问使用Matplotlib绘制二进制Numpy数组的边界
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Stack Overflow用户

提问于 2020-02-06 12:28:39

回答 1查看 743关注 0票数 3

我正在对一些图像使用图像分割，有时能够绘制分段的边界会很好。

我有一个用Matplotlib绘制的2D NumPy数组，我得到的最接近的是等值线绘制。这在数组中创建了角，但在其他方面是完美的。

Matplotlib的轮廓函数可以只绘制垂直/水平线吗，或者有其他方法可以做到这一点吗？

示例如下：

import matplotlib.pyplot as plt
import numpy as np


array = np.zeros((20, 20))
array[4:7, 3:8] = 1
array[4:7, 12:15] = 1
array[7:15, 7:15] = 1
array[12:14, 13:14] = 0

plt.imshow(array, cmap='binary')
plt.contour(array, levels=[0.5], colors='g')
plt.show()

python-3.x

numpy

matplotlib

回答 1

Stack Overflow用户

回答已采纳

发布于 2020-02-06 15:55:39

前段时间我写了一些函数来实现这一点，但我很高兴能想出如何能更快地完成它。

import numpy as np
import matplotlib.pyplot as plt
from matplotlib.collections import LineCollection


def get_all_edges(bool_img):
    """
    Get a list of all edges (where the value changes from True to False) in the 2D boolean image.
    The returned array edges has he dimension (n, 2, 2).
    Edge i connects the pixels edges[i, 0, :] and edges[i, 1, :].
    Note that the indices of a pixel also denote the coordinates of its lower left corner.
    """
    edges = []
    ii, jj = np.nonzero(bool_img)
    for i, j in zip(ii, jj):
        # North
        if j == bool_img.shape[1]-1 or not bool_img[i, j+1]:
            edges.append(np.array([[i, j+1],
                                   [i+1, j+1]]))
        # East
        if i == bool_img.shape[0]-1 or not bool_img[i+1, j]:
            edges.append(np.array([[i+1, j],
                                   [i+1, j+1]]))
        # South
        if j == 0 or not bool_img[i, j-1]:
            edges.append(np.array([[i, j],
                                   [i+1, j]]))
        # West
        if i == 0 or not bool_img[i-1, j]:
            edges.append(np.array([[i, j],
                                   [i, j+1]]))

    if not edges:
        return np.zeros((0, 2, 2))
    else:
        return np.array(edges)


def close_loop_edges(edges):
    """
    Combine thee edges defined by 'get_all_edges' to closed loops around objects.
    If there are multiple disconnected objects a list of closed loops is returned.
    Note that it's expected that all the edges are part of exactly one loop (but not necessarily the same one).
    """

    loop_list = []
    while edges.size != 0:

        loop = [edges[0, 0], edges[0, 1]]  # Start with first edge
        edges = np.delete(edges, 0, axis=0)

        while edges.size != 0:
            # Get next edge (=edge with common node)
            ij = np.nonzero((edges == loop[-1]).all(axis=2))
            if ij[0].size > 0:
                i = ij[0][0]
                j = ij[1][0]
            else:
                loop.append(loop[0])
                # Uncomment to to make the start of the loop invisible when plotting
                # loop.append(loop[1])
                break

            loop.append(edges[i, (j + 1) % 2, :])
            edges = np.delete(edges, i, axis=0)

        loop_list.append(np.array(loop))

    return loop_list


def plot_outlines(bool_img, ax=None, **kwargs):
    if ax is None:
        ax = plt.gca()

    edges = get_all_edges(bool_img=bool_img)
    edges = edges - 0.5  # convert indices to coordinates; TODO adjust according to image extent
    outlines = close_loop_edges(edges=edges)
    cl = LineCollection(outlines, **kwargs)
    ax.add_collection(cl)


array = np.zeros((20, 20))
array[4:7, 3:8] = 1
array[4:7, 12:15] = 1
array[7:15, 7:15] = 1
array[12:14, 13:14] = 0

plt.figure()
plt.imshow(array, cmap='binary')
plot_outlines(array.T, lw=5, color='r')