here's an issue: I want to find actual maximum width of boundary in segmented image with irregular shape (this)
below I post some example image I use for testing
So far I managed to obtain boundaries and skeleton line, but how do I measure distance between contours perpendicural to the skeleton line?
def get_skeleton(image_path):
im = cv2.imread(img_path , cv2.IMREAD_GRAYSCALE)
binary = im > filters.threshold_otsu(im)
skeleton = morphology.skeletonize(binary)
return skeleton
skeleton = get_skeleton(img_path)
plt.imshow(skeleton, cmap="gray")
def get_boundary(image_path):
reading_Img = cv2.imread(image_path, cv2.IMREAD_GRAYSCALE)
reading_Img = cv2.cvtColor(reading_Img,cv2.COLOR_BGR2RGB)
canny_Img = cv2.Canny(reading_Img,90,100)
contours,_ = cv2.findContours(canny_Img,cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)
canvas = np.zeros_like(reading_Img)
boundary = cv2.drawContours(canvas , contours, -1, (255, 0, 0), 1)
return boundary
boundary = get_boundary(img_path)
plt.imshow(boundary)
Sample input image
EDIT:
First of all thanks for your answer, I would like to add more detail on what I am trying to do.
So I made a segmentation model which detects cracks in concrete (they can be any shape, vertical, horizontal, diagonal, etc) and now I need to identify their max-width and draw a line that shows where it occurs.
I found that the medial axis returns the distance from the boundary and by filtering max value I was able to get the width (see colab below) and its coordinate on the medial axis. Now I need to draw a line connecting the width between boundaries, but I have no idea on how to find the coordinates of such a line.
I thought of an algorithm which starts at the point of max distance occurrence on medial axis and expands until it finds a boundary, but I don't know how to implement it.
This image shows what I need to have:
After I find x and y of points I will be able to calculate euclidean distance between 2 points
dist=sqrt((y2-y1)^2+(x2-x1)^2)
Please look at my code in colab notebook: https://colab.research.google.com/drive/1NvEyfrxpKGJ1kxjP48PGNB_UUSp6f6Ze?usp=sharing
sample input images:
https://imgur.com/ewTmH8M
https://imgur.com/JRAQCke
https://imgur.com/7QQFfAv
Starting with your approach using the medial axis function you can
interpolate the direction of the axis at the point that was found
derive the orthogonal from the direction
look where the orthogonal reaches the boundary.
The example below shows the principle and works with your example images. But I'm sure there will be some boundary conditions that are not yet considered. I leave it to you to get it robust against real live data.
import cv2
import numpy as np
from skimage.morphology import medial_axis
from skimage import img_as_ubyte
delta = 3 # delta index for interpolation
# get crack
im = cv2.imread("img.png", cv2.IMREAD_GRAYSCALE)
rgb = cv2.cvtColor(im, cv2.COLOR_GRAY2RGB) # rgb just for demo purpose
_, crack = cv2.threshold(im, 127, 255, cv2.THRESH_BINARY)
# get medial axis
medial, distance = medial_axis(im, return_distance=True)
med_img = img_as_ubyte(medial)
med_contours, _ = cv2.findContours(med_img, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_NONE)
cv2.drawContours(rgb, med_contours, -1, (255, 0, 0), 1)
med_pts = [v[0] for v in med_contours[0]]
# get point with maximal distance from medial axis
max_idx = np.argmax(distance)
max_pos = np.unravel_index(max_idx, distance.shape)
max_dist = distance[max_pos]
coords = np.array([max_pos[1], max_pos[0]])
print(f"max distance from medial axis to boundary = {max_dist} at {coords}")
# interpolate orthogonal of medial axis at coords
idx = next(i for i, v in enumerate(med_pts) if (v == coords).all())
px1, py1 = med_pts[(idx-delta) % len(med_pts)]
px2, py2 = med_pts[(idx+delta) % len(med_pts)]
orth = np.array([py1 - py2, px2 - px1]) * max(im.shape)
# intersect orthogonal with crack and get contour
orth_img = np.zeros(crack.shape, dtype=np.uint8)
cv2.line(orth_img, coords + orth, coords - orth, color=255, thickness=1)
gap_img = cv2.bitwise_and(orth_img, crack)
gap_contours, _ = cv2.findContours(gap_img, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
gap_pts = [v[0] for v in gap_contours[0]]
# determine the end points of the gap contour by negative dot product
n = len(gap_pts)
gap_ends = [
p for i, p in enumerate(gap_pts)
if np.dot(p - gap_pts[(i-1) % n], gap_pts[(i+1) % n] - p) < 0
]
print(f"Maximum gap found from {gap_ends[0]} to {gap_ends[1]}")
cv2.line(rgb, gap_ends[0], gap_ends[1], color=(0, 0, 255), thickness=1)
cv2.imwrite("test_out.png", rgb)
First thing I did was keep your images greyscale, there is no need to covert to 3 channels to find contours. Second was to convert the boundary image to a binary so that it is the same as the skeleton image. Then I simply added the two to get the both image.
