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，但边缘厚度不同。

我想通过python scipy.optimize.curve_fit函数将正弦函数拟合到此图像。但是，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，我希望有一个函数可以将条纹与它们成90度近似...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）边缘最亮的抓斗区域

2）沿垂直于条纹的线获取图像的值

3）计算最佳拟合曲线。

import cv2
import numpy as np
from matplotlib import pyplot as plt
import math

#function to rotate an image around a point
def 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 result


img = cv2.imread('sin.png', cv2.IMREAD_GRAYSCALE)


imgb = cv2.GaussianBlur(img, (7,7),0)
plt.imshow(imgb)
plt.show()

#grap bright values
v90 = np.percentile(imgb, 90)
#get mask for bright values
msk = imgb >= v90
msk = msk.astype(np.uint8)

plt.imshow(msk)
plt.show()

#get contours
cnts = 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 line
cv2.line(imc, pt1, pt2, 255, 2)

#put fat line on mask...will use this to sample from original image later
im4mask = 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] = 255
plt.imshow(im4maskrot)
plt.show()

imgbrot = rotateImage(imgb, mangle)
plt.imshow(imgbrot)
plt.show()

#gather values from original
ys, xs = np.where(im4maskrot == 255)
minx = np.min(xs)
miny = np.min(ys)
maxx = np.max(xs)
maxy = np.max(ys)
print 'x ', minx, maxx
print 'y ', miny, maxy
crop = imgbrot[miny:maxy, minx:maxx]
print crop.shape
plt.imshow(crop)
plt.show()

plt.plot(range(crop.shape[1]), np.mean(crop, axis=0))

#now time to fit a curve.
#first with a gaussian

from scipy import optimize

def 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 参数是与周期性相关的参数，两个值彼此之间非常接近（0.0644和0.0637）。知道了这一点，我选择了更简单的曲线拟合，因为起始参数更少。