如何仅对图像的一部分执行 2D 正弦拟合？

Question

我有一堆图像，看起来像下面的图像，边缘厚度各不相同。

我想通过 python scipy.optimize.curve_fit 函数将正弦函数拟合到这个图像。然而，如图所示，条纹图案仅限制在圆形区域内。那我该如何进行拟合呢？

Answer 1

我想如果我是你，我会想要一个函数来近似与它们成 90 度的条纹……这样我就可以抓住正弦函数的周期。这将是一个多步骤的解决方案。1) 抓住边缘最亮的区域2) 沿与条纹垂直的线获取图像的值3) 计算最佳拟合曲线。import cv2import numpy as npfrom matplotlib import pyplot as pltimport math#function to rotate an image around a pointdef rotateImage(image, angle):    image_center = tuple(np.array(image.shape[1::-1]) / 2)    rot_mat = cv2.getRotationMatrix2D(image_center, angle, 1.0)    result = cv2.warpAffine(image, rot_mat, image.shape[1::-1], flags=cv2.INTER_LINEAR)    return resultimg = cv2.imread('sin.png', cv2.IMREAD_GRAYSCALE)imgb = cv2.GaussianBlur(img, (7,7),0)plt.imshow(imgb)plt.show()#grap bright valuesv90 = np.percentile(imgb, 90)#get mask for bright valuesmsk = imgb >= v90msk = msk.astype(np.uint8)plt.imshow(msk)plt.show()#get contourscnts = cv2.findContours(msk, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)[-2]cnt = sorted(cnts, key=cv2.contourArea)c = np.vstack((cnt[-2], cnt[-1])) #combine 2 largest((cx, cy), radius) = cv2.minEnclosingCircle(c)cv2.circle(img, (int(cx), int(cy)), int(radius), 255, 2)plt.imshow(img)plt.show()#circle is just for show...use if you want.imc = img.copy()#take top 5 contours(4 would work too ymmv)angles = []for i in range(5):    c = cnt[-(i+1)]    ellipse = cv2.fitEllipse(c)    (x, y), (MA, ma), angle = cv2.fitEllipse(c)    cv2.ellipse(imc, ((x, y), (MA, ma), angle), 255, 2)    angles.append(angle)#average angle  of ellipse.mangle = np.mean(angles)#goal is create a line normal to the average angle of the ellipses.#the 0.6 factor here is to just grab the inner region...which will avoid the rapid fall-off of the envelop gaussian-like function.pt1 = (int(cx + .6*radius*math.cos(math.radians(mangle))), int(cy + .6*radius*math.sin(math.radians(mangle))))pt2 = (int(cx - .6*radius*math.cos(math.radians(mangle))), int(cy - .6*radius*math.sin(math.radians(mangle))))#show linecv2.line(imc, pt1, pt2, 255, 2)#put fat line on mask...will use this to sample from original image laterim4mask = np.zeros(imc.shape).astype(np.uint8)cv2.line(im4mask, pt1, pt2, 255, 9)plt.imshow(imc)plt.show()plt.imshow(im4mask)plt.show()#now do some rotating(to make the numpy slicing later easier)imnew = rotateImage(imc, mangle)plt.imshow(imnew)plt.show()im4maskrot = rotateImage(im4mask, mangle)im4maskrot[im4maskrot > 20] = 255plt.imshow(im4maskrot)plt.show()imgbrot = rotateImage(imgb, mangle)plt.imshow(imgbrot)plt.show()#gather values from originalys, xs = np.where(im4maskrot == 255)minx = np.min(xs)miny = np.min(ys)maxx = np.max(xs)maxy = np.max(ys)print 'x ', minx, maxxprint 'y ', miny, maxycrop = imgbrot[miny:maxy, minx:maxx]print crop.shapeplt.imshow(crop)plt.show()plt.plot(range(crop.shape[1]), np.mean(crop, axis=0))#now time to fit a curve.#first with a gaussianfrom scipy import optimizedef test_func(x, a, b, A, mu, sigma):    return A*np.exp(-(x-mu)**2/(2.*sigma**2)) + a * np.sin(b * x)params, params_covariance = optimize.curve_fit(test_func, np.arange(crop.shape[1]), np.mean(crop, axis=0), p0=[10, 1/15., 60, 2, 150], maxfev=200000000)print(params)plt.figure(figsize=(6, 4))plt.scatter(range(crop.shape[1]), np.mean(crop, axis=0), label='Data')plt.plot(np.arange(crop.shape[1]), test_func(np.arange(crop.shape[1]), params[0],     params[1], params[2], params[3], params[4]), label='Fitted function')plt.legend(loc='best')plt.show()#and without a gaussian...the result is close because of only grabbing a short region.def test_func(x, a, b, n):    return n + a * np.sin(b * x)params, params_covariance = optimize.curve_fit(test_func, np.arange(crop.shape[1]), np.mean(crop, axis=0), p0=[10, 1/15., 60], maxfev=200000000)print(params)plt.figure(figsize=(6, 4))plt.scatter(range(crop.shape[1]), np.mean(crop, axis=0), label='Data')plt.plot(np.arange(crop.shape[1]), test_func(np.arange(crop.shape[1]), params[0], params[1], params[2]), label='Fitted function')plt.legend(loc='best')plt.show()请注意，b参数与周期性有关，并且两个值彼此非常接近（0.0644 和 0.0637）。知道了这一点，我会选择更简单的曲线拟合，因为起始参数更少。