Find coordinates of contours - python

I'm trying to find the coordinates of each corner of a given rectangle.
So far I've been able to obtain two corners using the contours function, but when I look through the whole contour array there is an abundance of points to sift through. I'm only looking for the most extreme values (max and min x- and y-values)
import cv2
import numpy as np
#open image
img = cv2.imread('cnt-coords.jpg')
#convert to black and white
bw = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
#threshold image
ret, thresh = cv2.threshold(bw,127,255,0)
#find contours
im2, contours, hierarchy = cv2.findContours(thresh,cv2.RETR_TREE,cv2.CHAIN_APPROX_SIMPLE)
#obtain the first 2 points of contours
cntx1 = contours[0][0]
cntx = contours[0][1]
#convert coords to points
pt1 = (cntx1[0][0],cntx1[0][1])
pt2 = (cntx[0][0],cntx[0][1])
#draw circles on coordinates
cv2.circle(img,pt1,5,(0,255,0),-1)
cv2.circle(img,pt2, 5, (0,255,0),-1)
#display the image
cv2.imshow('f',img)
cv2.waitKey(0)
cv2.destroyAllWindows()
I sifted through what contours returns to me and its a huge array of points, which seems to be all the points along the diagonal of my image. Is there any way to simplify which points I'm receiving, in this case the corners of this rectangle/parallelogram?

findContours function returns pixels on the contour. You can try approxPolyDP function of OpenCV to find a polygonal approximation to the contour. The usage and explanation can be found here. In your case it would be something like the following:
epsilon = cv2.arcLength(contours[0],True)
approx = cv2.approxPolyDP(contours[0],epsilon,True)