I then clocked through each row (as you are looking for perpendicular distances) of the combined both image & looked for elements that where True i.e that are either boundary or skeleton pixels. I made a simplifying assumption at this point - I only searched for cases where there is a boundary followed by a single skeleton pixel then by a second boundary, I appreciate that this may not always be the case but I leave that particular headache for you to sort out.
After that its just a case of keeping track of he max & min distances recorded as you go through the image row by row. (edit: there maybe a cleaner way to do this than the way I've done it but hopefully you get the idea)
import numpy as np
import matplotlib.pyplot as plt
import cv2
from skimage import filters
from skimage import morphology
def get_skeleton(image_path):
im = cv2.imread(image_path , cv2.IMREAD_GRAYSCALE)
binary = im > filters.threshold_otsu(im)
skeleton = morphology.skeletonize(binary)
return skeleton
def get_boundary(image_path):
reading_Img = cv2.imread(image_path, cv2.IMREAD_GRAYSCALE)
canny_Img = cv2.Canny(reading_Img, 90, 100)
contours,_ = cv2.findContours(canny_Img,cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
canvas = np.zeros_like(reading_Img)
boundary = cv2.drawContours(canvas, contours, -1, (255, 0, 0), 1)
binary = boundary > filters.threshold_otsu(boundary)
return binary
skeleton = get_skeleton("LtqlM.png")
boundary = get_boundary("LtqlM.png")
both = skeleton + boundary
max_dist = 0
min_dist = 100000
for idx in range(both.shape[0]): # counting through rows
row = both[idx, :]
lines = np.where(row==True)[0]
if len(lines) == 3:
dist_1 = lines[1] - lines[0]
dist_2 = lines[2] - lines[1]
if (dist_1 > dist_2) and dist_1 > max_dist:
max_dist = dist_1
if (dist_2 > dist_1) and dist_2 > max_dist:
max_dist = dist_2
if (dist_1 < dist_2) and dist_1 < min_dist:
min_dist = dist_1
if (dist_2 < dist_1) and dist_2 < min_dist:
min_dist = dist_2
print("Maximum distance = ", max_dist)
print("Minimum distance = ", min_dist)
plt.imshow(both)
From the pixel with the largest distance, you can explore concentric square layers, until you find a background pixel. Then find the background pixel with the shortest Euclidean distance on the last layer. The second endpoint is symmetrical.
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
Edit: Quick Summary so far: I use the watershed algorithm but I have probably a problem with threshold. It didn't detect the brighter circles.
New: Fast radial symmetry transform approach which didn't quite work eiter (Edit 6).
I want to detect circles with different sizes. The use case is to detect coins on an image and to extract them solely. -> Get the single coins as single image files.
For this I used the Hough Circle Transform of open-cv:
(https://docs.opencv.org/2.4/doc/tutorials/imgproc/imgtrans/hough_circle/hough_circle.html)
import sys
import cv2 as cv
import numpy as np
def main(argv):
## [load]
default_file = "data/newcommon_1euro.jpg"
filename = argv[0] if len(argv) > 0 else default_file
# Loads an image
src = cv.imread(filename, cv.IMREAD_COLOR)
# Check if image is loaded fine
if src is None:
print ('Error opening image!')
print ('Usage: hough_circle.py [image_name -- default ' + default_file + '] \n')
return -1
## [load]
## [convert_to_gray]
# Convert it to gray
gray = cv.cvtColor(src, cv.COLOR_BGR2GRAY)
## [convert_to_gray]
## [reduce_noise]
# Reduce the noise to avoid false circle detection
gray = cv.medianBlur(gray, 5)
## [reduce_noise]
## [houghcircles]
rows = gray.shape[0]
circles = cv.HoughCircles(gray, cv.HOUGH_GRADIENT, 1, rows / 8,
param1=100, param2=30,
minRadius=0, maxRadius=120)
## [houghcircles]
## [draw]
if circles is not None:
circles = np.uint16(np.around(circles))
for i in circles[0, :]:
center = (i[0], i[1])
# circle center
cv.circle(src, center, 1, (0, 100, 100), 3)
# circle outline
radius = i[2]
cv.circle(src, center, radius, (255, 0, 255), 3)
## [draw]
## [display]
cv.imshow("detected circles", src)
cv.waitKey(0)
## [display]
return 0
if __name__ == "__main__":
main(sys.argv[1:])
I tried all parameters (rows, param1, param2, minRadius, and maxRadius) to optimize the results. This worked very well for one specific image but other images with different sized coins didn't work.
Examples:
Parameters
circles = cv.HoughCircles(gray, cv.HOUGH_GRADIENT, 1, rows / 16,
param1=100, param2=30,
minRadius=0, maxRadius=120)
With the same parameters:
Changed to rows/8
I also tried two other approaches of this thread: writing robust (color and size invariant) circle detection with opencv (based on Hough transform or other features)
The approach of fireant leads to this result:
The approach of fraxel didn't work either.
For the first approach: This happens with all different sizes and also the min and max radius.
How could I change the code, so that the coin size is not important or that it finds the parameters itself?
Thank you in advance for any help!
