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The goal is to find the line that represents the distance between the "hole" and the outer edge. Is Canny edge detection the best approach?

The goal is to find the line that represents the distance between the "hole" and the outer edge (transition from black to white). I was able to successfully binarize this photo and get a very clean black and white image. The next step would be to find the (almost) vertical line on it and calculate the perpendicular distance to the midpoint of this vertical line and the hole.
original picture
hole - zoomed in
ps: what I call "hole" is a shadow. I am shooting a laser into a hole. So the lines we can see is a steel and the black part without a line is a hole. The 2 white lines serve as a reference to measure the distance.
Is Canny edge detection the best approach? If so, what are good values for the A, B and C parameters? I can't tune it. I'm getting too much noise.

This is not complete; You have to take the time to reach the final result. But this idea might help you.
Preprocessors:
import os
import cv2
import numpy as np
Main code:
# Read original image
dir = os.path.abspath(os.path.dirname(__file__))
im = cv2.imread(dir+'/'+'im.jpg')
h, w = im.shape[:2]
print(w, h)
# Convert image to Grayscale
imGray = cv2.cvtColor(im, cv2.COLOR_BGR2GRAY)
cv2.imwrite(dir+'/im_1_grayscale.jpg', imGray)
# Eliminate noise and display laser light better
imHLine = imGray.copy()
imHLine = cv2.GaussianBlur(imHLine, (0, 9), 21) # 5, 51
cv2.imwrite(dir+'/im_2_h_line.jpg', imHLine)
# Make a BW mask to find the ROI of laser array
imHLineBW = cv2.threshold(imHLine, 22, 255, cv2.THRESH_BINARY)[1]
cv2.imwrite(dir+'/im_3_h_line_bw.jpg', imHLineBW)
# Remove noise with mask and extract just needed area
imHLineROI = imGray.copy()
imHLineROI[np.where(imHLineBW == 0)] = 0
imHLineROI = cv2.GaussianBlur(imHLineROI, (0, 3), 6)
imHLineROI = cv2.threshold(imHLineROI, 25, 255, cv2.THRESH_BINARY)[1] # 22
cv2.imwrite(dir+'/im_4_ROI.jpg', imHLineROI)
# Found laser array and draw box around it
cnts, _ = cv2.findContours(
imHLineROI, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
cnts.sort(key=lambda x: cv2.boundingRect(x)[0])
pts = []
for cnt in cnts:
x2, y2, w2, h2 = cv2.boundingRect(cnt)
if h2 < h/10:
cv2.rectangle(im, (x2, y2), (x2+w2, y2+h2), (0, 255, 0), 1)
pts.append({'x': x2, 'y': y2, 'w': w2, 'h': h2})
circle = {
'left': (pts[0]['x']+pts[0]['w'], pts[0]['y']+pts[0]['h']/2),
'right': (pts[1]['x'], pts[1]['y']+pts[1]['h']/2),
}
circle['center'] = calculateMiddlePint(circle['left'], circle['right'])
circle['radius'] = (circle['right'][0]-circle['left'][0])//2
# Draw line and Circle inside it
im = drawLine(im, circle['left'], circle['right'], color=(27, 50, 120))
im = cv2.circle(im, circle['center'], circle['radius'], (255, 25, 25), 3)
# Remove pepper/salt noise to find metal edge
imVLine = imGray.copy()
imVLine = cv2.medianBlur(imVLine, 17)
cv2.imwrite(dir+'/im_6_v_line.jpg', imVLine)
# Remove remove the shadows to find metal edge
imVLineBW = cv2.threshold(imVLine, 50, 255, cv2.THRESH_BINARY)[1]
cv2.imwrite(dir+'/im_7_v_bw.jpg', imVLineBW)
# Finding the right vertical edge of metal
y1, y2 = h/5, h-h/5
x1 = horizantalDistance(imVLineBW, y1)
x2 = horizantalDistance(imVLineBW, y2)
pt1, pt2 = (x1, y1), (x2, y2)
imVLineBW = drawLine(imVLineBW, pt1, pt2)
cv2.imwrite(dir+'/im_8_v_bw.jpg', imVLineBW)
# Draw lines
im = drawLine(im, pt1, pt2)
im = drawLine(im, calculateMiddlePint(pt1, pt2), circle['center'])
# Draw final image
cv2.imwrite(dir+'/im_8_output.jpg', im)
Extra functions:
Find the first white pixel in one line of picture:
# This function only processes on a horizontal line of the image
# Its job is to examine the pixels one by one from the right and
# report the distance of the first white pixel from the right of
# the image.
def horizantalDistance(im, y):
y = int(y)
h, w = im.shape[:2]
for i in range(0, w):
x = w-i-1
if im[y][x] == 255:
return x
return -1
To draw a line in opencv:
def drawLine(im, pt1, pt2, color=(128, 0, 200), thickness=2):
return cv2.line(
im,
pt1=(int(pt1[0]), int(pt1[1])),
pt2=(int(pt2[0]), int(pt2[1])),
color=color,
thickness=thickness,
lineType=cv2.LINE_AA # Anti-Aliased
)
To calculate middle point of two 2d points:
def calculateMiddlePint(p1, p2):
return (int((p1[0]+p2[0])/2), int((p1[1]+p2[1])/2))
Output:
Original image:
Eliminate noise and process to see the laser array better:
Find the laser area to extract the hole:
Work on another copy of image to find the right side of metal object:
Remove the shadows to better see the right edge:
The final output:
I first defined an ROI area. I changed the code later but did not change the names of the variables. If you were asked.

