Question

我有以下代码，用于根据分割步骤中创建的标签图像中的蒙版，用中值（或任何其他函数）值替换图像的每个值。感觉好像for循环可以被矢量化。这样做的最佳方法是什么？

我研究了为每个标签构建一个单独的索引数组，但最终没有看到它会如何帮助。

import numpy as np
from skimage.segmentation import slic
from skimage import data, io

def create_segment_image(original_image, labels_image):
    segment_image = np.zeros(original_image.shape, original_image.dtype)

    for label in np.unique(labels_image):
        segment_image[labels_image==label] = np.median(original_image[labels_image==label], axis=0)

    return segment_image

if __name__ == '__main__':

    original_image = data.astronaut()
    labels_image = slic(original_image, n_segments=1000, max_iter=10, compactness=7, sigma=1, convert2lab=True, enforce_connectivity=True, min_size_factor=0.1, max_size_factor=3, slic_zero=False)
    segment_image = create_segment_image(original_image, labels_image)
    # io.imsave('images/segment_image.png', segment_image)

Answer 1

我不知道矢量化最里面的循环。对median的调用每次计算不同数量的元素，这使得很难将所有调用放入单个数组中。

另一方面，就如何按标签选择元素而言，有一些相当低的成果。您可以在原始函数中找到每个标签的索引两次，只计算一次运行时约25％的索引数组

def create_segment_image_2(original_image, labels_image):
    segment_image = np.zeros(original_image.shape, original_image.dtype)
    for label in np.unique(labels_image):
        inds = np.where(labels_image == label)
        segment_image[inds] = np.median(original_image[inds], axis=0)    
    return segment_image

通过按标签对数组索引进行排序，然后利用该排序选择图像元素到中位数，可以获得更大的改进。用一种方式替换多次搜索可以提高20倍的速度。

def create_segment_image_3(original_image, labels_image):
    segment_image = np.zeros(original_image.shape, original_image.dtype)
    # sort the indices by their labels
    labelinds = np.argsort(labels_image, None)
    labels = np.unique(labels_image)
    # use the searchsorted to find the indices for each label
    rights = np.searchsorted(labels_image.flatten(), labels, side='right', sorter=labelinds)
    left = 0
    for right in rights:
        # choose our block of the image array
        inds = labelinds[left:right]
        # convert back to a two dimensional index array
        inds = [inds // original_image.shape[1], inds % original_image.shape[1]]
        segment_image[inds] = np.median(original_image[inds], axis=0)
        # update our boundaries
        left = right
    return segment_image

ipython中的基准测试

In [54]: %timeit create_segment_image(original_image, labels_image)
2.15 s ± 29.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [55]: %timeit create_segment_image_2(original_image, labels_image)
1.48 s ± 4.68 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [56]: %timeit create_segment_image_3(original_image, labels_image)
121 ms ± 561 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

确认我们的新解决方案与旧解决方案的结果相同。

In [57]: np.all(create_segment_image_2(original_image, labels_image) == create_segment_image(original_image, labels_image))
Out[57]: True

In [58]: np.all(create_segment_image_3(original_image, labels_image) == create_segment_image(original_image, labels_image))
Out[58]: True

在多个标签面具上的Numpy，矢量化功能

1 个答案: