 3D 数组,其中 coords[i, j] == [i, j]
。这里坐标需要切换
然后你所做的就是附加你要求的1
,然后使用@
coords_ext = np.empty((Y, X, 3))
coords_ext[...,[1,0]] = np.rollaxis(np.indices((Y, X)), 0, start=3)
coords_ext[...,2] = 1
# convert to column vectors and back for matmul broadcasting
result = (matrix @ coords_ext[...,None])[...,0]
# or alternatively, work with row vectors and do it in the other order
result = coords_ext @ matrix.T
借用 , the idea of creating 1D
arrays instead of the 2D
or 3D
meshes and using them with on-the-fly broadcasted
操作来提高内存效率,从而获得性能优势,这是一种方法 -
out = ys[:,None,None]*matrix[:,1] + xs[:,None]*matrix[:,0] + matrix[:,2]
如果您使用 xs
和 ys
覆盖 512x512
大小图像的所有索引,我们将使用 np.arange
创建它们,就像这样 -
ys = np.arange(512)
xs = np.arange(512)
运行时测试
函数定义-
def original_listcomp_app(matrix, X, Y): # Original soln with list-compr.
ys = np.arange(Y)
xs = np.arange(X)
out = np.array([[np.dot(matrix, np.array([x, y, 1])) for x in xs] \
for y in ys])
return out
def indices_app(matrix, X, Y): # @Eric's soln
coords_ext = np.empty((Y, X, 3))
coords_ext[...,[1,0]] = np.rollaxis(np.indices((Y, X)), 0, start=3)
coords_ext[...,2] = 1
result = np.matmul(coords_ext,matrix.T)
return result
def broadcasting_app(matrix, X, Y): # Broadcasting based
ys = np.arange(Y)
xs = np.arange(X)
out = ys[:,None,None]*matrix[:,1] + xs[:,None]*matrix[:,0] + matrix[:,2]
return out
时间和验证 -
In [242]: # Inputs
...: matrix = np.random.rand(3,3)
...: X,Y = 512, 512
...:
In [243]: out0 = original_listcomp_app(matrix, X, Y)
...: out1 = indices_app(matrix, X, Y)
...: out2 = broadcasting_app(matrix, X, Y)
...:
In [244]: np.allclose(out0, out1)
Out[244]: True
In [245]: np.allclose(out0, out2)
Out[245]: True
In [253]: %timeit original_listcomp_app(matrix, X, Y)
1 loops, best of 3: 1.32 s per loop
In [254]: %timeit indices_app(matrix, X, Y)
100 loops, best of 3: 16.1 ms per loop
In [255]: %timeit broadcasting_app(matrix, X, Y)
100 loops, best of 3: 9.64 ms per loop
我有以下关于计算刚性变换的更快方法的问题(是的,我知道我可以简单地使用库,但需要自己编写代码)。
我需要为给定图像中的每个 x,y 计算 x' 和 y'。我的主要瓶颈是所有坐标的点积(之后的插值不是问题)。目前我实现了三个选项:
列表理解
result = np.array([[np.dot(matrix, np.array([x, y, 1])) for x in xs] for y in ys])
简单双
for
循环result = np.zeros((Y, X, 3)) for x in xs: for y in ys: result[y, x, :] = np.dot(matrix, np.array([x, y, 1]))
np.ndenumerate
result = np.zeros((Y, X, 3)) for (y, x), value in np.ndenumerate(image): result[y, x, :] = np.dot(matrix, np.array([x, y, 1]))
512x512 图像中最快的方法是列表理解(比 np.ndenumerate
快 1.5 倍,比 double for 循环快 1.1 倍)。
有什么方法可以更快地做到这一点?
您可以使用 np.indices
和 np.rollaxis
生成 3D 数组,其中 coords[i, j] == [i, j]
。这里坐标需要切换
然后你所做的就是附加你要求的1
,然后使用@
coords_ext = np.empty((Y, X, 3))
coords_ext[...,[1,0]] = np.rollaxis(np.indices((Y, X)), 0, start=3)
coords_ext[...,2] = 1
# convert to column vectors and back for matmul broadcasting
result = (matrix @ coords_ext[...,None])[...,0]
# or alternatively, work with row vectors and do it in the other order
result = coords_ext @ matrix.T
借用 1D
arrays instead of the 2D
or 3D
meshes and using them with on-the-fly broadcasted
操作来提高内存效率,从而获得性能优势,这是一种方法 -
out = ys[:,None,None]*matrix[:,1] + xs[:,None]*matrix[:,0] + matrix[:,2]
如果您使用 xs
和 ys
覆盖 512x512
大小图像的所有索引,我们将使用 np.arange
创建它们,就像这样 -
ys = np.arange(512)
xs = np.arange(512)
运行时测试
函数定义-
def original_listcomp_app(matrix, X, Y): # Original soln with list-compr.
ys = np.arange(Y)
xs = np.arange(X)
out = np.array([[np.dot(matrix, np.array([x, y, 1])) for x in xs] \
for y in ys])
return out
def indices_app(matrix, X, Y): # @Eric's soln
coords_ext = np.empty((Y, X, 3))
coords_ext[...,[1,0]] = np.rollaxis(np.indices((Y, X)), 0, start=3)
coords_ext[...,2] = 1
result = np.matmul(coords_ext,matrix.T)
return result
def broadcasting_app(matrix, X, Y): # Broadcasting based
ys = np.arange(Y)
xs = np.arange(X)
out = ys[:,None,None]*matrix[:,1] + xs[:,None]*matrix[:,0] + matrix[:,2]
return out
时间和验证 -
In [242]: # Inputs
...: matrix = np.random.rand(3,3)
...: X,Y = 512, 512
...:
In [243]: out0 = original_listcomp_app(matrix, X, Y)
...: out1 = indices_app(matrix, X, Y)
...: out2 = broadcasting_app(matrix, X, Y)
...:
In [244]: np.allclose(out0, out1)
Out[244]: True
In [245]: np.allclose(out0, out2)
Out[245]: True
In [253]: %timeit original_listcomp_app(matrix, X, Y)
1 loops, best of 3: 1.32 s per loop
In [254]: %timeit indices_app(matrix, X, Y)
100 loops, best of 3: 16.1 ms per loop
In [255]: %timeit broadcasting_app(matrix, X, Y)
100 loops, best of 3: 9.64 ms per loop