Edit:
I tried the watershed algorithm of Open-cv, as suggested by Alexander Reynolds: https://docs.opencv.org/3.4/d3/db4/tutorial_py_watershed.html
import numpy as np
import cv2 as cv
from matplotlib import pyplot as plt
img = cv.imread('data/P1190263.jpg')
gray = cv.cvtColor(img,cv.COLOR_BGR2GRAY)
ret, thresh = cv.threshold(gray,0,255,cv.THRESH_BINARY_INV+cv.THRESH_OTSU)
# noise removal
kernel = np.ones((3,3),np.uint8)
opening = cv.morphologyEx(thresh,cv.MORPH_OPEN,kernel, iterations = 2)
# sure background area
sure_bg = cv.dilate(opening,kernel,iterations=3)
# Finding sure foreground area
dist_transform = cv.distanceTransform(opening,cv.DIST_L2,5)
ret, sure_fg = cv.threshold(dist_transform,0.7*dist_transform.max(),255,0)
# Finding unknown region
sure_fg = np.uint8(sure_fg)
unknown = cv.subtract(sure_bg,sure_fg)
# Marker labelling
ret, markers = cv.connectedComponents(sure_fg)
# Add one to all labels so that sure background is not 0, but 1
markers = markers+1
# Now, mark the region of unknown with zero
markers[unknown==255] = 0
markers = cv.watershed(img,markers)
img[markers == -1] = [255,0,0]
#Display:
cv.imshow("detected circles", img)
cv.waitKey(0)
It works very well on the test image of the open-cv website:
But it performs very bad on my own images:
I can't really think of a good reason why it's not working on my images?
Edit 2:
As suggested I looked at the intermediate images. The thresh looks not good in my opinion. Next, there is no difference between opening and dist_transform. The corresponding sure_fg shows the detected images.
thresh:
opening:
dist_transform:
sure_bg:
sure_fg:
Edit 3:
I tried all distanceTypes and maskSizes I could find, but the results were quite the same (https://www.tutorialspoint.com/opencv/opencv_distance_transformation.htm)
Edit 4:
Furthermore, I tried to change the (first) threshold function. I used different threshold values instead of the OTSU Function. The best one was with 160, but it was far from good:
In the tutorial it looks like this:
It seems like the coins are somehow too bright to be detected by this algorithm, but I don't know how to improve it?
Edit 5:
Changing the overall contrast and brightness of the image (with cv.convertScaleAbs) didn't improve the results. Increasing the contrast however should increase the "difference" between foreground and background, at least on the normal image. But it even got worse. The corresponding threshold image didn't improved (didn't get more white pixel).
Edit 6: I tried another approach, the fast radial symmetry transform (from here https://github.com/ceilab/frst_python)
import cv2
import numpy as np
def gradx(img):
img = img.astype('int')
rows, cols = img.shape
# Use hstack to add back in the columns that were dropped as zeros
return np.hstack((np.zeros((rows, 1)), (img[:, 2:] - img[:, :-2]) / 2.0, np.zeros((rows, 1))))
def grady(img):
img = img.astype('int')
rows, cols = img.shape
# Use vstack to add back the rows that were dropped as zeros
return np.vstack((np.zeros((1, cols)), (img[2:, :] - img[:-2, :]) / 2.0, np.zeros((1, cols))))
# Performs fast radial symmetry transform
# img: input image, grayscale
# radii: integer value for radius size in pixels (n in the original paper); also used to size gaussian kernel
# alpha: Strictness of symmetry transform (higher=more strict; 2 is good place to start)
# beta: gradient threshold parameter, float in [0,1]
# stdFactor: Standard deviation factor for gaussian kernel
# mode: BRIGHT, DARK, or BOTH
def frst(img, radii, alpha, beta, stdFactor, mode='BOTH'):
mode = mode.upper()
assert mode in ['BRIGHT', 'DARK', 'BOTH']
dark = (mode == 'DARK' or mode == 'BOTH')
bright = (mode == 'BRIGHT' or mode == 'BOTH')
workingDims = tuple((e + 2 * radii) for e in img.shape)
# Set up output and M and O working matrices
output = np.zeros(img.shape, np.uint8)
O_n = np.zeros(workingDims, np.int16)
M_n = np.zeros(workingDims, np.int16)
# Calculate gradients
gx = gradx(img)
gy = grady(img)
# Find gradient vector magnitude
gnorms = np.sqrt(np.add(np.multiply(gx, gx), np.multiply(gy, gy)))
# Use beta to set threshold - speeds up transform significantly
gthresh = np.amax(gnorms) * beta
# Find x/y distance to affected pixels
gpx = np.multiply(np.divide(gx, gnorms, out=np.zeros(gx.shape), where=gnorms != 0),
radii).round().astype(int);
gpy = np.multiply(np.divide(gy, gnorms, out=np.zeros(gy.shape), where=gnorms != 0),
radii).round().astype(int);
# Iterate over all pixels (w/ gradient above threshold)
for coords, gnorm in np.ndenumerate(gnorms):
if gnorm > gthresh:
i, j = coords
# Positively affected pixel
if bright:
ppve = (i + gpx[i, j], j + gpy[i, j])
O_n[ppve] += 1
M_n[ppve] += gnorm
# Negatively affected pixel
if dark:
pnve = (i - gpx[i, j], j - gpy[i, j])
O_n[pnve] -= 1
M_n[pnve] -= gnorm
# Abs and normalize O matrix
O_n = np.abs(O_n)
O_n = O_n / float(np.amax(O_n))
# Normalize M matrix
M_max = float(np.amax(np.abs(M_n)))
M_n = M_n / M_max
# Elementwise multiplication
F_n = np.multiply(np.power(O_n, alpha), M_n)
# Gaussian blur
kSize = int(np.ceil(radii / 2))
kSize = kSize + 1 if kSize % 2 == 0 else kSize
S = cv2.GaussianBlur(F_n, (kSize, kSize), int(radii * stdFactor))
return S
img = cv2.imread('data/P1190263.jpg')
gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
result = frst(gray, 60, 2, 0, 1, mode='BOTH')
cv2.imshow("detected circles", result)
cv2.waitKey(0)
I only get this nearly black output (it has some very dark grey shadows). I don't know what to change and would be thankful for help!