Related

how to fill the hollow lines opencv

I have an image like this:
after I applied some processings e.g. cv2.Canny(), it looks like this now:
As you can see that the black lines become hollow.
I have tried erosion and dilation, but if I do them many times, the 2 entrances will be closed(meaning become connected line or closed contour).
How could I make those lines solid like the below image while keep the 2 entrances not affected?
Update 1
I have tested the following answers with a few of photos, but the code seems customized to only be able to handle this one particular picture. Due to the restriction of SOF, I cannot upload photos larger than 2MB, so I uploaded them into my Microsoft OneDrive folder for your convenience to test.
https://1drv.ms/u/s!Asflam6BEzhjgbIhgkL4rt1NLSjsZg?e=OXXKBK
Update 2
I picked up #fmw42's post as answer as his answer is the most detailed one. It doesn't answer my question but points out the correct way to process maze which is my ultimate goal. I like his approach of answering questions, firstly tells you what each step should do so that you have a clear idea about how to do the task, then provide the full code example from beginning to end. Very helpful.
Due to the limitation of SOF, I can only pick up one answer. If multiple answers are allowed, I would also pick up Shamshirsaz.Navid's answer. His answer not only points to the correct direction to solve the issue, but also the explanation with visualization really works well for me~! I guess it works equally well for all people who are trying to understand why each line of code is needed. Also he follows up my questions in comments, this makes the SOF a bit interactive :)
The Threshold track bar in Ann Zen's answer is also a very useful tip for people to quickly find out a optimal value.
Here is one way to process the maze and rectify it in Python/OpenCV.
Read the input
Convert to gray
Threshold
Use morphology close to remove the thinnest (extraneous) black lines
Invert the threshold
Get the external contours
Keep on those contours that are larger than 1/4 of both the width and height of the input
Draw those contours as white lines on black background
Get the convex hull from the white contour lines image
Draw the convex hull as white lines on black background
Use GoodFeaturesToTrack to get the 4 corners from the white hull lines image
Sort the 4 corners by angle relative to the centroid so that they are ordered clockwise: top-left, top-right, bottom-right, bottom-left
Set these points as the array of conjugate control points for the input
Use 1/2 the dimensions of the input to define the array of conjugate control points for the output
Compute the perspective transformation matrix
Warp the input image using the perspective matrix
Save the results
Input:
import cv2
import numpy as np
import math
# load image
img = cv2.imread('maze.jpg')
hh, ww = img.shape[:2]
# convert to gray
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
# threshold
thresh = cv2.threshold(gray, 0, 255, cv2.THRESH_BINARY+cv2.THRESH_OTSU)[1]
# use morphology to remove the thin lines
kernel = cv2.getStructuringElement(cv2.MORPH_RECT , (5,1))
thresh = cv2.morphologyEx(thresh, cv2.MORPH_CLOSE, kernel)
# invert so that lines are white so that we can get contours for them
thresh_inv = 255 - thresh
# get external contours
contours = cv2.findContours(thresh_inv, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
contours = contours[0] if len(contours) == 2 else contours[1]
# keep contours whose bounding boxes are greater than 1/4 in each dimension
# draw them as white on black background
contour = np.zeros((hh,ww), dtype=np.uint8)
for cntr in contours:
x,y,w,h = cv2.boundingRect(cntr)
if w > ww/4 and h > hh/4:
cv2.drawContours(contour, [cntr], 0, 255, 1)
# get convex hull from contour image white pixels
points = np.column_stack(np.where(contour.transpose() > 0))
hull_pts = cv2.convexHull(points)
# draw hull on copy of input and on black background
hull = img.copy()
cv2.drawContours(hull, [hull_pts], 0, (0,255,0), 2)
hull2 = np.zeros((hh,ww), dtype=np.uint8)
cv2.drawContours(hull2, [hull_pts], 0, 255, 2)
# get 4 corners from white hull points on black background
num = 4
quality = 0.001
mindist = max(ww,hh) // 4
corners = cv2.goodFeaturesToTrack(hull2, num, quality, mindist)
corners = np.int0(corners)
for corner in corners:
px,py = corner.ravel()
cv2.circle(hull, (px,py), 5, (0,0,255), -1)
# get angles to each corner relative to centroid and store with x,y values in list
# angles are clockwise between -180 and +180 with zero along positive X axis (to right)
corner_info = []
center = np.mean(corners, axis=0)
centx = center.ravel()[0]
