我有一个由零组成的二维数组（称为标签）及其坐标列表（称为中心）。对于每个中心，我想在标签数组内围绕它的 5x5 簇中放置一个渐进数字。
如果两个中心非常接近，以至于相应的簇重叠，那么我希望标签数组优先考虑较低的标签值。换句话说，如果簇 2 与簇 1 重叠，那么我希望簇 1 为 5x5，簇 2 更小。
我已成功将此过程编码如下：
import numpy as np
labels = np.zeros((10,20))
centers = [(4,4), (7,7), (5,10), (5,17)]
for i, center in enumerate(centers, start=1):
# find the coordinates for the 5x5 cluster centered in center.
cluster = np.ix_(np.arange(center[0]-2, center[0]+3,1),
np.arange(center[1]-2, center[1]+3,1))
# create a temporary label array with all zeros
temp_label = np.zeros_like(labels)
# set the label value in the temporary array in the position corresponding to the cluster
temp_label[cluster] = i
# apply some boolean algebra
# (labels == 0) is a bool array corresponding to all positions that are not yet belonging to a label
# (labels == 0) * temp_label is like the temp_label, but only where the labels array is still free
# adding the labels back is ensuring that all previous labels are also counted.
labels = (labels == 0) * temp_label + labels
print(labels)
输出：
[[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 1. 1. 1. 1. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 1. 1. 1. 1. 1. 0. 3. 3. 3. 3. 3. 0. 0. 4. 4. 4. 4. 4.]
[0. 0. 1. 1. 1. 1. 1. 0. 3. 3. 3. 3. 3. 0. 0. 4. 4. 4. 4. 4.]
[0. 0. 1. 1. 1. 1. 1. 2. 2. 2. 3. 3. 3. 0. 0. 4. 4. 4. 4. 4.]
[0. 0. 1. 1. 1. 1. 1. 2. 2. 2. 3. 3. 3. 0. 0. 4. 4. 4. 4. 4.]
[0. 0. 0. 0. 0. 2. 2. 2. 2. 2. 3. 3. 3. 0. 0. 4. 4. 4. 4. 4.]
[0. 0. 0. 0. 0. 2. 2. 2. 2. 2. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 2. 2. 2. 2. 2. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]]
这个片段实际上正在做我想做的事情。簇 1 和 4 是完整的，而相互接触的簇 2 和 3 仅是部分的。为了实现这一点，我每次都必须传递一个临时数组。
更好的解决方案是什么？
您可以以相反的顺序编写补丁，以避免复杂（更昂贵）的基于掩码的计算。
使用 Numpy 编写非常小的补丁总是很昂贵，尽管肯定有更快的纯 Numpy 解决方案（也可能是复杂的解决方案）。这是因为 Numpy 针对计算大型数组进行了优化，并且由于昂贵的相对较大的临时数组（使代码更加受内存限制），任何向量化操作显然都不是最佳的。这个问题的关键就是使用基于编译的模块，如 Numba 或 Cython。
这是 Numba 中混合两种解决方案的解决方案：
import numpy as np
import numba as nb
def compute_naive(centers):
labels = np.zeros((10,20))
for i, center in enumerate(centers, start=1):
cluster = np.ix_(np.arange(center[0]-2, center[0]+3,1),
np.arange(center[1]-2, center[1]+3,1))
temp_label = np.zeros_like(labels)
temp_label[cluster] = i
labels = (labels == 0) * temp_label + labels
return labels
@nb.njit('(int32[:,::1],)')
def compute_numba(centers):
labels = np.zeros((10,20))
for i, center in enumerate(centers[::-1]):
cy, cx = center
yStart, yEnd = max(cy-2, 0), min(cy+3, labels.shape[0])
xStart, xEnd = max(cx-2, 0), min(cx+3, labels.shape[1])
for y in range(yStart, yEnd):
for x in range(xStart, xEnd):
labels[y, x] = centers.shape[0] - i
return labels
centers = [(4,4), (7,7), (5,10), (5,17)]
npCenters = np.array(centers, dtype=np.int32)
assert np.allclose(compute_naive(centers), compute_numba(npCenters))
%timeit -n 1000 labels = compute_numba(npCenters)
Numba 解决方案比这个非常小的示例中的初始代码快 >100 倍。它应该在更大规模的用例上更快。
np.ix_网格。
import numpy as np
from numpy.lib.stride_tricks import sliding_window_view as swv
labels = np.zeros((10,20))
centers = [(4,4), (7,7), (5,10), (5,17)]
H, W = 5, 5
解决方案：
view = swv(labels, (H, W), writeable=True)
for lbl, (r, c) in reversed(list(enumerate(centers, start=1))):
view[r - H//2, c - W//2] = lbl
虽然以下内容对我有用......但它可能是非法的，因为我们依赖于
view[...] = ...
的书写顺序，我不确定这是否一致。
rows, cols = np.transpose(centers[::-1])
values = np.arange(len(centers), 0, -1)
view[rows - H//2, cols - W//2] = values[:, None, None]
>>> print(labels)
[[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 1. 1. 1. 1. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 1. 1. 1. 1. 1. 0. 3. 3. 3. 3. 3. 0. 0. 4. 4. 4. 4. 4.]
[0. 0. 1. 1. 1. 1. 1. 0. 3. 3. 3. 3. 3. 0. 0. 4. 4. 4. 4. 4.]
[0. 0. 1. 1. 1. 1. 1. 2. 2. 2. 3. 3. 3. 0. 0. 4. 4. 4. 4. 4.]
[0. 0. 1. 1. 1. 1. 1. 2. 2. 2. 3. 3. 3. 0. 0. 4. 4. 4. 4. 4.]
[0. 0. 0. 0. 0. 2. 2. 2. 2. 2. 3. 3. 3. 0. 0. 4. 4. 4. 4. 4.]
[0. 0. 0. 0. 0. 2. 2. 2. 2. 2. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 2. 2. 2. 2. 2. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]]