I was doing a fun project: Solving a Sudoku from an input image using OpenCV (as in Google goggles etc). And I have completed the task, but at the end I found a little problem for which I came here.
I did the programming using Python API of OpenCV 2.3.1.
Below is what I did :
Read the image
Find the contours
Select the one with maximum area, ( and also somewhat equivalent to square).
Find the corner points.
e.g. given below:
(Notice here that the green line correctly coincides with the true boundary of the Sudoku, so the Sudoku can be correctly warped. Check next image)
warp the image to a perfect square
eg image:
Perform OCR ( for which I used the method I have given in Simple Digit Recognition OCR in OpenCV-Python )
And the method worked well.
Problem:
Check out this image.
Performing the step 4 on this image gives the result below:
The red line drawn is the original contour which is the true outline of sudoku boundary.
The green line drawn is approximated contour which will be the outline of warped image.
Which of course, there is difference between green line and red line at the top edge of sudoku. So while warping, I am not getting the original boundary of the Sudoku.
My Question :
How can I warp the image on the correct boundary of the Sudoku, i.e. the red line OR how can I remove the difference between red line and green line? Is there any method for this in OpenCV?
I have a solution that works, but you'll have to translate it to OpenCV yourself. It's written in Mathematica.
The first step is to adjust the brightness in the image, by dividing each pixel with the result of a closing operation:
src = ColorConvert[Import["http://davemark.com/images/sudoku.jpg"], "Grayscale"];
white = Closing[src, DiskMatrix[5]];
srcAdjusted = Image[ImageData[src]/ImageData[white]]
The next step is to find the sudoku area, so I can ignore (mask out) the background. For that, I use connected component analysis, and select the component that's got the largest convex area:
components =
ComponentMeasurements[
ColorNegate#Binarize[srcAdjusted], {"ConvexArea", "Mask"}][[All,
2]];
largestComponent = Image[SortBy[components, First][[-1, 2]]]
By filling this image, I get a mask for the sudoku grid:
mask = FillingTransform[largestComponent]
Now, I can use a 2nd order derivative filter to find the vertical and horizontal lines in two separate images:
lY = ImageMultiply[MorphologicalBinarize[GaussianFilter[srcAdjusted, 3, {2, 0}], {0.02, 0.05}], mask];
lX = ImageMultiply[MorphologicalBinarize[GaussianFilter[srcAdjusted, 3, {0, 2}], {0.02, 0.05}], mask];
I use connected component analysis again to extract the grid lines from these images. The grid lines are much longer than the digits, so I can use caliper length to select only the grid lines-connected components. Sorting them by position, I get 2x10 mask images for each of the vertical/horizontal grid lines in the image:
verticalGridLineMasks =
SortBy[ComponentMeasurements[
lX, {"CaliperLength", "Centroid", "Mask"}, # > 100 &][[All,
2]], #[[2, 1]] &][[All, 3]];
horizontalGridLineMasks =
SortBy[ComponentMeasurements[
lY, {"CaliperLength", "Centroid", "Mask"}, # > 100 &][[All,
2]], #[[2, 2]] &][[All, 3]];
Next I take each pair of vertical/horizontal grid lines, dilate them, calculate the pixel-by-pixel intersection, and calculate the center of the result. These points are the grid line intersections:
centerOfGravity[l_] :=
ComponentMeasurements[Image[l], "Centroid"][[1, 2]]
gridCenters =
Table[centerOfGravity[
ImageData[Dilation[Image[h], DiskMatrix[2]]]*
ImageData[Dilation[Image[v], DiskMatrix[2]]]], {h,
horizontalGridLineMasks}, {v, verticalGridLineMasks}];
The last step is to define two interpolation functions for X/Y mapping through these points, and transform the image using these functions:
fnX = ListInterpolation[gridCenters[[All, All, 1]]];
fnY = ListInterpolation[gridCenters[[All, All, 2]]];
transformed =
ImageTransformation[
srcAdjusted, {fnX ## Reverse[#], fnY ## Reverse[#]} &, {9*50, 9*50},
PlotRange -> {{1, 10}, {1, 10}}, DataRange -> Full]
All of the operations are basic image processing function, so this should be possible in OpenCV, too. The spline-based image transformation might be harder, but I don't think you really need it. Probably using the perspective transformation you use now on each individual cell will give good enough results.
