您好我现在想要绘制掩模,然后对图像进行FFT运算后得到幅度谱。然而,使用霍夫线等,我得到太多的线和面具失败。这里是链接到图像和口罩,我需要提前如何在FFT幅度谱上绘制掩模线
这里创建
https://www.dropbox.com/s/vgbmf3ju50uf821/fftimg4.jpg?dl=0 https://www.dropbox.com/s/ktl0yghan9t868o/image4.jpg?dl=0
由于是代码
import cv2
import numpy as np
from matplotlib import pyplot as plt
from scipy.misc import imshow, imsave, imread ,imsave
import mahotas
scaler = .5
img = imread("color2.jpeg")[:,:,:3]
mask = imread("image4.jpg")[:,:,:3]
img = cv2.resize(img, (0,0), fx=scaler, fy=scaler)
imggray = np.mean(img, -1)
imfft = np.fft.fft2(imggray)
mags = np.abs(np.fft.fftshift(imfft))
angles = np.angle(np.fft.fftshift(imfft))
visual = np.log(mags)
visual2 = (visual - visual.min())/(visual.max() - visual.min())*255
cv2.imshow('Visual 2', visual2.astype(np.uint8))
imsave('fftimg4.jpg',visual2)
height,width,depth = img.shape
masking = np.zeros((height,width))
vis = visual2.astype(np.uint8)
edges = cv2.Canny(vis,50,180)
cv2.imshow('Gradient', edges)
lines = cv2.HoughLines(edges,1,np.pi/180,200,200,10)
for rho,theta in lines[0]:
a = np.cos(theta)
b = np.sin(theta)
x0 = a*rho
y0 = b*rho
x1 = int(x0 + 1000*(-b))
y1 = int(y0 + 1000*(a))
x2 = int(x0 - 1000*(-b))
y2 = int(y0 - 1000*(a))
cv2.line(masking,(x1,y1),(x2,y2),(255,255,255),1)
cv2.imshow('HoughLines', masking)
mask = (np.mean(mask,-1) > 20)
visual[mask] = np.mean(visual)
newmagsshift = np.exp(visual)
newffts = newmagsshift * np.exp(1j*angles)
newfft = np.fft.ifftshift(newffts)
imrev = np.fft.ifft2(newfft)
newim2 = 255 - np.abs(imrev).astype(np.uint8)
imsave("fftimg2.jpg", newim2 )
cv2.imshow('Image without Lines', newim2.astype(np.uint8))
cv2.waitKey(0)
您好,欢迎来到Stack Overflow,如果您明确指出您的问题并提供显示您尝试解决方案的源代码,那么您将会收到更好的回应。 – reticentroot 2015-04-05 12:11:08
也许可以尝试减小Hough空间的粒度,这样可以减少输出。因此,不要说1度的粒度,在你的霍夫空间中对应于3或5度的“垃圾箱”更少。 – 2015-04-05 12:16:55
注意@ msanti这里是原始图像https://www.dropbox.com/s/kiox2jcnxtkpg15/color2.jpeg?dl=0 – 2015-04-05 13:03:16