centy = center.ravel()[1]
for corner in corners:
px,py = corner.ravel()
dx = px - centx
dy = py - centy
angle = (180/math.pi) * math.atan2(dy,dx)
corner_info.append([px,py,angle])
# function to define sort key as element 2 (i.e. angle)
def takeThird(elem):
return elem[2]
# sort corner_info on angle so result will be TL, TR, BR, BL order
corner_info.sort(key=takeThird)
# make conjugate control points
# get input points from corners
corner_list = []
for x, y, angle in corner_info:
corner_list.append([x,y])
print(corner_list)
# define input points from (sorted) corner_list
input = np.float32(corner_list)
# define output points from dimensions of image, say half of input image
width = ww // 2
height = hh // 2
output = np.float32([[0,0], [width-1,0], [width-1,height-1], [0,height-1]])
# compute perspective matrix
matrix = cv2.getPerspectiveTransform(input,output)
# do perspective transformation setting area outside input to black
result = cv2.warpPerspective(img, matrix, (width,height), cv2.INTER_LINEAR, borderMode=cv2.BORDER_CONSTANT, borderValue=(0,0,0))
# save output
cv2.imwrite('maze_thresh.jpg', thresh)
cv2.imwrite('maze_contour.jpg', contour)
cv2.imwrite('maze_hull.jpg', hull)
cv2.imwrite('maze_rectified.jpg', result)
# Display various images to see the steps
cv2.imshow('thresh', thresh)
cv2.imshow('contour', contour)
cv2.imshow('hull', hull)
cv2.imshow('result', result)
cv2.waitKey(0)
cv2.destroyAllWindows()
Thresholded Image after morphology:
Filtered Contours on black background:
Convex hull and 4 corners on input image:
Result from perspective warp:
You can try a simple threshold to detect the lines of the maze, as they are conveniently black:
import cv2
img = cv2.imread("maze.jpg")
gray = cv2.cvtColor(img, cv2.BGR2GRAY)
_, thresh = cv2.threshold(gray, 60, 255, cv2.THRESH_BINARY)
cv2.imshow("Image", thresh)
cv2.waitKey(0)
Output:
You can adjust the threshold yourself with trackbars:
import cv2
cv2.namedWindow("threshold")
cv2.createTrackbar("", "threshold", 0, 255, id)
img = cv2.imread("maze.jpg")
while True:
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
t = cv2.getTrackbarPos("", "threshold")
_, thresh = cv2.threshold(gray, t, 255, cv2.THRESH_BINARY)
cv2.imshow("Image", thresh)
if cv2.waitKey(1) & 0xFF == ord("q"): # If you press the q key
break
Canny is an edge detector. It detects the lines along which color changes. A line in your input image has two such transitions, one on each side. Therefore you see two parallel lines on each side of a line in the image. This answer of mine explains the difference between edges and lines.
So, you shouldn’t be using an edge detector to detect lines in an image.
If a simple threshold doesn't properly binarize this image, try using a local threshold ("adaptive threshold" in OpenCV). Another thing that works well for images like these is applying a top hat filter (for this image, it would be a closing(img) - img), where the structuring element is adjusted to the width of the lines you want to find. This will result in an image that is easy to threshold and will preserve all lines thinner than the structuring element.
Check this:
import cv2
import numpy as np
im=cv2.imread("test2.jpg",1)
#convert 2 gray
mask=cv2.cvtColor(im,cv2.COLOR_BGR2GRAY)
#convert 2 black and white
mask=cv2.threshold(mask,127,255,cv2.THRESH_BINARY)[1]
#remove thin lines and texts and then remake main lines
mask=cv2.dilate(mask,np.ones((5, 5), 'uint8'))
mask=cv2.erode(mask,np.ones((4, 4), 'uint8'))
#smooth lines
mask=cv2.medianBlur(mask,3)
#write output mask
cv2.imwrite("mask2.jpg",mask)
From now on, everything can be done. You can delete extra blobs, you can extract lines from the original image according to the mask, and things like that.
Median:
Median changes are not much for this project. And it can be safely removed. But I prefer it because it rounds the ends of the lines a bit. You have to zoom in a lot to see the pixels. But this technique is usually used to remove salt/pepper noise.
Erode Kernel:
In the case of the kernel, the larger the number, the thicker the lines. Well, this is not always good. Because it causes the path lines to stick to the arrow and later it becomes difficult to separate the paths from the arrow.
Update:
It does not matter if part of the Maze is cleared. The important thing is that from this mask you can draw a rectangle around this shape and create a new mask for this image.
Make a white rectangle around these paths in a new mask. Completely whiten the inside of the mask with FloodFill or any other technique. Now you have a new mask that can take the whole shape out of the original image. Now in the next step you can correct Perspective.