Nikie's answer solved my problem, but his answer was in Mathematica. So I thought I should give its OpenCV adaptation here. But after implementing I could see that OpenCV code is much bigger than nikie's mathematica code. And also, I couldn't find interpolation method done by nikie in OpenCV ( although it can be done using scipy, i will tell it when time comes.)
1. Image PreProcessing ( closing operation )
import cv2
import numpy as np
img = cv2.imread('dave.jpg')
img = cv2.GaussianBlur(img,(5,5),0)
gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
mask = np.zeros((gray.shape),np.uint8)
kernel1 = cv2.getStructuringElement(cv2.MORPH_ELLIPSE,(11,11))
close = cv2.morphologyEx(gray,cv2.MORPH_CLOSE,kernel1)
div = np.float32(gray)/(close)
res = np.uint8(cv2.normalize(div,div,0,255,cv2.NORM_MINMAX))
res2 = cv2.cvtColor(res,cv2.COLOR_GRAY2BGR)
Result :
2. Finding Sudoku Square and Creating Mask Image
thresh = cv2.adaptiveThreshold(res,255,0,1,19,2)
contour,hier = cv2.findContours(thresh,cv2.RETR_TREE,cv2.CHAIN_APPROX_SIMPLE)
max_area = 0
best_cnt = None
for cnt in contour:
area = cv2.contourArea(cnt)
if area > 1000:
if area > max_area:
max_area = area
best_cnt = cnt
cv2.drawContours(mask,[best_cnt],0,255,-1)
cv2.drawContours(mask,[best_cnt],0,0,2)
res = cv2.bitwise_and(res,mask)
Result :
3. Finding Vertical lines
kernelx = cv2.getStructuringElement(cv2.MORPH_RECT,(2,10))
dx = cv2.Sobel(res,cv2.CV_16S,1,0)
dx = cv2.convertScaleAbs(dx)
cv2.normalize(dx,dx,0,255,cv2.NORM_MINMAX)
ret,close = cv2.threshold(dx,0,255,cv2.THRESH_BINARY+cv2.THRESH_OTSU)
close = cv2.morphologyEx(close,cv2.MORPH_DILATE,kernelx,iterations = 1)
contour, hier = cv2.findContours(close,cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)
for cnt in contour:
x,y,w,h = cv2.boundingRect(cnt)
if h/w > 5:
cv2.drawContours(close,[cnt],0,255,-1)
else:
cv2.drawContours(close,[cnt],0,0,-1)
close = cv2.morphologyEx(close,cv2.MORPH_CLOSE,None,iterations = 2)
closex = close.copy()
Result :
4. Finding Horizontal Lines
kernely = cv2.getStructuringElement(cv2.MORPH_RECT,(10,2))
dy = cv2.Sobel(res,cv2.CV_16S,0,2)
dy = cv2.convertScaleAbs(dy)
cv2.normalize(dy,dy,0,255,cv2.NORM_MINMAX)
ret,close = cv2.threshold(dy,0,255,cv2.THRESH_BINARY+cv2.THRESH_OTSU)
close = cv2.morphologyEx(close,cv2.MORPH_DILATE,kernely)
contour, hier = cv2.findContours(close,cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)
for cnt in contour:
x,y,w,h = cv2.boundingRect(cnt)
if w/h > 5:
cv2.drawContours(close,[cnt],0,255,-1)
else:
cv2.drawContours(close,[cnt],0,0,-1)
close = cv2.morphologyEx(close,cv2.MORPH_DILATE,None,iterations = 2)
closey = close.copy()
Result :
Of course, this one is not so good.
5. Finding Grid Points
res = cv2.bitwise_and(closex,closey)
Result :
6. Correcting the defects
Here, nikie does some kind of interpolation, about which I don't have much knowledge. And i couldn't find any corresponding function for this OpenCV. (may be it is there, i don't know).
Check out this SOF which explains how to do this using SciPy, which I don't want to use : Image transformation in OpenCV
So, here I took 4 corners of each sub-square and applied warp Perspective to each.
For that, first we find the centroids.
contour, hier = cv2.findContours(res,cv2.RETR_LIST,cv2.CHAIN_APPROX_SIMPLE)
centroids = []
for cnt in contour:
mom = cv2.moments(cnt)
(x,y) = int(mom['m10']/mom['m00']), int(mom['m01']/mom['m00'])
cv2.circle(img,(x,y),4,(0,255,0),-1)
centroids.append((x,y))
But resulting centroids won't be sorted. Check out below image to see their order:
So we sort them from left to right, top to bottom.
centroids = np.array(centroids,dtype = np.float32)
c = centroids.reshape((100,2))
c2 = c[np.argsort(c[:,1])]
b = np.vstack([c2[i*10:(i+1)*10][np.argsort(c2[i*10:(i+1)*10,0])] for i in xrange(10)])
bm = b.reshape((10,10,2))
Now see below their order :
Finally we apply the transformation and create a new image of size 450x450.
output = np.zeros((450,450,3),np.uint8)
for i,j in enumerate(b):
ri = i/10
ci = i%10
if ci != 9 and ri!=9:
src = bm[ri:ri+2, ci:ci+2 , :].reshape((4,2))
dst = np.array( [ [ci*50,ri*50],[(ci+1)*50-1,ri*50],[ci*50,(ri+1)*50-1],[(ci+1)*50-1,(ri+1)*50-1] ], np.float32)
retval = cv2.getPerspectiveTransform(src,dst)
warp = cv2.warpPerspective(res2,retval,(450,450))
output[ri*50:(ri+1)*50-1 , ci*50:(ci+1)*50-1] = warp[ri*50:(ri+1)*50-1 , ci*50:(ci+1)*50-1].copy()
Result :
The result is almost same as nikie's, but code length is large. May be, better methods are available out there, but until then, this works OK.
