在大型3D Numpy数组中查找局部最大值

问题描述

我正在处理三维numpy数组中存在的一些大体积图像数据.我将用两个小的一维数组来说明我的任务.我有一张图片:

I'm processing some large volumetric image data that are present in three dimensional numpy arrays. I'll explain my task with two small 1D arrays. I have one image:

img = [5, 6, 70, 80, 3, 4, 80, 90]

以及该图像的一个分段和标记版本:

and one segmented and labeled version of that image:

labels = [0, 0, 1, 1, 0, 0, 2, 2]

labels中的每个数字代表img中的一个对象.两个阵列具有相同的尺寸.因此，在此示例中，img中有两个对象:

Each number in labels represents an object in img. Both arrays have the same dimensions. So in this example there's two objects in img:

[5，6， 70 ， 80 ，3，4， 80 ， 90 ]

[5, 6, 70, 80, 3, 4, 80, 90]

，现在我要尝试的是找到每个对象的最大值的位置，在这种情况下为3和7.目前，我遍历所有标签，创建一个img版本，该版本仅包含与当前标签对应的对象，然后寻找最大值:

and what I'm trying to do now is finding the location of the maximum value of each object, which in this case would be the 3 and 7. Currently I loop over all labels, create a version of img which contains only the object corresponding to the current label, and look for the maximum value:

for label in range(1, num_labels + 1):
    imgcp = np.copy(img)
    imgcp[labels != label] = 0
    max_pos = np.argmax(imgcp)
    max_coords = np.unravel_index(pos, imgcp.shape)

此方法的一个问题是，在每个步骤中复制img都容易产生内存错误.我觉得内存管理应该阻止这种情况，但是有没有更高的内存效率并且可能更快的方式来完成此任务?

One problem with this approach is that copying img in every step tends to create memory errors. I feel like memory management should prevent this, but is there a more memory efficient and possibly faster way to to this task?

推荐答案

这是使用argpartition的方法.

# small 2d example
>>> data = np.array([[0,1,4,0,0,2,1,0],[0,4,1,3,0,0,0,0]])
>>> segments = np.array([[0,1,1,0,0,2,2,0],[0,1,1,1,0,0,0,0]])
>>>
# discard zeros
>>> nz = np.where(segments)
>>> segc = segments[nz]
>>> dac = data[nz]

# count object sizes
>>> cnts = np.bincount(segc)
>>> bnds = np.cumsum(cnts)
# use counts to partition into objects
>>> idx = segc.argpartition(bnds[1:-1])
>>> dai = dac[idx]
# find maxima per object
>>> mx = np.maximum.reduceat(dai, bnds[:-1])
# find their positions
>>> am, = np.where(dai==mx.repeat(cnts[1:]))
# translate positions back to coordinate space
>>> im = idx[am]
>>> am = *(n[im] for n in nz),
>>>
>>>
# result
# coordinates, note that there are more points than objects because
# the maximum 4 occurs twice in object 1
>>> am
(array([1, 0, 0]), array([1, 2, 5]))
# maxima
>>> data[am]
array([4, 4, 2])
# labels
>>> segments[am]
array([1, 1, 2])

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