python + cv2 - determine radius of bright spot in image

I already have code that can detect the brightest point in an image (just gaussian blurring + finding the brightest pixel). I am working with photographs of sunsets, and right now can very easily get results like this:
My issue is that the radius of the circle is tied to how much gaussian blur i use - I would like to make it so that the radius reflects the size of the sun in the photo (I have a dataset of ~500 sunset photos I am trying to process).
Here is an image with no circle:
I don't even know where to start on this, my traditional computer vision knowledge is lacking.. If I don't get an answer I might try and do something like calculate the distance from the center of the circle to the nearest edge (using canny edge detection) - if there is a better way please let me know. Thank you for reading
Here is one way to get a representative circle in Python/OpenCV. It finds the minimum enclosing circle.
Read the input
Crop out the white on the right side
Convert to gray
Apply median filtering
Do Canny edge detection
Get the coordinates of all the white pixels (canny edges)
Compute minimum enclosing circle to get center and radius
Draw a circle with that center and radius on a copy of the input
Save the result
Input:
import cv2
import numpy as np
# read image as grayscale
img = cv2.imread('sunset.jpg')
hh, ww = img.shape[:2]
# shave off white region on right side
img = img[0:hh, 0:ww-2]
# convert to gray
gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
# median filter
median = cv2.medianBlur(gray, 3)
# do canny edge detection
canny = cv2.Canny(median, 100, 200)
# get canny points
# numpy points are (y,x)
points = np.argwhere(canny>0)
# get min enclosing circle
center, radius = cv2.minEnclosingCircle(points)
print('center:', center, 'radius:', radius)
# draw circle on copy of input
result = img.copy()
x = int(center[1])
y = int(center[0])
rad = int(radius)
cv2.circle(result, (x,y), rad, (255,255,255), 1)
# write results
cv2.imwrite("sunset_canny.jpg", canny)
cv2.imwrite("sunset_circle.jpg", result)
# show results
cv2.imshow("median", median)
cv2.imshow("canny", canny)
cv2.imshow("result", result)
cv2.waitKey(0)
Canny Edges:
Resulting Circle:
center: (265.5, 504.5) radius: 137.57373046875
Alternate
Fit ellipse to Canny points and then get the average of the two ellipse radii for the radius of the circle. Note a slight change in the Canny arguments to get only the top part of the sunset.
import cv2
import numpy as np
# read image as grayscale
img = cv2.imread('sunset.jpg')
hh, ww = img.shape[:2]
# shave off white region on right side
img = img[0:hh, 0:ww-2]
# convert to gray
gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
# median filter
median = cv2.medianBlur(gray, 3)
# do canny edge detection
canny = cv2.Canny(median, 100, 250)
# transpose canny image to compensate for following numpy points as y,x
canny_t = cv2.transpose(canny)
# get canny points
# numpy points are (y,x)
points = np.argwhere(canny_t>0)
# fit ellipse and get ellipse center, minor and major diameters and angle in degree
ellipse = cv2.fitEllipse(points)
(x,y), (d1,d2), angle = ellipse
print('center: (', x,y, ')', 'diameters: (', d1, d2, ')')
# draw ellipse
result = img.copy()
cv2.ellipse(result, (int(x),int(y)), (int(d1/2),int(d2/2)), angle, 0, 360, (0,0,0), 1)
# draw circle on copy of input of radius = half average of diameters = (d1+d2)/4
rad = int((d1+d2)/4)
xc = int(x)
yc = int(y)
print('center: (', xc,yc, ')', 'radius:', rad)
cv2.circle(result, (xc,yc), rad, (0,255,0), 1)
# write results
cv2.imwrite("sunset_canny_ellipse.jpg", canny)
cv2.imwrite("sunset_ellipse_circle.jpg", result)
# show results
cv2.imshow("median", median)
cv2.imshow("canny", canny)
cv2.imshow("result", result)
cv2.waitKey(0)
Canny Edge Image:
Ellipse and Circle drawn on Input:
Use Canny edge first. Then try either Hough circle or Hough ellipse on the edge image. These are brute force methods, so they will be slow, but they are resistant to non-circular or non-elliptical contours. You can easily filter results such that the detected result has a center near the brightest point. Also, knowing the estimated size of the sun will help with computation speed.
You can also look into using cv2.findContours and cv2.approxPolyDP to extract continuous contours from your images. You could filter by perimeter length and shape and then run a least squares fit, or Hough fit.
EDIT
It may be worth trying an intensity filter before the Canny edge detection. I suspect it will clean up the edge image considerably.