Regards
ARK.
You could try to use some kind of grid based modeling of you arbitrary warping. And since the sudoku already is a grid, that shouldn't be too hard.
So you could try to detect the boundaries of each 3x3 subregion and then warp each region individually. If the detection succeeds it would give you a better approximation.
I thought this was a great post, and a great solution by ARK; very well laid out and explained.
I was working on a similar problem, and built the entire thing. There were some changes (i.e. xrange to range, arguments in cv2.findContours), but this should work out of the box (Python 3.5, Anaconda).
This is a compilation of the elements above, with some of the missing code added (i.e., labeling of points).
'''
https://stackoverflow.com/questions/10196198/how-to-remove-convexity-defects-in-a-sudoku-square
'''
import cv2
import numpy as np
img = cv2.imread('test.png')
winname="raw image"
cv2.namedWindow(winname)
cv2.imshow(winname, img)
cv2.moveWindow(winname, 100,100)
img = cv2.GaussianBlur(img,(5,5),0)
winname="blurred"
cv2.namedWindow(winname)
cv2.imshow(winname, img)
cv2.moveWindow(winname, 100,150)
gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
mask = np.zeros((gray.shape),np.uint8)
kernel1 = cv2.getStructuringElement(cv2.MORPH_ELLIPSE,(11,11))
winname="gray"
cv2.namedWindow(winname)
cv2.imshow(winname, gray)
cv2.moveWindow(winname, 100,200)
close = cv2.morphologyEx(gray,cv2.MORPH_CLOSE,kernel1)
div = np.float32(gray)/(close)
res = np.uint8(cv2.normalize(div,div,0,255,cv2.NORM_MINMAX))
res2 = cv2.cvtColor(res,cv2.COLOR_GRAY2BGR)
winname="res2"
cv2.namedWindow(winname)
cv2.imshow(winname, res2)
cv2.moveWindow(winname, 100,250)
#find elements
thresh = cv2.adaptiveThreshold(res,255,0,1,19,2)
img_c, contour,hier = cv2.findContours(thresh,cv2.RETR_TREE,cv2.CHAIN_APPROX_SIMPLE)
max_area = 0
best_cnt = None
for cnt in contour:
area = cv2.contourArea(cnt)
if area > 1000:
if area > max_area:
max_area = area
best_cnt = cnt
cv2.drawContours(mask,[best_cnt],0,255,-1)
cv2.drawContours(mask,[best_cnt],0,0,2)
res = cv2.bitwise_and(res,mask)
winname="puzzle only"
cv2.namedWindow(winname)
cv2.imshow(winname, res)
cv2.moveWindow(winname, 100,300)
# vertical lines
kernelx = cv2.getStructuringElement(cv2.MORPH_RECT,(2,10))
dx = cv2.Sobel(res,cv2.CV_16S,1,0)
dx = cv2.convertScaleAbs(dx)
cv2.normalize(dx,dx,0,255,cv2.NORM_MINMAX)
ret,close = cv2.threshold(dx,0,255,cv2.THRESH_BINARY+cv2.THRESH_OTSU)
close = cv2.morphologyEx(close,cv2.MORPH_DILATE,kernelx,iterations = 1)
img_d, contour, hier = cv2.findContours(close,cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)
for cnt in contour:
x,y,w,h = cv2.boundingRect(cnt)
if h/w > 5:
cv2.drawContours(close,[cnt],0,255,-1)
else:
cv2.drawContours(close,[cnt],0,0,-1)
close = cv2.morphologyEx(close,cv2.MORPH_CLOSE,None,iterations = 2)
closex = close.copy()
winname="vertical lines"
cv2.namedWindow(winname)
cv2.imshow(winname, img_d)
cv2.moveWindow(winname, 100,350)
# find horizontal lines
kernely = cv2.getStructuringElement(cv2.MORPH_RECT,(10,2))
dy = cv2.Sobel(res,cv2.CV_16S,0,2)
dy = cv2.convertScaleAbs(dy)
cv2.normalize(dy,dy,0,255,cv2.NORM_MINMAX)
ret,close = cv2.threshold(dy,0,255,cv2.THRESH_BINARY+cv2.THRESH_OTSU)
close = cv2.morphologyEx(close,cv2.MORPH_DILATE,kernely)
img_e, contour, hier = cv2.findContours(close,cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)
for cnt in contour:
x,y,w,h = cv2.boundingRect(cnt)
if w/h > 5:
cv2.drawContours(close,[cnt],0,255,-1)
else:
cv2.drawContours(close,[cnt],0,0,-1)
close = cv2.morphologyEx(close,cv2.MORPH_DILATE,None,iterations = 2)
closey = close.copy()
winname="horizontal lines"
cv2.namedWindow(winname)
cv2.imshow(winname, img_e)
cv2.moveWindow(winname, 100,400)
# intersection of these two gives dots
res = cv2.bitwise_and(closex,closey)
winname="intersections"
cv2.namedWindow(winname)
cv2.imshow(winname, res)
cv2.moveWindow(winname, 100,450)
# text blue
textcolor=(0,255,0)
# points green
pointcolor=(255,0,0)
# find centroids and sort
img_f, contour, hier = cv2.findContours(res,cv2.RETR_LIST,cv2.CHAIN_APPROX_SIMPLE)
centroids = []
for cnt in contour:
mom = cv2.moments(cnt)
