我正在使用 numba 在 python 中编写一个函数来标记 2D 或 3D 数组中的对象，这意味着输入数组中具有相同值的所有正交连接单元将在输出数组中获得从 1 到 N 的唯一标签，其中 N 是正交连接组的数量。它与 scipy.ndimage.label 等功能非常相似。以及 scikit-image 等库中的类似函数，但这些函数标记所有正交连接的非零单元格组，因此它将合并具有不同值的连接组，这是我不想要的。例如，给定以下输入:
[0 0 7 7 0 0
0 0 7 0 0 0
0 0 0 0 0 7
0 6 6 0 0 7
0 0 4 4 0 0]
scipy 函数将返回
[0 0 1 1 0 0
0 0 1 0 0 0
0 0 0 0 0 3
0 2 2 0 0 3
0 0 2 2 0 0]
请注意，6 和 4 已合并到标签 2 中。我希望将它们标记为单独的组，例如:
[0 0 1 1 0 0
0 0 1 0 0 0
0 0 0 0 0 4
0 2 2 0 0 4
0 0 3 3 0 0]
我asked this about a year ago并一直在接受的答案中使用该解决方案，但是我正在努力优化代码的运行时并重新审视这个问题。
对于我通常使用的数据大小，链接的解决方案需要大约 1 分 30 秒才能运行。我编写了以下递归算法，该算法作为常规 python 运行大约需要 30 秒，而 numba 的 JIT 在 1-2 秒内运行(旁注，我讨厌那个相邻的函数，任何让它不那么困惑同时仍然与 numba 兼容的提示将不胜感激):
@numba.njit
def adjacent(idx, shape):
coords = []
if len(shape) > 2:
if idx[0] < shape[0] - 1:
coords.append((idx[0] + 1, idx[1], idx[2]))
if idx[0] > 0:
coords.append((idx[0] - 1, idx[1], idx[2]))
if idx[1] < shape[1] - 1:
coords.append((idx[0], idx[1] + 1, idx[2]))
if idx[1] > 0:
coords.append((idx[0], idx[1] - 1, idx[2]))
if idx[2] < shape[2] - 1:
coords.append((idx[0], idx[1], idx[2] + 1))
if idx[2] > 0:
coords.append((idx[0], idx[1], idx[2] - 1))
else:
if idx[0] < shape[0] - 1:
coords.append((idx[0] + 1, idx[1]))
if idx[0] > 0:
coords.append((idx[0] - 1, idx[1]))
if idx[1] < shape[1] - 1:
coords.append((idx[0], idx[1] + 1))
if idx[1] > 0:
coords.append((idx[0], idx[1] - 1))
return coords
@numba.njit
def apply_label(labels, decoded_image, current_label, idx):
labels[idx] = current_label
for aidx in adjacent(idx, labels.shape):
if decoded_image[aidx] == decoded_image[idx] and labels[aidx] == 0:
apply_label(labels, decoded_image, current_label, aidx)
@numba.njit
def label_image(decoded_image):
labels = np.zeros_like(decoded_image, dtype=np.uint32)
current_label = 0
for idx in zip(*np.where(decoded_image >= 0)):
if labels[idx] == 0:
current_label += 1
apply_label(labels, decoded_image, current_label, idx)
return labels, current_label
这适用于某些数据，但在其他数据上崩溃，我发现问题是当有非常大的对象要标记时，就会达到递归限制。我尝试重写 label_image 以不使用递归，但现在使用 numba 需要大约 10 秒。与我开始的地方相比仍然有很大的改进，但似乎应该可以获得与递归版本相同的性能。这是我的迭代版本:
@numba.njit
def label_image(decoded_image):
labels = np.zeros_like(decoded_image, dtype=np.uint32)
current_label = 0
for idx in zip(*np.where(decoded_image >= 0)):
if labels[idx] == 0:
current_label += 1
idxs = [idx]
while idxs:
cidx = idxs.pop()
if labels[cidx] == 0:
labels[cidx] = current_label
for aidx in adjacent(cidx, labels.shape):
if labels[aidx] == 0 and decoded_image[aidx] == decoded_image[idx]:
idxs.append(aidx)
return labels, current_label
有什么办法可以改善这个问题吗？
最佳答案
Can this recursive function be turned into an iterative function with similar performance?