Opencv not finding all contours

I'm trying to find the contours of this image, but the method findContours only returns 1 contour, the contour is highlighted in image 2. I'm trying to find all external contours like these circles where the numbers are inside. What am i doing wrong? What can i do to accomplish it?
image 1:
image 2:
Below is the relevant portion of my code.
thresh = cv2.threshold(image, 0, 255,
cv2.THRESH_BINARY_INV | cv2.THRESH_OTSU)[1]
cnts = cv2.findContours(thresh.copy(), cv2.RETR_EXTERNAL,
cv2.CHAIN_APPROX_SIMPLE)
When i change cv2.RETR_EXTERNAL to cv2.RETR_LIST it seems to detect the same contour twice or something like this. Image 3 shows when the border of circle is first detected and then it is detected again as shows image 4. I'm trying to find only outer borders of these circles. How can i accomplish that?
image 3
image 4
The problem is the flag cv2.RETR_EXTERNAL that you used in the function call. As described in the OpenCV documentation, this only returns the external contour.
Using the flag cv2.RETR_LIST you get all contours in the image. Since you try to detect rings, this list will contain the inner and the outer contour of these rings.
To filter the outer boundary of the circles, you could use cv2.contourArea() to find the larger of two overlapping contours.
I am not sure this is really what you expect nevertheless in case like this there is many way to help findContours to do its job.
Here is a way I use frequently.
Convert your image to gray
Ig = cv2.cvtColor(I,cv2.COLOR_BGR2GRAY)
Thresholding
The background and foreground values looklike quite uniform in term of colours but locally they are not so I apply an thresholding based on Otsu's method in order to binarise the intensities.
_,It = cv2.threshold(Ig,0,255,cv2.THRESH_OTSU)
Sobel magnitude
In order to extract only the contours I process the magnitude of the Sobel edges detector.
sx = cv2.Sobel(It,cv2.CV_32F,1,0)
sy = cv2.Sobel(It,cv2.CV_32F,0,1)
m = cv2.magnitude(sx,sy)
m = cv2.normalize(m,None,0.,255.,cv2.NORM_MINMAX,cv2.CV_8U)
thinning (optional)
I use the thinning function which is implemented in ximgproc.
The interest of the thining is to reduce the contours thickness to as less pixels as possible.
m = cv2.ximgproc.thinning(m,None,cv2.ximgproc.THINNING_GUOHALL)
Final Step findContours
_,contours,hierarchy = cv2.findContours(m,cv2.RETR_TREE,cv2.CHAIN_APPROX_SIMPLE)
disp = cv2.merge((m,m,m)
disp = cv2.drawContours(disp,contours,-1,hierarchy=hierarchy,color=(255,0,0))
Hope it help.
I think an approach based on SVM or a CNN might be more robust.
You can find an example here.
This one may also be interesting.
-EDIT-
I found a let say easier way to reach your goal.
Like previously after loading the image applying a threshold ensure that the image is binary.
By reversing the image using a bitwise not operation the contours become white over a black background.
Applying cv2.connectedComponentsWithStats return (among others) a label matrix in which each connected white region in the source has been assign a unique label.
Then applying findContours based on the labels it is possible give the external contours for every areas.
import numpy as np
import cv2
from matplotlib import pyplot as plt
I = cv2.imread('/home/smile/Downloads/ext_contours.png',cv2.IMREAD_GRAYSCALE)
_,I = cv2.threshold(I,0.,255.,cv2.THRESH_OTSU)
I = cv2.bitwise_not(I)
_,labels,stats,centroid = cv2.connectedComponentsWithStats(I)
result = np.zeros((I.shape[0],I.shape[1],3),np.uint8)
for i in range(0,labels.max()+1):
mask = cv2.compare(labels,i,cv2.CMP_EQ)
_,ctrs,_ = cv2.findContours(mask,cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)
result = cv2.drawContours(result,ctrs,-1,(0xFF,0,0))
plt.figure()
plt.imshow(result)
P.S. Among the outputs return by the function findContours there is a hierachy matrix.
It is possible to reach the same result by analyzing that matrix however it is a little bit more complex as explain here.
Instead of finding contours, I would suggest applying the Hough circle transform using the appropriate parameters.
Finding contours poses a challenge. Once you invert the binary image the circles are in white. OpenCV finds contours both along the outside and the inside of the circle. Moreover since there are letters such as 'A' and 'B', contours will again be found along the outside of the letters and within the holes. You can find contours using the appropriate hierarchy criterion but it is still tedious.
Here is what I tried by finding contours and using hierarchy:
Code:
#--- read the image, convert to gray and obtain inverse binary image ---
img = cv2.imread('keypad.png', 1)
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
_, binary = cv2.threshold(gray, 0, 255, cv2.THRESH_BINARY_INV|cv2.THRESH_OTSU)
#--- find contours ---
_, contours, hierarchy = cv2.findContours(binary, cv2.RETR_CCOMP, cv2.CHAIN_APPROX_NONE)
#--- copy of original image ---
img2 = img.copy()
#--- select contours having a parent contour and append them to a list ---
l = []
for h in hierarchy[0]:
if h[0] > -1 and h[2] > -1:
l.append(h[2])
#--- draw those contours ---
for cnt in l:
if cnt > 0:
cv2.drawContours(img2, [contours[cnt]], 0, (0,255,0), 2)
cv2.imshow('img2', img2)
For more info on contours and their hierarchical relationship please refer this
UPDATE
I have a rather crude way to ignore unwanted contours. Find the average area of all the contours in list l and draw those that are above the average:
Code:
img3 = img.copy()
a = 0
for j, i in enumerate(l):
a = a + cv2.contourArea(contours[i])
mean_area = int(a/len(l))
for cnt in l:
if (cnt > 0) & (cv2.contourArea(contours[cnt]) > mean_area):
cv2.drawContours(img3, [contours[cnt]], 0, (0,255,0), 2)
cv2.imshow('img3', img3)
You can select only the outer borders by this function:
def _select_contours(contours, hierarchy):
"""select contours of the second level"""
# find the border of the image, which has no father
father_i = None
for i, h in enumerate(hierarchy):
if h[3] == -1:
father_i = i
break
# collect its sons
new_contours = []
for c, h in zip(contours, hierarchy):
if h[3] == father_i:
new_contours.append(c)
return new_contours
Note that you should use cv2.RETR_TREE in cv2.findContours() to get the contours and hierarchy.