(x,y) = int(mom['m10']/mom['m00']), int(mom['m01']/mom['m00'])
cv2.circle(img,(x,y),4,(0,255,0),-1)
centroids.append((x,y))
# sorting
centroids = np.array(centroids,dtype = np.float32)
c = centroids.reshape((100,2))
c2 = c[np.argsort(c[:,1])]
b = np.vstack([c2[i*10:(i+1)*10][np.argsort(c2[i*10:(i+1)*10,0])] for i in range(10)])
bm = b.reshape((10,10,2))
# make copy
labeled_in_order=res2.copy()
for index, pt in enumerate(b):
cv2.putText(labeled_in_order,str(index),tuple(pt),cv2.FONT_HERSHEY_DUPLEX, 0.75, textcolor)
cv2.circle(labeled_in_order, tuple(pt), 5, pointcolor)
winname="labeled in order"
cv2.namedWindow(winname)
cv2.imshow(winname, labeled_in_order)
cv2.moveWindow(winname, 100,500)
# create final
output = np.zeros((450,450,3),np.uint8)
for i,j in enumerate(b):
ri = int(i/10) # row index
ci = i%10 # column index
if ci != 9 and ri!=9:
src = bm[ri:ri+2, ci:ci+2 , :].reshape((4,2))
dst = np.array( [ [ci*50,ri*50],[(ci+1)*50-1,ri*50],[ci*50,(ri+1)*50-1],[(ci+1)*50-1,(ri+1)*50-1] ], np.float32)
retval = cv2.getPerspectiveTransform(src,dst)
warp = cv2.warpPerspective(res2,retval,(450,450))
output[ri*50:(ri+1)*50-1 , ci*50:(ci+1)*50-1] = warp[ri*50:(ri+1)*50-1 , ci*50:(ci+1)*50-1].copy()
winname="final"
cv2.namedWindow(winname)
cv2.imshow(winname, output)
cv2.moveWindow(winname, 600,100)
cv2.waitKey(0)
cv2.destroyAllWindows()
I want to add that above method works only when sudoku board stands straight, otherwise height/width (or vice versa) ratio test will most probably fail and you will not be able to detect edges of sudoku. (I also want to add that if lines that are not perpendicular to the image borders, sobel operations (dx and dy) will still work as lines will still have edges with respect to both axes.)
To be able to detect straight lines you should work on contour or pixel-wise analysis such as contourArea/boundingRectArea, top left and bottom right points...
Edit: I managed to check whether a set of contours form a line or not by applying linear regression and checking the error. However linear regression performed poorly when slope of the line is too big (i.e. >1000) or it is very close to 0. Therefore applying the ratio test above (in most upvoted answer) before linear regression is logical and did work for me.
To remove undected corners I applied gamma correction with a gamma value of 0.8.
The red circle is drawn to show the missing corner.
The code is:
gamma = 0.8
invGamma = 1/gamma
table = np.array([((i / 255.0) ** invGamma) * 255
for i in np.arange(0, 256)]).astype("uint8")
cv2.LUT(img, table, img)
This is in addition to Abid Rahman's answer if some corner points are missing.
I'm trying to split an image into several sub-images with opencv by identifying templates of the original image and then copy the regions where I matched those templates. I'm a TOTAL newbie to opencv! I've identified the sub-images using:
result = cv2.matchTemplate(img, template, cv2.TM_CCORR_NORMED)
After some cleanup I get a list of tuples called points in which I iterate to show the rectangles. tw and th is the template width and height respectively.
for pt in points:
re = cv2.rectangle(img, pt, (pt[0] + tw, pt[1] + th), 0, 2)
print('%s, %s' % (str(pt[0]), str(pt[1])))
count+=1
What I would like to accomplish is to save the octagons (https://dl.dropbox.com/u/239592/region01.png) into separated files.
How can I do this? I've read something about contours but I'm not sure how to use it. Ideally I would like to contour the octagon.
Thanks a lot for your help!
If template matching is working for you, stick to it. For instance, I considered the following template:
Then, we can pre-process the input in order to make it a binary one and discard small components. After this step, the template matching is performed. Then it is a matter of filtering the matches by means of discarding close ones (I've used a dummy method for that, so if there are too many matches you could see it taking some time). After we decide which points are far apart (and thus identify different hexagons), we can do minor adjusts to them in the following manner:
Sort by y-coordinate;
If two adjacent items start at a y-coordinate that is too close, then set them both to the same y-coord.
