基于相似度和区分度的边缘检测

78次阅读
没有评论

共计 2606 个字符,预计需要花费 7 分钟才能阅读完成。

提醒:本文最后更新于 2024-08-30 15:12,文中所关联的信息可能已发生改变,请知悉!

区分度分布

基于相似度和区分度的边缘检测

5 : 0.4680251736111111
10 : 0.7441905381944445
15 : 0.8501150173611111
20 : 0.8981987847222223
25 : 0.9249197048611111
30 : 0.9417599826388889
35 : 0.9525086805555556
40 : 0.9603667534722222
45 : 0.9662131076388889
50 : 0.9709288194444444
55 : 0.9748697916666667
60 : 0.9779774305555555
65 : 0.9807183159722223
70 : 0.9829730902777778
75 : 0.9847005208333334
80 : 0.9863433159722222
85 : 0.9875368923611111
90 : 0.9887651909722223
95 : 0.9897873263888889

区分度和相似度结合

相似度 > 50

基于相似度和区分度的边缘检测

相似度 > 50, 区分度 > 30

基于相似度和区分度的边缘检测

相似度 > 50, 区分度 > 20

基于相似度和区分度的边缘检测

代码示例

# utf-8
# python3.8

import cv2 as cv
import numpy as np

def get_unit_similarity(a_4, b_4, k):
    # a_4 = sorted(a_4)
    # b_4 = sorted(b_4)
    result = 0
    for i in range(4):
        result = result + (abs(int(a_4[i]) - int(b_4[i]))) ** k
    result = result ** (1/k)
    return int(result)

def input_flag(flag, x, y, similarity):
    for i in range(4):
        if flag[x][y][i] == 0:
            flag[x][y][i] = similarity
            break
    return flag

def get_channels_similarity(channel, k, rank):
    row, col = channel.shape
    flag = np.zeros((row - 1, col - 1, 4))
    result = np.zeros((row - 1, col - 1))
    for i in range(row - 2):
        for j in range(col - 2):
            array = [channel[i][j], channel[i][j + 1], channel[i + 1][j + 1], channel[i + 1][j]]
            r_array = [channel[i][j + 1], channel[i + 1][j + 1], channel[i][j + 2], channel[i + 1][j + 2]]
            d_array = [channel[i + 1][j + 1], channel[i + 1][j], channel[i + 2][j + 1], channel[i + 2][j]]
            r_similarity = get_unit_similarity(array, r_array, k)
            d_similarity = get_unit_similarity(array, d_array, k)
            flag = input_flag(flag, i, j, r_similarity)
            flag = input_flag(flag, i, j, d_similarity)
            flag = input_flag(flag, i+1, j, r_similarity)
            flag = input_flag(flag, i, j+1, d_similarity)
            # result[i][j] = (r_similarity + d_similarity) / 2
    for i in range(row-2):
        for j in range(col-2):
            flag4 = sorted(flag[i][j])
            result[i][j] = flag4[4-rank]
    return result

def get_qufendu(array):
    return int(max(array)) - int(min(array))

def get_channels_qufendu(channel):
    row, col = channel.shape
    result = np.zeros((row - 1, col - 1))
    for i in range(row-1):
        for j in range(col-1):
            array = [channel[i][j], channel[i][j + 1], channel[i + 1][j + 1], channel[i + 1][j]]
            qufendu = get_qufendu(array)
            result[i][j] = qufendu
    return result

def mark(channel, similarity, qufendu):
    row, col = similarity.shape
    for i in range(row):
        for j in range(col):
            if similarity[i][j] > 50 and qufendu[i][j] > 20:
                channel[i][j] = 0
            else:
                channel[i][j] = 255
    return channel

def mark_similarity(img, k, rank):
    p = cv.imread(img, 1)
    b, g, r = cv.split(p)
    b_similarity = get_channels_similarity(b, k, rank)
    b_qufendu = get_channels_qufendu(b)
    b = mark(b, b_similarity, b_qufendu)
    g_similarity = get_channels_similarity(g, k, rank)
    g_qufendu = get_channels_qufendu(g)
    g = mark(g, g_similarity, g_qufendu)
    r_similarity = get_channels_similarity(r, k, rank)
    r_qufendu = get_channels_qufendu(r)
    r = mark(r, r_similarity, r_qufendu)
    p = cv.merge((b, g, r))
    cv.imwrite(img[0:-4] + '_mark.png', p)

if __name__ == '__main__':
    img_addr = '8068.jpg'
    mark_similarity(img_addr, 1, 1)
正文完
 0
icvuln
版权声明:本站原创文章,由 icvuln 于2022-05-09发表,共计2606字。
转载说明:除特殊说明外本站文章皆由CC-4.0协议发布,转载请注明出处。
评论(没有评论)