[英]Calculate gradient only in a masked area
我有一个非常大的阵列,只有几个感兴趣的小区域。 我需要计算这个数组的梯度,但出于性能原因,我需要将这个计算限制在这些感兴趣的领域。
我做不到这样的事情:
phi_grad0[mask] = np.gradient(phi[mask], axis=0)
由于花式索引的工作原理, phi[mask]
只是成为被遮罩像素的一维数组,丢失了空间信息,使得渐变计算变得毫无价值。
np.gradient
会处理np.ma.masked_array
,但性能会np.ma.masked_array
一个数量级:
import numpy as np
from timeit_context import timeit_context
phi = np.random.randint(low=-100, high=100, size=[100, 100])
phi_mask = np.random.randint(low=0, high=2, size=phi.shape, dtype=np.bool)
with timeit_context('full array'):
for i2 in range(1000):
phi_masked_grad1 = np.gradient(phi)
with timeit_context('masked_array'):
phi_masked = np.ma.masked_array(phi, ~phi_mask)
for i1 in range(1000):
phi_masked_grad2 = np.gradient(phi_masked)
这会产生以下输出:
[full array] finished in 143 ms
[masked_array] finished in 1961 ms
我认为它是因为在masked_array
运行的操作没有矢量化,但我不确定。
有没有办法限制np.gradient
以达到更好的性能?
这个timeit_context
是一个方便的计时器,如果有人感兴趣的话:
from contextlib import contextmanager
import time
@contextmanager
def timeit_context(name):
"""
Use it to time a specific code snippet
Usage: 'with timeit_context('Testcase1'):'
:param name: Name of the context
"""
start_time = time.time()
yield
elapsed_time = time.time() - start_time
print('[{}] finished in {} ms'.format(name, int(elapsed_time * 1000)))
不完全是答案,但这是我为我的情况设法修补的,这非常有效:
我得到条件为真的像素的1D索引(在这种情况下,条件< 5
):
def get_indices_1d(image, band_thickness):
return np.where(image.reshape(-1) < 5)[0]
这给了我一个带有这些索引的数组。
然后我以不同的方式手动计算这些位置的梯度:
def gradient_at_points1(image, indices_1d):
width = image.shape[1]
size = image.size
# Using this instead of ravel() is more likely to produce a view instead of a copy
raveled_image = image.reshape(-1)
res_x = 0.5 * (raveled_image[(indices_1d + 1) % size] - raveled_image[(indices_1d - 1) % size])
res_y = 0.5 * (raveled_image[(indices_1d + width) % size] - raveled_image[(indices_1d - width) % size])
return [res_y, res_x]
def gradient_at_points2(image, indices_1d):
indices_2d = np.unravel_index(indices_1d, dims=image.shape)
# Even without doing the actual deltas this is already slower, and we'll have to check boundary conditions, etc
res_x = 0.5 * (image[indices_2d] - image[indices_2d])
res_y = 0.5 * (image[indices_2d] - image[indices_2d])
return [res_y, res_x]
def gradient_at_points3(image, indices_1d):
width = image.shape[1]
raveled_image = image.reshape(-1)
res_x = 0.5 * (raveled_image.take(indices_1d + 1, mode='wrap') - raveled_image.take(indices_1d - 1, mode='wrap'))
res_y = 0.5 * (raveled_image.take(indices_1d + width, mode='wrap') - raveled_image.take(indices_1d - width, mode='wrap'))
return [res_y, res_x]
def gradient_at_points4(image, indices_1d):
width = image.shape[1]
raveled_image = image.ravel()
res_x = 0.5 * (raveled_image.take(indices_1d + 1, mode='wrap') - raveled_image.take(indices_1d - 1, mode='wrap'))
res_y = 0.5 * (raveled_image.take(indices_1d + width, mode='wrap') - raveled_image.take(indices_1d - width, mode='wrap'))
return [res_y, res_x]
我的测试数组看起来像这样:
a = np.random.randint(-10, 10, size=[512, 512])
# Force edges to not pass the condition
a[:, 0] = 99
a[:, -1] = 99
a[0, :] = 99
a[-1, :] = 99
indices = get_indices_1d(a, 5)
mask = a < 5
然后我可以运行这些测试:
with timeit_context('full gradient'):
for i in range(100):
grad1 = np.gradient(a)
with timeit_context('With masked_array'):
for im in range(100):
ma = np.ma.masked_array(a, mask)
grad6 = np.gradient(ma)
with timeit_context('gradient at points 1'):
for i1 in range(100):
grad2 = gradient_at_points1(image=a, indices_1d=indices)
with timeit_context('gradient at points 2'):
for i2 in range(100):
grad3 = gradient_at_points2(image=a, indices_1d=indices)
with timeit_context('gradient at points 3'):
for i3 in range(100):
grad4 = gradient_at_points3(image=a, indices_1d=indices)
with timeit_context('gradient at points 4'):
for i4 in range(100):
grad5 = gradient_at_points4(image=a, indices_1d=indices)
这给出了以下结果:
[full gradient] finished in 576 ms
[With masked_array] finished in 3455 ms
[gradient at points 1] finished in 421 ms
[gradient at points 2] finished in 451 ms
[gradient at points 3] finished in 112 ms
[gradient at points 4] finished in 102 ms
正如您所看到的,方法4是迄今为止最好的(但不关心它消耗多少内存)。
这可能只是因为我的2D数组相对较小(512x512)。 也许对于更大的阵列,这不是真的。
另一个需要注意的是, ndarray.take(indices, mode='wrap')
会在图像边缘周围做一些奇怪的事情(一行将'循环'到下一个等等)以保持良好的性能,所以如果边缘对于您的应用程序可能需要在边缘周围填充1个像素的输入数组。
仍然非常有趣的是masked_array
的速度有多慢。 在循环外部拉构造函数ma = np.ma.masked_array(a, mask)
不会影响时间,因为masked_array
本身只保留对数组及其掩码的引用
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