将其转换为迭代函数很简单，考虑到它只是一个简单的深度优先搜索(您也可以使用队列而不是堆栈来使用广度优先搜索，两者都可以)。只需使用堆栈来跟踪要访问的节点即可。这是适用于任意数量维度的通用解决方案:
def label_image(decoded_image):
shape = decoded_image.shape
labels = np.zeros_like(decoded_image, dtype=np.uint32)
current_label = 0
for idx in zip(*np.where(decoded_image > 0)):
if labels[idx] == 0:
current_label += 1
stack = [idx]
while stack:
top = stack.pop()
labels[top] = current_label
for i in range(0, len(shape)):
if top[i] > 0:
neighbor = list(top)
neighbor[i] -= 1
neighbor = tuple(neighbor)
if decoded_image[neighbor] == decoded_image[idx] and labels[neighbor] == 0:
stack.append(neighbor)
if top[i] < shape[i] - 1:
neighbor = list(top)
neighbor[i] += 1
neighbor = tuple(neighbor)
if decoded_image[neighbor] == decoded_image[idx] and labels[neighbor] == 0:
stack.append(neighbor)
return labels
尽管从元组的第 i 个分量中添加或减去一个很尴尬(我将在此处查看临时列表)，并且 numba 不接受它(类型错误)。一种简单的解决方案是显式编写 2d 和 3d 版本，这可能会极大地提高性能:
@numba.njit
def label_image_2d(decoded_image):
w, h = decoded_image.shape
labels = np.zeros_like(decoded_image, dtype=np.uint32)
current_label = 0
for idx in zip(*np.where(decoded_image > 0)):
if labels[idx] == 0:
current_label += 1
stack = [idx]
while stack:
x, y = stack.pop()
if decoded_image[x, y] != decoded_image[idx] or labels[x, y] != 0:
continue # already visited or not part of this group
labels[x, y] = current_label
if x > 0: stack.append((x-1, y))
if x+1 < w: stack.append((x+1, y))
if y > 0: stack.append((x, y-1))
if y+1 < h: stack.append((x, y+1))
return labels
@numba.njit
def label_image_3d(decoded_image):
w, h, l = decoded_image.shape
labels = np.zeros_like(decoded_image, dtype=np.uint32)
current_label = 0
for idx in zip(*np.where(decoded_image > 0)):
if labels[idx] == 0:
current_label += 1
stack = [idx]
while stack:
x, y, z = stack.pop()
if decoded_image[x, y, z] != decoded_image[idx] or labels[x, y, z] != 0:
continue # already visited or not part of this group
labels[x, y, z] = current_label
if x > 0: stack.append((x-1, y, z))
if x+1 < w: stack.append((x+1, y, z))
if y > 0: stack.append((x, y-1, z))
if y+1 < h: stack.append((x, y+1, z))
if z > 0: stack.append((x, y, z-1))
if z+1 < l: stack.append((x, y, z+1))
return labels
def label_image(decoded_image):
dim = len(decoded_image.shape)
if dim == 2:
return label_image_2d(decoded_image)
assert dim == 3
return label_image_3d(decoded_image)
另请注意，迭代解决方案不受堆栈限制的影响:np.full((100,100,100), 1) 在迭代解决方案中工作得很好，但在递归解决方案中失败(段错误)如果使用 numba)。
做一个非常基本的基准测试
for i in range(1, 10000):
label_image(np.full((20,20,20), i))
(多次迭代以尽量减少 JIT 的影响，也可以进行一些预热运行，然后开始测量时间或类似的操作)
迭代解决方案似乎快了几倍(在我的机器上大约是 5 倍，见下文)。您可能可以优化递归解决方案并使其达到可比较的速度，例如通过避免临时 coords 列表或将 np.where 更改为 > 0。
我不知道 numba 可以如何优化压缩的 np.where。为了进一步优化，您可以考虑(和基准测试)使用显式嵌套的 for x in range(0, w): for y in range(0, h): 循环。
为了与 Nick 提出的合并策略保持竞争力，我进一步优化了这一策略，挑选了一些容易实现的目标:
- 使用
continue而不是np.where将zip转换为显式循环。 - 将
decoded_image[idx]存储在局部变量中(理想情况下应该不重要，但不会造成伤害)。 - 重用堆栈。这可以防止不必要的(重新)分配和 GC 压力。还可以考虑为堆栈提供初始容量(分别为
w*h或w*h*l)。
@numba.njit
def label_image_2d(decoded_image):
w, h = decoded_image.shape
labels = np.zeros_like(decoded_image, dtype=np.uint32)
current_label = 0
stack = []
for sx in range(0, w):
for sy in range(0, h):
start = (sx, sy)
image_label = decoded_image[start]
if image_label <= 0 or labels[start] != 0:
continue
current_label += 1
stack.append(start)
while stack:
x, y = stack.pop()
if decoded_image[x, y] != image_label or labels[x, y] != 0:
