I used opencv's minAreaRect to deskew the mnist digits.It worked well for most of the digits but,in some cases the minAreaRect was not detected correctly and it lead to further skewing of the digits.
Images with which this code worked:
Input image:
minAreaRect Image:
deskewed image:
But,for this the didn't work well:
Input image:
minAreaRect Image:
deskewed image:
I want to mention here that I did use: #coords = np.column_stack(np.where(thresh>0)) but,this didn't work at all.
Please suggest a solution using minAreaRect(Preferred) function of opencv.
And I've tested with several images and I do understand that the problem is with the formation of the min Area Rectangle,in the second example it is clear that the min Area rectangle is not visible(because it passess through the digit itself).
Here goes the code:
import numpy as np
import cv2
image=cv2.imread('MNIST/mnist_png/testing/9/73.png')#for 4##5032,6780 #8527,2436,1391
gray=cv2.cvtColor(image,cv2.COLOR_BGR2GRAY)## for 9 problem with 4665,8998,73,7
gray=cv2.bitwise_not(gray)
Gblur=cv2.blur(gray,(5,5))
thresh=cv2.threshold(Gblur,0,255,cv2.THRESH_BINARY | cv2.THRESH_OTSU)[1]
#cv2.imshow("gray_thresh_blur",thresh)
#Finding Contours will be used to draw the min area rectangle
_,contours,_=cv2.findContours(thresh.copy(),cv2.RETR_LIST,cv2.CHAIN_APPROX_NONE)
cnt1 = contours[0]
cnt=cv2.convexHull(contours[0])
angle = cv2.minAreaRect(cnt)[-1]
print("Actual angle is:"+str(angle))
rect = cv2.minAreaRect(cnt)
p=np.array(rect[1])
#print(p[0])
if p[0] < p[1]:
print("Angle along the longer side:"+str(rect[-1] + 180))
act_angle=rect[-1]+180
else:
print("Angle along the longer side:"+str(rect[-1] + 90))
act_angle=rect[-1]+90
#act_angle gives the angle with bounding box
if act_angle < 90:
angle = (90 + angle)
print("angleless than -45")
# otherwise, just take the inverse of the angle to make
# it positive
else:
angle=act_angle-180
print("grter than 90")
# rotate the image to deskew it
(h, w) = image.shape[:2]
center = (w // 2, h // 2)
M = cv2.getRotationMatrix2D(center, angle, 1.0)
rotated = cv2.warpAffine(image,M,(w,h),flags=cv2.INTER_CUBIC,borderMode=cv2.BORDER_REPLICATE)
box = cv2.boxPoints(rect)
print(box)
box = np.int0(box)
print(box)
p=cv2.drawContours(thresh,[box],0,(0,0,255),1)
print("contours"+str(p))
cv2.imwrite("post/MinAreaRect9.png",p)
cv2.imwrite("post/Input_9.png", image)
cv2.imwrite('post/Deskewed_9.png', rotated)
A few points to take note of:
Most of OpenCV's functions work with white foreground and black background. So comment out this line:
gray=cv2.bitwise_not(gray)
Make sure you're computing the EXTERNEL contours of the letters. This means that you need to ignore all the child contours. For this use cv2.RETR_EXTERNAL.
contours=cv2.findContours(thresh.copy(),cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)[1]
Finally make sure you're assigning the correct angle to find rotation matrix.
With these changes:
Related
I was using this code to detect the rectangle on the photo, at first it was working well, untill i realized that i would have an object that is also a square in the middle :
Question:
How can i properly detect the 4 corners like on the first result picture without detecting the corner of the thing in the middle of the square. Thanks a lot.
Code:
import numpy as np
import cv2
img = cv2.imread('Photos/lastBoard.png')
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
canny = cv2.Canny(gray, 100, 200)
corners = cv2.goodFeaturesToTrack(gray, 25, 0.01, 50)
corner_list = []
for corner in corners:
x, y = corner.ravel()
if(y < 700 and (50 < x < 800 )):
corner_list.append([int(x), int(y)])
cv2.circle(img, (x, y), 5, (36, 255, 12), -1)
cv2.imshow("yo", img)
cv2.waitKey(0)
My man, it breaks my heart you aren't using the techniques and processing we covered in your last question. You have already plenty of functions you could re-use. The rectangle you are trying to segment has a unique color (kind of green) and has a defined area and aspect ratio! Look all the things you have on the table, they are smaller than the rectangle! Plus, the rectangle is almost a square! That means that its aspect ratio is close to 1.0. If you somehow segment the rectangle, approximating its corners should be relativity easy.