You can try one of binarized Canny, or straight binarized image (Otsu). In both cases, you will obtain a quasi-straight line from which you can extract the right-most white pixel on every row or so.
Then use line fitting to these points (depending on image quality the fitting needs to be robust or not).
It would be wise to calibrate the method against ground truth, because the edge your are looking after borders a gradient area, and the location of the edge might be offset by several pixels.

Remove and measure a line openCV

Links to all images at the bottom
I have drawn a line over an arrow which captures the angle of that arrow. I would like to then remove the arrow, keep only the line, and use cv2.minAreaRect to determine the angle. So far I've got everything to work except removing the original arrow, which results in an incorrect angle generated by the cv2.minAreaRect bounding box.
Really, I just want the bold black line running through the arrow to use to measure the angle, not the arrow itself. if anyone has an idea to make this work, or a simpler way, please let me know. Thanks
Code:
import numpy as np
import cv2
image = cv2.imread("templates/a_15.png")
image = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)
ret, thresh = cv2.threshold(image, 127, 255, 0)
contours, hierarchy = cv2.findContours(thresh, 1, 2)
cont = contours[0]
rows,cols = image.shape[:2]
[vx,vy,x,y] = cv2.fitLine(cont, cv2.DIST_L2,0,0.01,0.01)
leftish = int((-x*vy/vx) + y)
rightish = int(((cols-x)*vy/vx)+y)
line = cv2.line(image,(cols-1,rightish),(0,leftish),(0,255,0),10)
# thresholding
thresh = cv2.threshold(line, 0, 255, cv2.THRESH_BINARY | cv2.THRESH_OTSU)[1]
# compute rotated bounding box based on all pixel values > 0 and
# use coordinates to compute a rotated bounding box of those coordinates
coordinates = np.column_stack(np.where(thresh > 0))
w = coordinates[0]
h = coordinates[1]
# Compute minimum rotated angle that contains entire image.
# Return angle values in the range [-90, 0).
# As the rectangle rotates clockwise, angle values increase towards 0.
# Once 0 is reached, angle is set back to -90 degrees.
angle = cv2.minAreaRect(coordinates)[-1]
# for angles less than -45 degrees, add 90 degrees to angle to take the inverse.
if angle < - 45:
angle = -(90 + angle)
else:
angle = -angle
# rotate image
(h, w) = image.shape[:2]
center = (w // 2, h // 2) # image center
RM = cv2.getRotationMatrix2D(center, angle, 1.0)
rotated = cv2.warpAffine(image, RM, (w, h),
flags=cv2.INTER_CUBIC, borderMode=cv2.BORDER_REPLICATE)
# correction angle for validation
cv2.putText(rotated, "Angle {:.2f} degrees".format(angle),
(10, 30), cv2.FONT_HERSHEY_DUPLEX, 0.9, (0, 255, 0), 2)
# output
print("[INFO] angle: {:.3f}".format(angle))
cv2.imshow("Line", line)
cv2.imshow("Input", image)
cv2.imshow("Rotated", rotated)
cv2.waitKey(0)
Images
original
current results
goal