Now you can sort this point list in an appropriate order such that the crops are done in raster order. The cropping part is easily achieved using slicing provided by numpy.
import sys
import cv2
import numpy
outbasename = 'hexagon_%02d.png'
img = cv2.imread(sys.argv[1])
template = cv2.cvtColor(cv2.imread(sys.argv[2]), cv2.COLOR_BGR2GRAY)
theight, twidth = template.shape[:2]
# Binarize the input based on the saturation and value.
hsv = cv2.cvtColor(img, cv2.COLOR_BGR2HSV)
saturation = hsv[:,:,1]
value = hsv[:,:,2]
value[saturation > 35] = 255
value = cv2.threshold(value, 0, 255, cv2.THRESH_OTSU)[1]
# Pad the image.
value = cv2.copyMakeBorder(255 - value, 3, 3, 3, 3, cv2.BORDER_CONSTANT, value=0)
# Discard small components.
img_clean = numpy.zeros(value.shape, dtype=numpy.uint8)
contours, _ = cv2.findContours(value, cv2.RETR_LIST, cv2.CHAIN_APPROX_SIMPLE)
for i, c in enumerate(contours):
area = cv2.contourArea(c)
if area > 500:
cv2.drawContours(img_clean, contours, i, 255, 2)
def closest_pt(a, pt):
if not len(a):
return (float('inf'), float('inf'))
d = a - pt
return a[numpy.argmin((d * d).sum(1))]
match = cv2.matchTemplate(img_clean, template, cv2.TM_CCORR_NORMED)
# Filter matches.
threshold = 0.8
dist_threshold = twidth / 1.5
loc = numpy.where(match > threshold)
ptlist = numpy.zeros((len(loc[0]), 2), dtype=int)
count = 0
print "%d matches" % len(loc[0])
for pt in zip(*loc[::-1]):
cpt = closest_pt(ptlist[:count], pt)
dist = ((cpt[0] - pt[0]) ** 2 + (cpt[1] - pt[1]) ** 2) ** 0.5
if dist > dist_threshold:
ptlist[count] = pt
count += 1
# Adjust points (could do for the x coords too).
ptlist = ptlist[:count]
view = ptlist.ravel().view([('x', int), ('y', int)])
view.sort(order=['y', 'x'])
for i in xrange(1, ptlist.shape[0]):
prev, curr = ptlist[i - 1], ptlist[i]
if abs(curr[1] - prev[1]) < 5:
y = min(curr[1], prev[1])
curr[1], prev[1] = y, y
# Crop in raster order.
view.sort(order=['y', 'x'])
for i, pt in enumerate(ptlist, start=1):
cv2.imwrite(outbasename % i,
img[pt[1]-2:pt[1]+theight-2, pt[0]-2:pt[0]+twidth-2])
print 'Wrote %s' % (outbasename % i)
If you want only the contours of the hexagons, then crop on img_clean instead of img (but then it is pointless to sort the hexagons in raster order).
Here is a representation of the different regions that would be cut for your two examples without modifying the code above:
I am sorry, I didn't understand from your question on how do you relate matchTemplate and Contours.
Anyway, below is a small technique using contours. It is on the assumption that your other images are also like the one you provided. I am not sure if it works with your other images. But I think it would help to get a startup. Try this yourself and make necessary adjustments and modifications.
What I did :
1 - I needed the edge of octagons . So Thresholded Image using Otsu and apply dilation and erosion (or use any method you like that works well for all your images, beware of the edges in left edge of image).
2 - Then found contours (More about contours : http://goo.gl/r0ID0
3 - For each contours, find its convex hull, find its area(A) & perimeter(P)
4 - For a perfect octagon, P*P/A = 13.25 approximately. I used it here and cut it and saved it.
5 - You can see cropping it also removes some edges of octagon. If you want it, adjust the cropping dimension.
Code :
import cv2
import numpy as np
img = cv2.imread('region01.png')
gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
ret,thresh = cv2.threshold(gray,0,255,cv2.THRESH_BINARY_INV+cv2.THRESH_OTSU)
thresh = cv2.dilate(thresh,None,iterations = 2)
thresh = cv2.erode(thresh,None)
contours,hierarchy = cv2.findContours(thresh,cv2.RETR_LIST,cv2.CHAIN_APPROX_SIMPLE)
number = 0
for cnt in contours:
hull = cv2.convexHull(cnt)
area = cv2.contourArea(hull)
P = cv2.arcLength(hull,True)
if ((area != 0) and (13<= P**2/area <= 14)):
#cv2.drawContours(img,[hull],0,255,3)
x,y,w,h = cv2.boundingRect(hull)
number = number + 1
roi = img[y:y+h,x:x+w]
cv2.imshow(str(number),roi)
cv2.imwrite("1"+str(number)+".jpg",roi)
cv2.imshow('img',img)
cv2.waitKey(0)
cv2.destroyAllWindows()
Those 6 octagons will be stored as separate files.
Hope it helps !!!