This is valuable info, because it allows you to trace your action plan. I see you are using cv2.goodFeaturesToTrack to detect the corners of everything. That's OK, but it could be simplified. I propose a plan of action very similar to last time:
Try to segment the rectangle using its color, let's compute an
HSV-based mask
Let's clean the mask from noise using an area filter and some morphology
Find contours - we are looking for the biggest green contour, the rectangle.
The contour of interest has defined features. Use the area and aspect ratio to filter garbage contours.
Once you have the contour/blob of interest, approximate its corners.
Let's see the code:
# imports:
import numpy as np
import cv2
# image path
path = "D://opencvImages//"
fileName = "table1.jpg"
# Reading an image in default mode:
inputImage = cv2.imread(path + fileName)
inputCopy = inputImage.copy()
# The HSV mask values:
lowerValues = np.array([58, 151, 25])
upperValues = np.array([86, 255, 75])
# Convert the image to HSV:
hsvImage = cv2.cvtColor(inputImage, cv2.COLOR_BGR2HSV)
# Create the HSV mask
mask = cv2.inRange(hsvImage, lowerValues, upperValues)
The first steps aim to create the HSV mask. Very similar to last time, I've defined the HSV range of interest already and applied exactly the same stuff as before. You could (and should) explore more exotic techniques latter, but let's stick with what we know works for the time being, as the project surely is due soon. This is the result:
You see how the mask is pretty nice already? Only the green puck and the rectangle survived the thresholding. It doesn't matter that the rectangle is not complete, because we're gonna approximate its contour with a bounding rectangle! Alright, let's clean this bad boy a little bit better. Use a filterArea (this is exactly the same function we saw last time) and then a closing (dilate followed by erode) just to get a nice mask:
# Run a minimum area filter:
minArea = 50
mask = areaFilter(minArea, mask)
# Pre-process mask:
kernelSize = 3
structuringElement = cv2.getStructuringElement(cv2.MORPH_RECT, (kernelSize, kernelSize))
iterations = 2
mask = cv2.morphologyEx(mask, cv2.MORPH_DILATE, structuringElement, None, None, iterations, cv2.BORDER_REFLECT101)
mask = cv2.morphologyEx(mask, cv2.MORPH_ERODE, structuringElement, None, None, iterations, cv2.BORDER_REFLECT101)
This is the filtered mask, the noise is mostly gone:
Now, let's find contours and filtered based on area and aspect ratio, just like last time. The parameters, however, are different, because our target is not the plucks, but the rectangle:
# Find the big contours/blobs on the filtered image:
contours, hierarchy = cv2.findContours(mask, cv2.RETR_CCOMP, cv2.CHAIN_APPROX_SIMPLE)
# Store the poly approximation and bound
contoursPoly = [None] * len(contours)
# Store the corners of the square here:
detectedCorners = []
# Look for the outer bounding boxes:
for _, c in enumerate(contours):
# Approximate the contour to a polygon:
contoursPoly = cv2.approxPolyDP(c, 3, True)
# Convert the polygon to a bounding rectangle:
boundRect = cv2.boundingRect(contoursPoly)
# Get the bounding rect's data:
rectX = boundRect[0]
rectY = boundRect[1]
rectWidth = boundRect[2]
rectHeight = boundRect[3]
# Calculate the rect's area:
rectArea = rectWidth * rectHeight
# Calculate the aspect ratio:
aspectRatio = rectWidth / rectHeight
delta = abs(1.0 - aspectRatio)
# Set the min threshold values to identify the
# blob of interest:
minArea = 2500
epsilon = 0.2
Alright, so far so good, I hope. As you see I approximated the contour to a 4-vertex polygon and then computed its bounding rectangle. This approximation should fit very nicely to our blob of interest. Now, apply the contour filter and use the bounding rectangle data to approximate the corners. I approximate each corner, one by one, and store them in the
detectedCorners array. Then, we can draw 'em. Here, still inside the for loop:
# Is this bounding rectangle we
# are looking for?
if rectArea > minArea and delta < epsilon:
# Compute the corners/vertices:
# Corner 1 (top left)
corner1 = (rectX, rectY)
detectedCorners.append(corner1)
# Corner 2 (top right)
corner2 = (rectX + rectWidth, rectY)
detectedCorners.append(corner2)
# Corner 3 (bottom left)
corner3 = (rectX, rectY + rectHeight)
detectedCorners.append(corner3)
# Corner 4 (bottom right)
corner4 = (rectX + rectWidth, rectY + rectHeight)
detectedCorners.append(corner4)
# Draw the corner points:
for p in detectedCorners:
color = (0, 0, 255)
cv2.circle(inputCopy, (p[0], p[1]), 5, color, -1)
cv2.imshow("Square Corners", inputCopy)
cv2.waitKey(0)
Here are the results for both images. The approximated corners are the red dots:
Here's the definition and implementation of the areaFilter function:
def areaFilter(minArea, inputImage):
# Perform an area filter on the binary blobs:
componentsNumber, labeledImage, componentStats, componentCentroids = \
cv2.connectedComponentsWithStats(inputImage, connectivity=4)
# Get the indices/labels of the remaining components based on the area stat
# (skip the background component at index 0)
remainingComponentLabels = [i for i in range(1, componentsNumber) if componentStats[i][4] >= minArea]
# Filter the labeled pixels based on the remaining labels,
# assign pixel intensity to 255 (uint8) for the remaining pixels
filteredImage = np.where(np.isin(labeledImage, remainingComponentLabels) == True, 255, 0).astype('uint8')
return filteredImage
I have a dataset of x-ray images that i am trying to clean by rotating the images so the arm is vertical and cropping the image of any excess space. Here are some examples from the dataset:
I am currently working out the best way to work out the angle of the x-ray and rotate the image based on that.
My curent approach is to detect the line of the side of the rectangle that the scan is in using the hough transform, and rotate the image based on that.
I tried to run the hough transform on the output of a canny edge detector but this doesnt work so well for images where the edge of the rectangle is blurred like in the first image.
I cant use cv's box detection as sometimes the rectangle around the scan has an edge off screen.
So i currently use adaptive thresholding to find the edge of the box and then median filter it and try to find the longest line in this, but sometimes the wrong line is the longest and the image gets rotated completley wrong.
Adaptive thresholding is used due to the fact that soem scans have different brightnesses.
The current implementation i have is:
def get_lines(img):
#threshold
thresh = cv2.adaptiveThreshold(img, 255, cv2.ADAPTIVE_THRESH_GAUSSIAN_C, cv2.THRESH_BINARY_INV, 15, 4.75)
median = cv2.medianBlur(thresh, 3)
# detect lines
lines = cv2.HoughLines(median, 1, np.pi/180, 175)
return sorted(lines, key=lambda x: x[0][0], reverse=True)
def rotate(image, angle):
(h, w) = image.shape[:2]
(cX, cY) = (w // 2, h // 2)
M = cv2.getRotationMatrix2D((cX, cY), angle, 1.0)
cos = np.abs(M[0, 0])
sin = np.abs(M[0, 1])
nW = int((h * sin) + (w * cos))
nH = int((h * cos) + (w * sin))
M[0, 2] += (nW / 2) - cX
M[1, 2] += (nH / 2) - cY
return cv2.warpAffine(image, M, (nW, nH))
def fix_rotation(input):
lines = get_lines(input)
rho, theta = lines[0][0]
return rotate_bound(input, theta*180/np.pi)
and produces the following results:
When it goes wrong:
I was wondering if there are any better techniques to usein order to improve the performance of this and what the best way to go about cropping the images after they have been rotated would be?
The idea is to use the blob of the arm itself and fit an ellipse around it. Then, extract its major axis. I quickly tested the idea in Matlab – not OpenCV. Here's what I did, you should be able to use OpenCV's equivalent functions to achieve similar outputs.
First, compute the threshold value of your input via Otsu. Then add some bias to the threshold value to find a better segmentation and use this value to threshold the image.
In pseudo-code:
//the bias value
threshBias = 0.4;
//get the binary threshold via otsu:
thresholdLevel = graythresh( grayInput, “otsu” );
//add bias to the original value
thresholdLevel = thresholdLevel - threshSensitivity * thresholdLevel;
//get the fixed binary image:
thresholdLevel = imbinarize( grayInput, thresholdLevel );
After small blob filtering, this is the output:
Now, get the contours/blobs and fit an ellipse for each contour. Check out the OpenCV example here: https://docs.opencv.org/3.4.9/de/d62/tutorial_bounding_rotated_ellipses.html
You end up with two ellipses:
We are looking for the biggest ellipse, the one with the biggest area and the biggest major and minor axis. I used the width and height of each ellipse to filter the results. The target ellipse is then colored in green. Finally, I get the major axis of the target ellipse, here colored in yellow:
Now, to implement these ideas in OpenCV you have these options:
Use fitEllipse to find the ellipses. The return value of this
function is a RotatedRect object. The data stored here are the
vertices of the ellipse.
