在Python中计算位图中两点之间的最短路径 [英] Calculating the shortest path between two points in a bitmap in python

问题描述

我有一个黑白位图图像，显示在这里:

I have a black and white bitmap image shown here:

图片大小为200,158.

The image size is 200,158.

我想选择落在白色路径上的两个点，并计算仅跟随白色像素的两点之间的最短距离.我不确定如何解决这个问题.我想创建一个使用两组x，y坐标调用的函数，它只返回沿着白色像素的最短路径的像素数.

I want to pick two points that fall on the white path and calculate the shortest distance between those two points following only the white pixels. I am not sure how to approach this problem. I want to create a function that I call with 2 sets of x,y coordinates and it returns the number of pixels following the shortest route along the white pixels only.

任何指针将不胜感激.

推荐答案

如评论中所述，此问题可以减少为 Dijkstra .

As stated in the comments, this problem can be reduced to Dijkstra.

解决方案背后的关键概念是将图像表示为图形，然后使用最短路径算法的预制实现.

The key concept behind the solution is to represent the image as a graph and then use a pre-made implementation of the shortest-path algorithm.

首先，观察4x4大小的图像的幼稚表现形式:

Firstly, observe a naive representation of an image of size 4x4:

T F F T
T T F T
F T T F
T T T T

其中T是一个白色的点，而F是一个黑色的点.在这种情况下，路径是相邻白点之间的一组移动.

Where T is a white dot and F is a black one. In this case, a path is a set of moves between adjacent white points.

假设一个图是一组节点{1, 2, ..., 16}，我们可以将每个点(i, j)映射到数字i * 4 + j.在图中，边缘是相邻点的反射，这意味着如果(i1, j1)和(i2, j2)在图像中相邻，则i1 * 4 + j1和i2 * 4 + j2在图中是相邻的.

Assuming a graph would be a set of nodes {1, 2, ..., 16}, we can map every point (i, j) to the number i * 4 + j. In the graph, the edges are a reflection of neighboring points, meaning that if (i1, j1) and (i2, j2) are adjacent in the image, then i1 * 4 + j1 and i2 * 4 + j2 are adjacent in the graph.

这时，我们有了一个可以计算最短路径的图.

At this point, we have a graph on which we can compute the shortest path.

幸运的是，python为图像加载和最短路径实现提供了简单的实现.以下代码处理路径计算以可视化结果:

Luckily, python provides easy implementation to the image loading and to the shortest path implementation. The following code handles the path computation the visualizes the result:

import itertools

from scipy import misc
from scipy.sparse.dok import dok_matrix
from scipy.sparse.csgraph import dijkstra

# Load the image from disk as a numpy ndarray
original_img = misc.imread('path_t_image')

# Create a flat color image for graph building:
img = original_img[:, :, 0] + original_img[:, :, 1] + original_img[:, :, 2]


# Defines a translation from 2 coordinates to a single number
def to_index(y, x):
    return y * img.shape[1] + x


# Defines a reversed translation from index to 2 coordinates
def to_coordinates(index):
    return index / img.shape[1], index % img.shape[1]


# A sparse adjacency matrix.
# Two pixels are adjacent in the graph if both are painted.
adjacency = dok_matrix((img.shape[0] * img.shape[1],
                        img.shape[0] * img.shape[1]), dtype=bool)

# The following lines fills the adjacency matrix by
directions = list(itertools.product([0, 1, -1], [0, 1, -1]))
for i in range(1, img.shape[0] - 1):
    for j in range(1, img.shape[1] - 1):
        if not img[i, j]:
            continue

        for y_diff, x_diff in directions:
            if img[i + y_diff, j + x_diff]:
                adjacency[to_index(i, j),
                          to_index(i + y_diff, j + x_diff)] = True

# We chose two arbitrary points, which we know are connected
source = to_index(14, 47)
target = to_index(151, 122)

# Compute the shortest path between the source and all other points in the image
_, predecessors = dijkstra(adjacency, directed=False, indices=[source],
                           unweighted=True, return_predecessors=True)

# Constructs the path between source and target
pixel_index = target
pixels_path = []
while pixel_index != source:
    pixels_path.append(pixel_index)
    pixel_index = predecessors[0, pixel_index]


# The following code is just for debugging and it visualizes the chosen path
import matplotlib.pyplot as plt

for pixel_index in pixels_path:
    i, j = to_coordinates(pixel_index)
    original_img[i, j, 0] = original_img[i, j, 1] = 0

plt.imshow(original_img)
plt.show()

免责声明:

• 我在图像处理方面没有经验，所以我怀疑解决方案中的每个步骤.
• 该解决方案假定一个非常幼稚的邻接谓词.对于这部分，在计算几何中可能有一些更好的方法.

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