continue # already visited or not part of this group
labels[x, y] = current_label
if x > 0: stack.append((x-1, y))
if x+1 < w: stack.append((x+1, y))
if y > 0: stack.append((x, y-1))
if y+1 < h: stack.append((x, y+1))
return labels
@numba.njit
def label_image_3d(decoded_image):
w, h, l = decoded_image.shape
labels = np.zeros_like(decoded_image, dtype=np.uint32)
current_label = 0
stack = []
for sx in range(0, w):
for sy in range(0, h):
for sz in range(0, l):
start = (sx, sy, sz)
image_label = decoded_image[start]
if image_label <= 0 or labels[start] != 0:
continue
current_label += 1
stack.append(start)
while stack:
x, y, z = stack.pop()
if decoded_image[x, y, z] != image_label or labels[x, y, z] != 0:
continue # already visited or not part of this group
labels[x, y, z] = current_label
if x > 0: stack.append((x-1, y, z))
if x+1 < w: stack.append((x+1, y, z))
if y > 0: stack.append((x, y-1, z))
if y+1 < h: stack.append((x, y+1, z))
if z > 0: stack.append((x, y, z-1))
if z+1 < l: stack.append((x, y, z+1))
return labels
然后，我拼凑了一个基准来比较四种方法(原始递归、旧迭代、新迭代、基于合并)，并将它们放入四个不同的模块中:
import numpy as np
import timeit
import rec
import iter_old
import iter_new
import merge
shape = (100, 100, 100)
n = 20
for module in [rec, iter_old, iter_new, merge]:
print(module)
label_image = module.label_image
# Trigger compilation of 2d & 3d functions
label_image(np.zeros((1, 1)))
label_image(np.zeros((1, 1, 1)))
i = 0
def test_full():
global i
i += 1
label_image(np.full(shape, i))
print("single group:", timeit.timeit(test_full, number=n))
print("random (few groups):", timeit.timeit(
lambda: label_image(np.random.randint(low = 1, high = 10, size = shape)),
number=n))
print("random (many groups):", timeit.timeit(
lambda: label_image(np.random.randint(low = 1, high = 400, size = shape)),
number=n))
print("only groups:", timeit.timeit(
lambda: label_image(np.arange(np.prod(shape)).reshape(shape)),
number=n))
这会输出类似的内容
<module 'rec' from '...'>
single group: 32.39212468900041
random (few groups): 14.648884047001047
random (many groups): 13.304533919001187
only groups: 13.513677138000276
<module 'iter_old' from '...'>
single group: 10.287227957000141
random (few groups): 17.37535468200076
random (many groups): 14.506630064999626
only groups: 13.132202609998785
<module 'iter_new' from '...'>
single group: 7.388022166000155
random (few groups): 11.585243002000425
random (many groups): 9.560101995000878
only groups: 8.693653742000606
<module 'merge' from '...'>
single group: 14.657021331999204
random (few groups): 14.146574055999736
random (many groups): 13.412314713001251
only groups: 12.642367746000673
在我看来，改进的迭代方法可能会更好。请注意，最初的基本基准似乎是递归变体的最坏情况。一般来说差异不会那么大。
测试的阵列非常小(20立方)。如果我使用较大的数组 (100³) 和较小的 n (20) 进行测试，我大致会得到以下结果(省略了 rec，因为由于堆栈限制，它会出现段错误):
<module 'iter_old' from '...'>
single group: 3.5357716739999887
random (few groups): 4.931695729999774
random (many groups): 3.4671142009992764
only groups: 3.3023930709987326
<module 'iter_new' from '...'>
single group: 2.45903080700009
random (few groups): 2.907660342001691
random (many groups): 2.309699692999857
only groups: 2.052835552000033
<module 'merge' from '...'>
single group: 3.7620838259990705
random (few groups): 3.3524249689999124
random (many groups): 3.126650959999097
only groups: 2.9456547739991947
迭代方法似乎仍然更有效。
关于python - 这个递归函数能否变成具有类似性能的迭代函数？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/76956654/