Instead of fitting an ellipse you could try using minAreaRect, which
finds a rotated rectangle of the minimum area enclosing a blob.
You can use image moments to calculate the rotation angle.
Using opencv moments function, calculate the second order central moments to construct a covariance matrix and then obtain the orientation as shown here in the Image moment wiki page.
Obtain the normalized central moments nu20, nu11 and nu02 from opencv moments. Then the orientation is calculated as
0.5 * arctan(2 * nu11/(nu20 - nu02))
Please refer the given link for details.
You can use the raw image itself or the preprocessed one for the calculation of orientation. See which one gives you better accuracy and use it.
As for the bounding-box, once you rotate the image, assuming you used the preprocessed one, get all the non-zero pixel coordinates of the rotated image and calculate their upright bounding-box using opencv boundingRect.
I have some hundreds of images (scanned documents), most of them are skewed. I wanted to de-skew them using Python.
Here is the code I used:
import numpy as np
import cv2
from skimage.transform import radon
filename = 'path_to_filename'
# Load file, converting to grayscale
img = cv2.imread(filename)
I = cv2.cvtColor(img, COLOR_BGR2GRAY)
h, w = I.shape
# If the resolution is high, resize the image to reduce processing time.
if (w > 640):
I = cv2.resize(I, (640, int((h / w) * 640)))
I = I - np.mean(I) # Demean; make the brightness extend above and below zero
# Do the radon transform
sinogram = radon(I)
# Find the RMS value of each row and find "busiest" rotation,
# where the transform is lined up perfectly with the alternating dark
# text and white lines
r = np.array([np.sqrt(np.mean(np.abs(line) ** 2)) for line in sinogram.transpose()])
rotation = np.argmax(r)
print('Rotation: {:.2f} degrees'.format(90 - rotation))
# Rotate and save with the original resolution
M = cv2.getRotationMatrix2D((w/2,h/2),90 - rotation,1)
dst = cv2.warpAffine(img,M,(w,h))
cv2.imwrite('rotated.jpg', dst)
This code works well with most of the documents, except with some angles: (180 and 0) and (90 and 270) are often detected as the same angle (i.e it does not make difference between (180 and 0) and (90 and 270)). So I get a lot of upside-down documents.
Here is an example:
The resulted image that I get is the same as the input image.
Is there any suggestion to detect if an image is upside down using Opencv and Python?
PS: I tried to check the orientation using EXIF data, but it didn't lead to any solution.
EDIT:
It is possible to detect the orientation using Tesseract (pytesseract for Python), but it is only possible when the image contains a lot of characters.
For anyone who may need this:
import cv2
import pytesseract
print(pytesseract.image_to_osd(cv2.imread(file_name)))
If the document contains enough characters, it is possible for Tesseract to detect the orientation. However, when the image has few lines, the orientation angle suggested by Tesseract is usually wrong. So this can not be a 100% solution.
Python3/OpenCV4 script to align scanned documents.
Rotate the document and sum the rows. When the document has 0 and 180 degrees of rotation, there will be a lot of black pixels in the image:
Use a score keeping method. Score each image for it's likeness to a zebra pattern. The image with the best score has the correct rotation. The image you linked to was off by 0.5 degrees. I omitted some functions for readability, the full code can be found here.
# Rotate the image around in a circle
angle = 0
while angle <= 360:
# Rotate the source image
img = rotate(src, angle)
# Crop the center 1/3rd of the image (roi is filled with text)
h,w = img.shape
buffer = min(h, w) - int(min(h,w)/1.15)
roi = img[int(h/2-buffer):int(h/2+buffer), int(w/2-buffer):int(w/2+buffer)]
# Create background to draw transform on
bg = np.zeros((buffer*2, buffer*2), np.uint8)
# Compute the sums of the rows
row_sums = sum_rows(roi)
# High score --> Zebra stripes
score = np.count_nonzero(row_sums)
scores.append(score)
# Image has best rotation
if score <= min(scores):
# Save the rotatied image
print('found optimal rotation')
best_rotation = img.copy()
k = display_data(roi, row_sums, buffer)
if k == 27: break
# Increment angle and try again
angle += .75
cv2.destroyAllWindows()
How to tell if the document is upside down? Fill in the area from the top of the document to the first non-black pixel in the image. Measure the area in yellow. The image that has the smallest area will be the one that is right-side-up:
# Find the area from the top of page to top of image
_, bg = area_to_top_of_text(best_rotation.copy())
right_side_up = sum(sum(bg))
# Flip image and try again
best_rotation_flipped = rotate(best_rotation, 180)
_, bg = area_to_top_of_text(best_rotation_flipped.copy())
upside_down = sum(sum(bg))
# Check which area is larger
if right_side_up < upside_down: aligned_image = best_rotation
else: aligned_image = best_rotation_flipped
# Save aligned image
cv2.imwrite('/home/stephen/Desktop/best_rotation.png', 255-aligned_image)
cv2.destroyAllWindows()
Assuming you did run the angle-correction already on the image, you can try the following to find out if it is flipped:
Project the corrected image to the y-axis, so that you get a 'peak' for each line. Important: There are actually almost always two sub-peaks!