Here's a possible solution. The main idea is to identify de "tip" and the "tail" of the arrow approximating some key points. After you have identified both ends, you can draw a line joining both points. It is also an advantage to know which of the endpoints is the tip, because that way you can measure the angle from a constant point.
There's more than one way to achieve this. I choose something that I have applied in the past: I will use this approach to identify the endpoints of the overall shape. My assumption is that the tip will yield more points than the tail. After that, I'll cluster all the endpoints in two groups: tip and tail. I can use K-Means for that, as it will return the mean centers for both clusters. After that, we have our tip and tail points that can be joined easily with a line. These are the steps:
Convert the image to grayscale
Get the skeleton of the image, to normalize the shape to a width of 1 pixel
Apply the method described in the link to get the arrow's endpoints
Divide the endpoints in two clusters and use K-Means to get their centers
Join both endpoints with a line
Let's see the code:
# imports:
import cv2
import numpy as np
# image path
path = "D://opencvImages//"
fileName = "CoXeb.png"
# Reading an image in default mode:
inputImage = cv2.imread(path + fileName)
# Grayscale conversion:
grayscaleImage = cv2.cvtColor(inputImage, cv2.COLOR_BGR2GRAY)
grayscaleImage = 255 - grayscaleImage
# Extend the borders for the skeleton:
extendedImg = cv2.copyMakeBorder(grayscaleImage, 5, 5, 5, 5, cv2.BORDER_CONSTANT)
# Store a deep copy of the crop for results:
grayscaleImageCopy = cv2.cvtColor(extendedImg, cv2.COLOR_GRAY2BGR)
# Compute the skeleton:
skeleton = cv2.ximgproc.thinning(extendedImg, None, 1)
The first step is to get the skeleton of the arrow. As I said, this step is needed prior to the convolution-based method that identifies the endpoints of a shape. Computing the skeleton normalizes the shape to a one pixel width. However, sometimes, if the shape is too close to the "canvas" borders, the skeleton could show some artifacts. This is avoided with a border extension. The skeleton of the arrow is this:
Check that image out. If we identify the endpoints, the tip will exhibit at least 3 points, while the tail at least 1. That's handy - the tip will always have more points than the tail. If only we could detect those points... Luckily, we can:
# Threshold the image so that white pixels get a value of 0 and
# black pixels a value of 10:
_, binaryImage = cv2.threshold(skeleton, 128, 10, cv2.THRESH_BINARY)
# Set the end-points kernel:
h = np.array([[1, 1, 1],
[1, 10, 1],
[1, 1, 1]])
# Convolve the image with the kernel:
imgFiltered = cv2.filter2D(binaryImage, -1, h)
# Extract only the end-points pixels, those with
# an intensity value of 110:
binaryImage = np.where(imgFiltered == 110, 255, 0)
# The above operation converted the image to 32-bit float,
# convert back to 8-bit uint
binaryImage = binaryImage.astype(np.uint8)
This endpoint detecting method convolves the skeleton with a special kernel that identifies endpoints. It returns a binary image where all the endpoints have the value 110. After thresholding this mid-result, we get this image, which represents the arrow endpoints:
Nice, as you see, we can group the points in two clusters and get their cluster centers. Sounds like a job for K-Means, because that's exactly what it does. We first need to treat our data, though, because K-Means operates on defined-shaped arrays of float data:
# Find the X, Y location of all the end-points
# pixels:
Y, X = binaryImage.nonzero()
# Reshape the arrays for K-means
Y = Y.reshape(-1,1)
X = X.reshape(-1,1)
Z = np.hstack((X, Y))
# K-means operates on 32-bit float data:
floatPoints = np.float32(Z)
# Set the convergence criteria and call K-means:
criteria = (cv2.TERM_CRITERIA_EPS + cv2.TERM_CRITERIA_MAX_ITER, 10, 1.0)
ret, label, center = cv2.kmeans(floatPoints, 2, None, criteria, 10, cv2.KMEANS_RANDOM_CENTERS)
# Set the cluster count, find the points belonging
# to cluster 0 and cluster 1:
cluster1Count = np.count_nonzero(label)
cluster0Count = np.shape(label)[0] - cluster1Count
print("Elements of Cluster 0: "+str(cluster0Count))
print("Elements of Cluster 1: " + str(cluster1Count))
The last two lines prints the endpoints that are assigned to Cluster 0 Cluster 1, respectively. That outputs this:
Elements of Cluster 0: 3
Elements of Cluster 1: 2
Just as expected - well, kinda. Seems that Cluster 0 is the tip and cluster 2 the tail! But the tail actually got 2 points. If you look the image of the skeleton closely, you'll see there's a small bifurcation at the tail. That's why we, in reality, got two points instead of just one. Alright, let's get the center points and draw them on the original input:
# Look for the cluster of max number of points
# That cluster will be the tip of the arrow:
maxCluster = 0
if cluster1Count > cluster0Count:
maxCluster = 1
# Check out the centers of each cluster:
matRows, matCols = center.shape
# Store the ordered end-points here:
orderedPoints = [None] * 2
# Let's identify and draw the two end-points
# of the arrow:
for b in range(matRows):
# Get cluster center:
pointX = int(center[b][0])
pointY = int(center[b][1])
# Get the "tip"
if b == maxCluster:
color = (0, 0, 255)
orderedPoints[0] = (pointX, pointY)
# Get the "tail"
else:
color = (255, 0, 0)
orderedPoints[1] = (pointX, pointY)
# Draw it:
cv2.circle(grayscaleImageCopy, (pointX, pointY), 3, color, -1)
cv2.imshow("End-Points", grayscaleImageCopy)
cv2.waitKey(0)
This is the resulting image:
The tip always gets drawn in red while the tail is drawn in blue. Very cool, let's store these points in the orderedPoints list and draw the final line in a new "canvas", with dimension same as the original image:
# Store the tip and tail points:
p0x = orderedPoints[1][0]
p0y = orderedPoints[1][1]
p1x = orderedPoints[0][0]
p1y = orderedPoints[0][1]
# Create a new "canvas" (image) using the input dimensions:
imageHeight, imageWidth = binaryImage.shape[:2]
newImage = np.zeros((imageHeight, imageWidth), np.uint8)
newImage = 255 - newImage
# Draw a line using the detected points:
(x1, y1) = orderedPoints[0]
(x2, y2) = orderedPoints[1]
lineColor = (0, 0, 0)
cv2.line(newImage , (x1, y1), (x2, y2), lineColor, thickness=2)
cv2.imshow("Detected Line", newImage)
cv2.waitKey(0)
The line overlaid on the original image and the new image containing only the line:

