使用 python 代码从图像中估计重叠纤维的纤维长度

Question

我需要帮助来根据图像估算准确的纤维长度。我在 python 中开发了代码，通过它可以在一定程度上估计长度。
我使用 Skan 开源库从 skeletonized 纤维图像中获取纤维段的直径和长度。我在追踪重叠点或交叉点处的光纤以进行长度估算方面面临挑战。目前估计的长度比实际图像小得多，因为它只估计从光纤端点到连接点的段的长度。如果有人可以帮助估计所有重叠纤维的长度，那将很有帮助。分享 code 和原图供参考。

import cv2
import numpy as np
from matplotlib import pyplot as plt
from skimage.morphology import skeletonize
from skimage import morphology

img00 = cv2.imread(r'original_img.jpg')
img_01 = cv2.cvtColor(img00, cv2.COLOR_BGR2GRAY)
img0 = cv2.cvtColor(img00, cv2.COLOR_BGR2GRAY)

i_size = min(np.size(img_01,1),600) # image size for imshow
# Creating kernel
kernel = np.ones((2, 2), np.uint8)
  
# Using cv2.dialate() method 
img01 = cv2.dilate(img0, kernel, iterations=2)
cv2.imwrite('Img1_Filtered.jpg',img01)

ret,thresh1 = cv2.threshold(img01,245,255,cv2.THRESH_BINARY)
thresh = (thresh1/255).astype(np.uint8)
cv2.imwrite('Img2_Binary.jpg',thresh1)

# skeleton based on Lee's method
skeleton1 = (skeletonize(thresh, method='lee')/255).astype(bool)
skeleton1 = morphology.remove_small_objects(skeleton1, 100, connectivity=2)

# fiber Detection through skeletonization and its characterization
from skan import draw, Skeleton, summarize
spacing_nm = 1   # pixel

fig, ax = plt.subplots()
draw.overlay_skeleton_2d(img_01, skeleton1, dilate=1, axes=ax);

from skan.csr import skeleton_to_csgraph
pixel_graph, coordinates0 = skeleton_to_csgraph(skeleton1, spacing=spacing_nm)

skel_analysis = Skeleton(skeleton1, spacing=spacing_nm,source_image=img00)
branch_data = summarize(skel_analysis)
branch_data.hist(column='branch-distance', bins=100);
draw.overlay_euclidean_skeleton_2d(img_01, branch_data,skeleton_color_source='branch-type');

from scipy import ndimage
dd = ndimage.distance_transform_edt(thresh)
radii = np.multiply(dd, skeleton1);
Fiber_D_mean = np.mean(2*radii[radii>0]);
criteria = 2 * Fiber_D_mean; # Remove branches smaller than this length for characterization

aa = branch_data[(branch_data['branch-distance']>criteria)];
CNT_L_count, CNT_L_mean, CNT_L_stdev = aa['branch-distance'].describe().loc[['count','mean','std']]
print("Fiber Length (px[enter image description here][1])  : Count, Average, Stdev:",int(CNT_L_count),round(CNT_L_mean,2),round(CNT_L_stdev,2))

Answer 1

从骨架开始，我将按以下步骤进行：

将骨架转换为路径图
为每对路径识别有效的路口
计算每条相邻路径之间的角度
合并几乎笔直穿过交叉路口的路径

这是一个可以在骨架中找到重叠纤维的草图。我留给您来优化它，使其对现实生活中的图像具有鲁棒性以及如何从结果中得出统计数据。

import cv2
import numpy as np
from skimage import morphology, graph
from skan import Skeleton

MAX_JUNCTION = 4    # maximal size of junctions
MAX_ANGLE = 80      # maximal angle in junction
DELTA = 3           # distance from endpoint to inner point to estimate direction at endpoint


def angle(v1, v2):
    rad = np.arctan2(v2[0], v2[1]) - np.arctan2(v1[0], v1[1])
    return np.abs((np.rad2deg(rad) % 360) - 180)


img = cv2.imread('img.jpg')
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)

# dilate and threshold
kernel = np.ones((2, 2), np.uint8)
dilated = cv2.dilate(gray, kernel, iterations=1)
ret, thresh = cv2.threshold(dilated, 245, 255, cv2.THRESH_BINARY)

# skeletonize
skeleton = morphology.skeletonize(thresh, method='lee')
skeleton = morphology.remove_small_objects(skeleton.astype(bool), 100, connectivity=2)

# split skeleton into paths, for each path longer than MAX_JUNCTION get list of point coordinates
g = Skeleton(skeleton)
lengths = np.array(g.path_lengths())
paths = [list(np.array(g.path_coordinates(i)).astype(int)) for i in range(g.n_paths) if lengths[i] > MAX_JUNCTION]

# get endpoints of path and vector to inner point to estimate direction at endpoint
endpoints = [[p[0], np.subtract(p[0], p[DELTA]), i] for i, p in enumerate(paths)] +\
            [[p[-1], np.subtract(p[-1], p[-1 - DELTA]), i] for i, p in enumerate(paths)]

# get each pair of distinct endpoints with the same junction and calculate deviation of angle
angles = []
costs = np.where(skeleton, 1, 255)  # cost array for route_through_array

for i1 in range(len(endpoints)):
    for i2 in range(i1 + 1, len(endpoints)):
        e1, d1, p1 = endpoints[i1]
        e2, d2, p2 = endpoints[i2]
        if p1 != p2:
            p, c = graph.route_through_array(costs, e1, e2)       # check connectivity of endpoints at junction
            if c <= MAX_JUNCTION:
                deg = angle(d1, d2)                               # get deviation of directions at junction
                if deg <= MAX_ANGLE:
                    angles.append((deg, i1, i2, p))

# merge paths, with least deviation of angle first
angles.sort(key=lambda a: a[0])

for deg, i1, i2, p in angles:
    e1, e2 = endpoints[i1], endpoints[i2]
    if e1 and e2:
        p1, p2 = e1[2], e2[2]
        paths[p1] = paths[p1] + paths[p2] + p   # merge path 2 into path 1, add junction from route_through_array
        for i, e in enumerate(endpoints):       # switch path 2 at other endpoint to new merged path 1
            if e and e[2] == p2:
                endpoints[i][2] = p1
        paths[p2], endpoints[i1], endpoints[i2] = [], [], []    # disable merged path and endpoints

# display results
for p in paths:
    if p:
        img1 = img.copy()
        for v in p:
            img1[v[0], v[1]] = [0, 0, 255]
        cv2.imshow(f'fiber', img1)
        cv2.waitKey(0)

cv2.destroyAllWindows()

使用 python 代码从图像中估计重叠纤维的纤维长度

Estimation fiber length of overlapping fibers from image using python code

python

opencv

image-processing

scikit-image