Smooth this projection by convolving with a gaussian in order to get rid of fine structure, noise, etc.
For each peak, check if the stronger sub-peak is on top or at the bottom.
Calculate the fraction of peaks that have sub-peaks on the bottom side. This is your scalar value that gives you the confidence that the image is oriented correctly.
The peak finding in step 3 is done by finding sections with above average values. The sub-peaks are then found via argmax.
Here's a figure to illustrate the approach; A few lines of you example image
Blue: Original projection
Orange: smoothed projection
Horizontal line: average of the smoothed projection for the whole image.
here's some code that does this:
import cv2
import numpy as np
# load image, convert to grayscale, threshold it at 127 and invert.
page = cv2.imread('Page.jpg')
page = cv2.cvtColor(page, cv2.COLOR_BGR2GRAY)
page = cv2.threshold(page, 127, 255, cv2.THRESH_BINARY_INV)[1]
# project the page to the side and smooth it with a gaussian
projection = np.sum(page, 1)
gaussian_filter = np.exp(-(np.arange(-3, 3, 0.1)**2))
gaussian_filter /= np.sum(gaussian_filter)
smooth = np.convolve(projection, gaussian_filter)
# find the pixel values where we expect lines to start and end
mask = smooth > np.average(smooth)
edges = np.convolve(mask, [1, -1])
line_starts = np.where(edges == 1)[0]
line_endings = np.where(edges == -1)[0]
# count lines with peaks on the lower side
lower_peaks = 0
for start, end in zip(line_starts, line_endings):
line = smooth[start:end]
if np.argmax(line) < len(line)/2:
lower_peaks += 1
print(lower_peaks / len(line_starts))
this prints 0.125 for the given image, so this is not oriented correctly and must be flipped.
Note that this approach might break badly if there are images or anything not organized in lines in the image (maybe math or pictures). Another problem would be too few lines, resulting in bad statistics.
Also different fonts might result in different distributions. You can try this on a few images and see if the approach works. I don't have enough data.
You can use the Alyn module. To install it:
pip install alyn
Then to use it to deskew images(Taken from the homepage):
from alyn import Deskew
d = Deskew(
input_file='path_to_file',
display_image='preview the image on screen',
output_file='path_for_deskewed image',
r_angle='offest_angle_in_degrees_to_control_orientation')`
d.run()
Note that Alyn is only for deskewing text.
I am using OpenCV for a robot vision project - navigating a maze. I can detect the lines where the walls of the maze meet the floor. And now need to use these detected lines to calculate which way the robot should turn.
In order to work out which way the robot should move I believe the solution is to calculate the angle of the walls in relation to the position of the robot. However where both walls are found how do I select which points to use as a reference.
I understand that I can use the python atan2 formula to calculate the angle between two points but after that I am completely lost.