It sounds like you want to measure the angle of the line but because you are measuring a line you drew in the original image, you must now filter out the original image to get an accurate measure of the line...which you drew with coordinates you know the endpoints of?
I guess:
make a better filter?
draw the line in a blank image and detect angle there?
determine the angle from the known coordinates?
Since you were asking for just a line, I tried that...just made a blank image, drew your detected line on it and then used that downstream...
blankIm = np.ones((height, width, channels), dtype=np.uint8)
blankIm.fill(255)
line = cv2.line(blankIm,(cols-1,rightish),(0,leftish),(0,255,0),10)

Can I randomly add a small image (13x12) to an array of pixels (random pixels of the same size of my small image)

I have 20 small images (that I want to put in a target area of a background image (13x12). I have already marked my target area with a circle, I have the coordinates of the circle in two arrays of pixels. Now I want to know how I can randomly add my 20 small images in random area in my arrays of pixels which are basically the target area (the drawn circle).
In my code, I was trying for just one image, if it works, I'll pass the folder of my 20 small images.
# Depencies importation
import cv2
import numpy as np
# Saving directory
saving_dir = "../Saved_Images/"
# Read the background image
bgimg = cv2.imread("../Images/background.jpg")
# Resizing the bacground image
bgimg_resized = cv2.resize(bgimg, (2050,2050))
# Read the image that will be put in the background image (exemple of 1)
# I'm just trying with one, if it works, I'll pass the folder of the 20
small_img = cv2.imread("../Images/small.jpg")
# Convert the resized background image to gray
bgimg_gray = cv2.cvtColor(bgimg, cv2.COLOR_BGR2GRAY)
# Convert the grayscale image to a binary image
ret, thresh = cv2.threshold(bgimg_gray,127,255,0)
# Determine the moments of the binary image
M = cv2.moments(thresh)
# calculate x,y coordinate of center
cX = int(M["m10"] / M["m00"])
cY = int(M["m01"] / M["m00"])
# Drawing the circle in the background image
circle = cv2.circle(bgimg, (cX, cY), 930, (0,0,255), 9)
print(circle) # This returns None
# Getting the coordinates of the circle
combined = bgimg[:,:,0] + bgimg[:,:,1] + bgimg[:,:,2]
rows, cols = np.where(combined >= 0)
# I have those pixels in rows and cols, but I don't know
# How to randomly put my small image in those pixel
# Saving the new image
cv2.imwrite(saving_dir+"bgimg"+".jpg", bgimg)
cv2.namedWindow('image', cv2.WINDOW_NORMAL)
cv2.resizeWindow("Test", 1000, 1200)
# Showing the images
cv2.imshow("image", bgimg)
# Waiting for any key to stop the program execution
cv2.waitKey(0)
In the expected results, the small images must be placed in the background image randomly.

If you have the center and the radius of your circle, you can easily generate random coordinates by randomly choosing an angle theta from [0, 2*pi], calculating corresponding x and y values by cos(theta) and sin(theta) and scaling these by some random chosen scaling factors from [0, radius]. I prepared some code for you, see below.
I omitted a lot of code from yours (reading, preprocessing, saving) to focus on the relevant parts (see how to create a minimal, complete, and verifiable example). Hopefully, you can integrate the main idea of my solution into your code on your own. If not, I will provide further explanations.
import cv2
import numpy as np
# (Artificial) Background image (instead of reading an actual image...)
bgimg = 128 * np.ones((401, 401, 3), np.uint8)
# Circle parameters (obtained somehow...)
center = (200, 200)
radius = 100
# Draw circle in background image
cv2.circle(bgimg, center, radius, (0, 0, 255), 3)
# Shape of small image (known before-hand...?)
(w, h) = (13, 12)
for k in range(200):
# (Artificial) Small image (instead of reading an actual image...)
smallimg = np.uint8(np.add(128 * np.random.rand(w, h, 3), (127, 127, 127)))
# Select random angle theta from [0, 2*pi]
theta = 2 * np.pi * np.random.rand()
# Select random distance factors from center
factX = (radius - w/2) * np.random.rand()
factY = (radius - h/2) * np.random.rand()
# Calculate random coordinates for small image from angle and distance factors
(x, y) = np.uint16(np.add((np.cos(theta) * factX - w/2, np.sin(theta) * factY - h/2), center))
# Replace (rather than "add") determined area in background image with small image
bgimg[x:x+smallimg.shape[0], y:y+smallimg.shape[1]] = smallimg
cv2.imshow("bgimg", bgimg)
cv2.waitKey(0)
The exemplary output:
Caveat: I haven't paid attention, if the small images might violate the circle boundary. Therefore, some additional checks or limitations to the scaling factors must be added.
EDIT: I edited my above code. To take the below comment into account, I shift the small image by (width/2, height/2), and limit the radius scale factor accordingly, so that the circle boundary isn't violated, neither top/left nor bottom/right.
Before, it was possible, that the boundary is violated in the bottom/right part (n = 200):
After the edit, this should be prevented (n = 20000):
The touching of the red line in the image is due to the line's thickness. For "safety reasons", one could add another 1 pixel distance.

