有没有办法加快 numpy.where 的循环速度？

想象一下，您有一个分割图，其中每个对象都由唯一索引标识，例如看起来与此类似：

对于每个对象，我想保存它覆盖的像素，但我只能提出标准for循环至今。不幸的是，对于包含数千个单独对象的较大图像，这结果非常慢——至少对于我的真实数据来说是这样。我能以某种方式加快速度吗？

import numpy as np
import matplotlib as mpl
import matplotlib.pyplot as plt
from skimage.draw import random_shapes


# please ignore that this does not always produce 20 objects each with a
# unique color. it is simply a quick way to produce data that is similar to
# my problem that can also be visualized.
segmap, labels = random_shapes(
    (100, 100), 20, min_size=6, max_size=20, multichannel=False,
    intensity_range=(0, 20), num_trials=100,
)
segmap = np.ma.masked_where(segmap == 255, segmap)

object_idxs = np.unique(segmap)[:-1]
objects = np.empty(object_idxs.size, dtype=[('idx', 'i4'), ('pixels', 'O')])

# important bit here:

# this I can vectorize
objects['idx'] = object_idxs
# but this I cannot. and it takes forever.
for i in range(object_idxs.size):
    objects[i]['pixels'] = np.where(segmap == i)

# just plotting here
fig, ax = plt.subplots(constrained_layout=True)
image = ax.imshow(
    segmap, cmap='tab20', norm=mpl.colors.Normalize(vmin=0, vmax=20)
)
fig.colorbar(image)
fig.show()

Using np.where循环中的算法效率不高，因为时间复杂度 is O(s n m) where s = object_idxs.size and n, m = segmap.shape。该操作可以在O(n m).

使用 Numpy 的一种解决方案是首先选择所有对象像素位置，然后根据它们的关联对象对它们进行排序segmap，最后根据对象的数量来分割它们。这是代码：

background = np.max(segmap)
mask = segmap != background
objects = segmap[mask]
uniqueObjects, counts = np.unique(objects, return_counts=True)
ordering = np.argsort(objects)
i, j = np.where(mask)
indices = np.vstack([i[ordering], j[ordering]])
indicesPerObject = np.split(indices, counts.cumsum()[:-1], axis=1)

objects = np.empty(uniqueObjects.size, dtype=[('idx', 'i4'), ('pixels', 'O')])
objects['idx'] = uniqueObjects
for i in range(uniqueObjects.size):
    # Use `tuple(...)` to get the exact same type as the initial code here
    objects[i]['pixels'] = tuple(indicesPerObject[i])
# In case the conversion to tuple is not required, the loop can also be accelerated:
# objects['pixels'] = indicesPerObject