检测测试盒opencv上的线条 [英] Detecting lines on test cassettes opencv

问题描述

(C 处有一条实线，T 处有一条暗线)

(There is a solid line at C and a faint line at T)

我想检测 T 处的线.目前我正在使用 opencv 定位二维码并旋转图像，直到二维码直立.然后我通过使用二维码的坐标计算出C和T标记的大概位置.然后我的代码将沿 y 轴向下扫描并检测绿色和蓝色值是否存在差异.

I want to detect the line at T. Currently I am using opencv to locate the qr code and rotate the image until the qr code is upright. Then I calculate the approximate location of the C and T mark by using the coordinates of the qr code. Then my code will scan along the y axis down and detect there are difference in the Green and Blue values.

我的问题是，即使T线像图中那样微弱，也应该被认为是阳性的.我怎样才能进行更好的检测?

My problem is, even if the T line is as faint as shown, it should be regarded as positive. How could I make a better detection?

推荐答案

我只裁剪了白色条带，因为我假设您已经找到了它的方法.由于我们正在寻找红色，因此我更改为 LAB 色彩空间并查看a"颜色.频道.

I cropped out just the white strip since I assume you have a way of finding it already. Since we're looking for red, I changed to the LAB colorspace and looked on the "a" channel.

注意:为方便查看，条带的所有图像都已转置(np.transpose)，代码中不是这样的.

Note: all images of the strip have been transposed (np.transpose) for viewing convenience, it's not that way in the code.

A 频道

我做了一个线性重构来提高对比度

I did a linear reframe to improve the contrast

图像非常嘈杂.同样，我不确定这是来自相机还是 jpg 压缩.我平均每一行以消除一些废话.

The image is super noisy. Again, I'm not sure if this is from the camera or the jpg compression. I averaged each row to smooth out some of the nonsense.

我绘制了强度(x-vals 是行索引)

I graphed the intensities (x-vals were the row index)

使用均值滤波器平滑图形

Use a mean filter to smooth out the graph

我运行了一个登山者算法来寻找山峰和山谷

I ran a mountain climber algorithm to look for peaks and valleys

然后我过滤了爬升大于 10 的山峰(第二高的山峰有 25.5 的爬升，第三高是 4.4).

And then I filtered for peaks with a climb greater than 10 (the second highest peak has a climb of 25.5, the third highest is 4.4).

使用这些峰值，我们可以确定有两条线，它们(大约)在这里:

Using these peaks we can determine that there are two lines and they are (about) here:

import cv2
import numpy as np
import matplotlib.pyplot as plt

# returns direction of gradient
# 1 if positive, -1 if negative, 0 if flat
def getDirection(one, two):
    dx = two - one;
    if dx == 0:
        return 0;
    if dx > 0:
        return 1;
    return -1;

# detects and returns peaks and valleys
def mountainClimber(vals, minClimb):
    # init trackers
    last_valley = vals[0];
    last_peak = vals[0];
    last_val = vals[0];
    last_dir = getDirection(vals[0], vals[1]);

    # get climbing
    peak_valley = []; # index, height, climb (positive for peaks, negative for valleys)
    for a in range(1, len(vals)):
        # get current direction
        sign = getDirection(last_val, vals[a]);
        last_val = vals[a];

        # if not equal, check gradient
        if sign != 0:
            if sign != last_dir:
                # change in gradient, record peak or valley
                # peak
                if last_dir > 0:
                    last_peak = vals[a];
                    climb = last_peak - last_valley;
                    climb = round(climb, 2);
                    peak_valley.append([a, vals[a], climb]);
                else:
                    # valley
                    last_valley = vals[a];
                    climb = last_valley - last_peak;
                    climb = round(climb, 2);
                    peak_valley.append([a, vals[a], climb]);

                # change direction
                last_dir = sign;

    # filter out very small climbs
    filtered_pv = [];
    for dot in peak_valley:
        if abs(dot[2]) > minClimb:
            filtered_pv.append(dot);
    return filtered_pv;

# run an mean filter over the graph values
def meanFilter(vals, size):
    fil = [];
    filtered_vals = [];
    for val in vals:
        fil.append(val);

        # check if full
        if len(fil) >= size:
            # pop front
            fil = fil[1:];
            filtered_vals.append(sum(fil) / size);
    return filtered_vals;

# averages each row (also gets graph values while we're here)
def smushRows(img):
    vals = [];
    h,w = img.shape[:2];
    for y in range(h):
        ave = np.average(img[y, :]);
        img[y, :] = ave;
        vals.append(ave);
    return vals;

# linear reframe [min1, max1] -> [min2, max2]
def reframe(img, min1, max1, min2, max2):
    copy = img.astype(np.float32);
    copy -= min1;
    copy /= (max1 - min1);
    copy *= (max2 - min2);
    copy += min2;
    return copy.astype(np.uint8);

# load image
img = cv2.imread("strip.png");

# resize
scale = 2;
h,w = img.shape[:2];
h = int(h*scale);
w = int(w*scale);
img = cv2.resize(img, (w,h));

# lab colorspace
lab = cv2.cvtColor(img, cv2.COLOR_BGR2LAB);
l,a,b = cv2.split(lab);

# stretch contrast
low = np.min(a);
high = np.max(a);
a = reframe(a, low, high, 0, 255);

# smush and get graph values
vals = smushRows(a);

# filter and round values
mean_filter_size = 20;
filtered_vals = meanFilter(vals, mean_filter_size);
for ind in range(len(filtered_vals)):
    filtered_vals[ind] = round(filtered_vals[ind], 2);

# get peaks and valleys
pv = mountainClimber(filtered_vals, 1);

# pull x and y values
pv_x = [ind[0] for ind in pv];
pv_y = [ind[1] for ind in pv];

# find big peaks
big_peaks = [];
for dot in pv:
    if dot[2] > 10: # climb filter size
        big_peaks.append(dot);
print(big_peaks);

# make plot points for the two best
tops_x = [dot[0] for dot in big_peaks];
tops_y = [dot[1] for dot in big_peaks];

# plot
x = [index for index in range(len(filtered_vals))];
fig, ax = plt.subplots()
ax.plot(x, filtered_vals);
ax.plot(pv_x, pv_y, 'og');
ax.plot(tops_x, tops_y, 'vr');
plt.show();

# draw on original image
h,w = img.shape[:2];
for dot in big_peaks:
    y = int(dot[0] + mean_filter_size / 2.0); # adjust for mean filter cutting
    cv2.line(img, (0, y), (w,y), (100,200,0), 2);

# show
cv2.imshow("a", a);
cv2.imshow("strip", img);
cv2.waitKey(0);

我想知道为什么这些线条看起来如此偏离，然后我意识到我忘记考虑 meanFilter 减小列表大小(它从正面和背面切割)这一事实.我已经更新以考虑到这一点.

I was wondering why the lines seemed so off, then I realized that I forgot to account for the fact that the meanFilter reduces the size of the list (it cuts from the front and back). I've updated to take that into account.

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