Detect specific corners in image - OpenCV - Python

I am new using OpenCV and I am having a problem detecting corners in a image. Supposing I have an image-grid like this one where there are multiple corners.
I'd like to catch just the 4 extreme corners, highlighted in red.
Using cv2.cornerHarris() I've managed to highlight all the possible corners really accurately but I can't find a way to adjust these values and let just the 4 extreme remain.
This is my code:
import cv2
import numpy as np
filename = 'start.jpg'
img = cv2.imread(filename)
gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
gray = np.float32(gray)
dst = cv2.cornerHarris(gray,2,29,0.04)
#result is dilated for marking the corners, not important
dst = cv2.dilate(dst,None)
# Threshold for an optimal value, it may vary depending on the image.
img[dst>0.01*dst.max()]=[0,0,255]
cv2.imshow('dst',img)
if cv2.waitKey(0) & 0xff == 27:
cv2.destroyAllWindows()
I've got this code from the official OpenCV website and I've adjusted few values.
Anyone be able to help? Am I in the completely wrong direction? If I've not been clear what I'd like to have is the output-image the I've attached in my question while my code is just giving me all the possible corners in the grid.
Grid does not start from coordinates x,y = 0,0
Thank you in advance guys
I have another solution using contours.
Convert image to grayscale
Obtain binary image
Perform morphological close operation
Find contour present externally
contours, hierarchy = cv2.findContours(th, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
Obtain the bounding rectangle around this contour. You will get the coordinates of this rectangle, which are the corners of the grid
cnts = contours[0]
x,y,w,h = cv2.boundingRect(cnts)
Draw those corners on the grid
cv2.circle(im,(x,y), 3, (0,0,255), -1)
cv2.circle(im,(x+w,y), 3, (0,0,255), -1)
cv2.circle(im,(x+w,y+h), 3, (0,0,255), -1)
cv2.circle(im,(x,y+h), 3, (0,0,255), -1)
cv2.imshow("Corners of grid", img)
There is no easy solution if the grid is in a generic position in the image. The grid extremal corners might even not be detected by the Harris filter, depending on the quality of the image itself. For a robust solution you will need to explicitly model the topology of the graph of detected corners (i.e. find out which corner is neighbor of which ones in the grid). Here is an example of how this is done for the case of a checkerboard.
For a somewhat less robust solution, you could compute the convex hull of the detected corner distribution, then fit 4 lines to it. The lines' intersections would then yield the extremal corners.

Circular contour detection in an image python opencv

I am trying to have the circle detected in the following image.
So I did color thresholding and finally got this result.
Because of the lines in the center being removed, the circle is split into many small parts, so if I do contour detection on this, it can only give me each contour separately.
But is there a way I can somehow combine the contours so I could get a circle instead of just pieces of it?
Here is my code for color thresholding:
blurred = cv2.GaussianBlur(img, (9,9), 9)
ORANGE_MIN = np.array((12, 182, 221),np.uint8)
ORANGE_MAX = np.array((16, 227, 255),np.uint8)
hsv_disk = cv2.cvtColor(blurred,cv2.COLOR_BGR2HSV)
disk_threshed = cv2.inRange(hsv_disk, ORANGE_MIN, ORANGE_MAX)
The task is much easier when performed with the red plane only.
I guess there was problem with the thresholds for color segmentation, So the idea here was to generate a binary mask. By inspection your region of interest seems to be brighter than the other regions of input image, so thresholding can simply be done on a grayScale image to simplify the context. Note: You may change this step as per your requirement. After satisfying with the threshold output, you may use cv2.convexHull() to get the convex shape of your contour.
Also keep in mind to select the largest contour and ignore the small contours. The following code can be used to generate the required output:
import cv2
import numpy as np
# Loading the input_image
img = cv2.imread("/Users/anmoluppal/Downloads/3xGG4.jpg")
# Converting the input image to grayScale
img_gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
# Thresholding the image to get binary mask.
ret, img_thresh = cv2.threshold(img_gray, 145, 255, cv2.THRESH_BINARY)
# Dilating the mask image
kernel = np.ones((3,3),np.uint8)
dilation = cv2.dilate(img_thresh,kernel,iterations = 3)
# Getting all the contours
_, contours, __ = cv2.findContours(dilation, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_NONE)
# Finding the largest contour Id
largest_contour_area = 0
largest_contour_area_idx = 0
for i in xrange(len(contours)):
if (cv2.contourArea(contours[i]) > largest_contour_area):
largest_contour_area = cv2.contourArea(contours[i])
largest_contour_area_idx = i
# Get the convex Hull for the largest contour
hull = cv2.convexHull(contours[largest_contour_area_idx])
# Drawing the contours for debugging purposes.
img = cv2.drawContours(img, [hull], 0, [0, 255, 0])
cv2.imwrite("./garbage.png", img)

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