Image processing - eliminate arc-like smears

I am dealing with this kind of image
(upper is post-processed)
(lower is raw)
So, first I converted the grayscale image into pure black and white binary image. I am interested in detecting the white blobs, and want to get rid of the arc-like smears in the corners. How can I do that?
I general, I know that my targets are almost circular in shape, not too big, but I want to encode something that automatically gets rid of everything else, like the lighter arcs in the upper left and right corners.
How would I do this in python, ideally skimage?

You can just detect circle of the right size with skimage's methods hough_circle and hough_circle_peaks and cut it out.
Here I adapted my previous answer to your other question to do this:
# skimage version 0.14.0
import math
import numpy as np
import matplotlib.pyplot as plt
from skimage import color
from skimage.io import imread
from skimage.transform import hough_circle, hough_circle_peaks
from skimage.feature import canny
from skimage.draw import circle
from skimage.util import img_as_ubyte
INPUT_IMAGE = 'dish1.png' # input image name
BEST_COUNT = 1 # how many circles to detect (one dish)
MIN_RADIUS = 100 # min radius of the Petri dish
MAX_RADIUS = 122 # max radius of the Petri dish (in pixels)
LARGER_THRESH = 1.2 # circle is considered significantly larger than another one if its radius is at least so much bigger
OVERLAP_THRESH = 0.1 # circles are considered overlapping if this part of the smaller circle is overlapping
def circle_overlap_percent(centers_distance, radius1, radius2):
'''
Calculating the percentage area overlap between circles
See Gist for comments:
https://gist.github.com/amakukha/5019bfd4694304d85c617df0ca123854
'''
R, r = max(radius1, radius2), min(radius1, radius2)
if centers_distance >= R + r:
return 0.0
elif R >= centers_distance + r:
return 1.0
R2, r2 = R**2, r**2
x1 = (centers_distance**2 - R2 + r2 )/(2*centers_distance)
x2 = abs(centers_distance - x1)
y = math.sqrt(R2 - x1**2)
a1 = R2 * math.atan2(y, x1) - x1*y
if x1 <= centers_distance:
a2 = r2 * math.atan2(y, x2) - x2*y
else:
a2 = math.pi * r2 - a2
overlap_area = a1 + a2
return overlap_area / (math.pi * r2)
def circle_overlap(c1, c2):
d = math.sqrt((c1[0]-c2[0])**2 + (c1[1]-c2[1])**2)
return circle_overlap_percent(d, c1[2], c2[2])
def inner_circle(cs, c, thresh):
'''Is circle `c` is "inside" one of the `cs` circles?'''
for dc in cs:
# if new circle is larger than existing -> it's not inside
if c[2] > dc[2]*LARGER_THRESH: continue
# if new circle is smaller than existing one...
if circle_overlap(dc, c)>thresh:
# ...and there is a significant overlap -> it's inner circle
return True
return False
# Load picture and detect edges
image = imread(INPUT_IMAGE, 1)
image = img_as_ubyte(image)
edges = canny(image, sigma=3, low_threshold=10, high_threshold=50)
# Detect circles of specific radii
hough_radii = np.arange(MIN_RADIUS, MAX_RADIUS, 2)
hough_res = hough_circle(edges, hough_radii)
# Select the most prominent circles (in order from best to worst)
accums, cx, cy, radii = hough_circle_peaks(hough_res, hough_radii)
# Determine BEST_COUNT circles to be drawn
drawn_circles = []
for crcl in zip(cy, cx, radii):
# Do not draw circles if they are mostly inside better fitting ones
if not inner_circle(drawn_circles, crcl, OVERLAP_THRESH):
# A good circle found: exclude smaller circles it covers
i = 0
while i<len(drawn_circles):
if circle_overlap(crcl, drawn_circles[i]) > OVERLAP_THRESH:
t = drawn_circles.pop(i)
else:
i += 1
# Remember the new circle
drawn_circles.append(crcl)
# Stop after have found more circles than needed
if len(drawn_circles)>BEST_COUNT:
break
drawn_circles = drawn_circles[:BEST_COUNT]
# Draw circle and cut it out
colors = [(250, 0, 0), (0, 250, 0), (0, 0, 250)]
fig, ax = plt.subplots(ncols=1, nrows=3, figsize=(10, 4))
color_image = color.gray2rgb(image)
black_image = np.zeros_like(image)
for center_y, center_x, radius in drawn_circles[:1]:
circy, circx = circle(center_y, center_x, radius, image.shape)
color = colors.pop(0)
color_image[circy, circx] = color
black_image[circy, circx] = image[circy, circx]
colors.append(color)
# Output
ax[0].imshow(image, cmap=plt.cm.gray) # original image
ax[1].imshow(color_image) # detected circle
ax[2].imshow(black_image, cmap=plt.cm.gray) # cutout
plt.show()
Output:
Again, as in my previous answer, most of the code here is doing "hierarchy" computation to find the biggest best fitting circle.