Here is my code:
# https://towardsdatascience.com/finding-driving-lane-line-live-with-opencv-f17c266f15db
# Testing edge detection for maze
import cv2
import numpy as np
import math
image = cv2.imread("/Users/BillHarvey/Documents/Electronics_and_Robotics/Robot_Vision_Project/mazeme/maze1.png")
gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)
kernel_size = 5
blur_gray = cv2.GaussianBlur(gray,(kernel_size,kernel_size),0)
low_threshold = 50
high_threshold = 150
edges = cv2.Canny(blur_gray, low_threshold, high_threshold)
# create a mask of the edges image using cv2.filpoly()
mask = np.zeros_like(edges)
ignore_mask_color = 255
# define the Region of Interest (ROI) - source code sets as a trapezoid for roads
imshape = image.shape
vertices = np.array([[(0,imshape[0]),(100, 420), (1590, 420),(imshape[1],imshape[0])]], dtype=np.int32)
cv2.fillPoly(mask, vertices, ignore_mask_color)
masked_edges = cv2.bitwise_and(edges, mask)
# mybasic ROI bounded by a blue rectangle
#ROI = cv2.rectangle(image,(0,420),(1689,839),(0,255,0),3)
# define the Hough Transform parameters
rho = 2 # distance resolution in pixels of the Hough grid
theta = np.pi/180 # angular resolution in radians of the Hough grid
threshold = 15 # minimum number of votes (intersections in Hough grid cell)
min_line_length = 40 #minimum number of pixels making up a line
max_line_gap = 30 # maximum gap in pixels between connectable line segments
# make a blank the same size as the original image to draw on
line_image = np.copy(image)*0
# run Hough on edge detected image
lines = cv2.HoughLinesP(masked_edges, rho, theta, threshold, np.array([]),min_line_length, max_line_gap)
for line in lines:
for x1,y1,x2,y2 in line:
cv2.line(line_image,(x1,y1),(x2,y2),(255,0,0),10)
angle = math.atan2(x2-x1, y2-y1)
angle = angle * 180 / 3.14
print("Angle = ", angle)
# draw the line on the original image
lines_edges = cv2.addWeighted(image, 0.8, line_image, 1, 0)
#return lines_edges
#cv2.imshow("original", image)
#cv2.waitKey(0)
#cv2.imshow("edges", edges)
#cv2.waitKey(0)
cv2.imshow("detected", lines_edges)
cv2.waitKey(0)
cv2.imwrite("lanes_detected.jpg", lines_edges)
cv2.destroyAllWindows()
I have added the athn2 forumla in the piece of code that draws blue lines where HoughLinesP has detected lines.
And to convert the results (angle) to degrees I found this formula:
angle = angle * 180 / 3.14
The following piece of code:
print("Angle = ", angle)
Prints 13 angles which may or may not equate to the lines in the pic, do they? To avoid getting a - degrees I had to do x2-x1, y2-y1 rather than the other way around which I have seen elsewhere.
I do apologise for my fundental lack of python and mathematical knowledge but any help would be gratefully received.
I have some code which detects circular shapes but I am unable to understand how it works.
From this code:
How can i find the radius and center point of the circle?
What is the behaviour of `cv2.approxPolyDP' for detecting circles?
Now find the contours in the segmented mask
contours, hierarchy = cv2.findContours(mask.copy(), cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_NONE)
Sorting the contours w.r.t contour rect X
contours.sort(key = lambda x:cv2.boundingRect(x)[0])
for contour in contours:
approx = cv2.approxPolyDP(contour, 0.01*cv2.arcLength(contour,True), True)
if len(approx) > 8:
# Find the bounding rect of contour.
contour_bounding_rect = cv2.boundingRect(contour)
mid_point = contour_bounding_rect[0] + contour_bounding_rect[2]/2, contour_bounding_rect[1] + contour_bounding_rect[3]/2
print mid_point[1]/single_element_height, ", ",
So I have figured out the answer to your first question: determining the center and radius of circles in the image.
Initially I am finding all the contours present in the image. Then using a for loop, I found the center and radius using cv2.minEnclosingCircle for every contour in the image. I printed them in the console screen.
contours,hierarchy = cv2.findContours(thresh,2,1)
print len(contours)
cnt = contours
for i in range (len(cnt)):
(x,y),radius = cv2.minEnclosingCircle(cnt[i])
center = (int(x),int(y))
radius = int(radius)
cv2.circle(img,center,radius,(0,255,0),2)
print 'Circle' + str(i) + ': Center =' + str(center) + 'Radius =' + str(radius)
To answer your second question on cv2.approxPolyDP(); this function draws an approximate contour around the object in the image based on a parameter called 'epsilon'. Higher the value of 'epsilon', the contour is roughly approximated. For a lower value of epsilon, the contour grazes almost every edge of the object in the image. Visit THIS PAGE for a better understanding.
Hope this helped!! :)
Don't think approxPolyDP is the right way to go here.
If you have an image where you only have circles and you want to find center and radius, try minEnclosingCircle()
If you have an image where you have various shapes and you want to find the circles, try Hough transform (may take a long time) or fitEllipse() where you check if the bounding box it returns is square.
See documentation for both these functions