10 Pin Bowling score capture

I want to use OCR to capture the bowling scores from the monitor at the lances. I had a look at this sudoku solver, as I think its pretty similar - numbers and grids right? It has trouble finding the horizontal lines. Has anyone got any tips for pre-processing this image to make it easier to detect the lines (or numbers!). Also any tips for how to deal with the split (the orange ellipse around some of the 8's int he image)?
So far I have got the outline of the score area and cropped it.
import matplotlib
matplotlib.use('TkAgg')
from skimage import io
import numpy as np
import matplotlib.pyplot as plt
from skimage import measure
from skimage.color import rgb2gray
# import pytesseract
from matplotlib.path import Path
from qhd import *
def polygonArea(poly):
"""
Return area of an unclosed polygon.
:see: https://stackoverflow.com/a/451482
:param poly: (n,2)-array
"""
# we need a plain list for the following operations
if isinstance(poly, np.ndarray):
poly = poly.tolist()
segments = zip(poly, poly[1:] + [poly[0]])
return 0.5 * abs(sum(x0*y1 - x1*y0
for ((x0, y0), (x1, y1)) in segments))
filename = 'good.jpg'
image = io.imread(filename)
image = rgb2gray(image)
# Find contours at a constant value of 0.8
contours = measure.find_contours(image, 0.4)
# Display the image and plot all contours found
fig, ax = plt.subplots()
c = 0
biggest = None
biggest_size = 0
for n, contour in enumerate(contours):
curr_size = polygonArea(contour)
if curr_size > biggest_size:
biggest = contour
biggest_size = curr_size
biggest = qhull2D(biggest)
# Approximate that so we just get a rectangle.
biggest = measure.approximate_polygon(biggest, 500)
# vertices of the cropping polygon
yc = biggest[:,0]
xc = biggest[:,1]
xycrop = np.vstack((xc, yc)).T
# xy coordinates for each pixel in the image
nr, nc = image.shape
ygrid, xgrid = np.mgrid[:nr, :nc]
xypix = np.vstack((xgrid.ravel(), ygrid.ravel())).T
# construct a Path from the vertices
pth = Path(xycrop, closed=False)
# test which pixels fall within the path
mask = pth.contains_points(xypix)
# reshape to the same size as the image
mask = mask.reshape(image.shape)
# create a masked array
masked = np.ma.masked_array(image, ~mask)
# if you want to get rid of the blank space above and below the cropped
# region, use the min and max x, y values of the cropping polygon:
xmin, xmax = int(xc.min()), int(np.ceil(xc.max()))
ymin, ymax = int(yc.min()), int(np.ceil(yc.max()))
trimmed = masked[ymin:ymax, xmin:xmax]
plt.imshow(trimmed, cmap=plt.cm.gray), plt.title('trimmed')
plt.show()
https://imgur.com/LijB85I is an example of how the score is displayed.