I have worked on a project to detect bad points generated by machine for several week and can not find any good solution. I wonder if you guys can give me some clues on it.
The corrupted image is shown as follows. Bad points are very bright or dark points. These points has the following characteristics:
relatively bigger or smaller intensity.
they are mostly one or two pixels together.
What I have tried:
I regard them as harris conner and detect them using bigger gradients. However, some point in the edge have big gradients also. In addition, threshold for gradient is not easy to fixed. Smaller threshold introduces false positive and bigger threshold introduces false negative.
Since bad points has bigger or smaller intensity compared to its local region, I calculate the average intensity except the center point and compared it with the center point. However, some normal points which have bigger or smaller intensity may be misclassified by this method. Also threshold for the difference between average and center point is also hard to fixed.
I also tried to extract some features for points and classify them as bad or good points. Although my classifier achieve 96% accuracy, this can misclassify many points because points are numerous in image.(6000,000)
I wonder if there is some deep learning point detection modes? I would like to try them to see if they can achieve 99.99...% accuracy.
Moreover, below examples are corrupted image and normal image. Though they are very obvious for human eyes, I can not think of a perfect method to distinguish them by computer.
Normal image with some bright pixels:
corrupted image with two bad points:
I will appreciate if you can give me some clues on this issue. Thank you so much!
You may have a try with a median filter with a small radius (3x3 or 5x5). Then detect the salt-and-noise pepper when the difference with the original image is large.
Leveraging Numpy to detect pixel intensity in the range [0,255] with minimum/maximum thresholds could work. The idea is to create a mask of all pixels greater than some threshold for bright points and lower than another threshold for dark points. The detected points are colored in green
import cv2
import numpy as np
image = cv2.imread('1.png')
gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)
mask = ((gray >= 200) | (gray <= 100))
image[mask] = [36,255,12]
cv2.imshow('image', image)
cv2.waitKey()
Related
I'm working on a project which involves determining the volume of a transparent liquid (or air if it proves easier) in a confined space.
The images I'm working with are a background image of the container without any liquid and a foreground image which may be also be empty in rare cases, but most times is partly filled with some amount of liquid.
While it may seem like a pretty straightforward smooth and threshold approach, it proves somewhat more difficult.
I'm working with a set with tons of these image pairs of background and foreground images, and I can't seem to find an approach that is robust enough to be applied to all images in the set.
My work so far involves smoothing and thresholding the image and applying closing to wrap it up.
bg_image = cv.imread("bg_image", 0)
fg_image = cv.imread("fg_image", 0)
blur_fg = cv.GaussianBlur(fg_image, (5, 5), sigmaX=0, sigmaY=0)
thresholded_image = cv.threshold(blur_fg, 186, 255, cv.THRESH_BINARY_INV)[1]
kernel = np.ones((4,2),np.uint8)
closing = cv.morphologyEx(thresholded_image, cv.MORPH_CLOSE, kernel)
The results vary, here is an example when it goes well:
In other examples, it doesn't go as well:
Aside from that, I have also tried:
Subtraction of the background and foreground images
Contrast stretching
Histogram equalization
Other thresholding techniques such as Otsu
The main issue is that the pixel intensities in air and liquid sometime overlap (and pretty low contrast in general), causing inaccurate estimations. I am leaning towards utilizing the edge that occurs between the liquid and air but I'm not really sure how..
I don't want to overflow with information here so I'm leaving it at that. I am grateful for any suggestions and can provide more information if necessary.
EDIT:
Here are some sample images to play around with.
Here is an approach whereby you calculate the mean of each column of pixels in your image, then calculate the gradient of the means:
#!/usr/bin/env python3
import cv2
import numpy as np
import matplotlib.pyplot as plt
filename = 'fg1.png'
# Load image as greyscale and calculate means of each column of pixels
im = cv2.imread(filename, cv2.IMREAD_GRAYSCALE)
means = np.mean(im, axis=0)
# Calculate the gradient of the means
y = np.gradient(means)
# Plot the gradient of the means
xdata = np.arange(0, y.shape[0])
plt.plot(xdata, y, 'bo') # blue circles
plt.title(f'Gradient of Column Means for "{filename}"')
plt.xlabel('x')
plt.ylabel('Gradient of Column Means')
plt.grid(True)
plt.show()
If you just plot the means of all columns, without taking the gradient, you get this:
Suppose that the Canny edge detector successfully detects an edge in an image. The edge is then rotated by θ, where the relationship between a point on the original edge (x,y)(x,y) and a point on the rotated edge (x′,y′)(x′,y′) is defined as x′ = xcosθ; y′ = xsinθ;
Will the rotated edge be detected using the same Canny edge detector?
(I think we should find answer considering that the detection of an edge by the Canny edge detector depends only on the magnitude of its derivative.)
The answer is both yes and no, and which one you go for depends on how literally you take the question.
First of all, we're dealing with a rectangular grid, so given an integer location (x,y), the corresponding point (x',y') in a rotated image is highly likely not an integer location. And considering that the output of Canny is a set of points, and not a smooth function that can be interpolated, it would be difficult to establish a correspondence between the set resulting from the rotated and the one resulting from the original image.
Think for example about the number of pixels on a discrete line of a given length at 0 degrees and at 45 degrees. (Hint: the line at 45 degrees has sqrt(2) times fewer pixels.)
But if you take the question more generally and interpret it as "will an edge that is detected in the original image also be detected after rotating the image by θ degrees?" then the answer is yes, in theory.
Of course practice is always a bit different than theory. The details of the implementation matter here. And there is always numerical imprecision to contend with.
Let's start by assuming the rotation is computed correctly, with a precise interpolation scheme (cubic, Lanczos) and not rounded after to uint8 or something (i.e. we're computing using floating-point values).
If you read the original paper by Canny, you'll see he proposes using Gaussian derivatives as the best compromise between compact support and computational precision. I have seen few implementations that actually do. Typically I see a convolution with a Gaussian and then Sobel derivatives. Especially for smaller sigmas (less smoothing) the difference can be quite large. Gaussian derivatives are rotationally invariant, Sobel derivatives are not.
The next step in the algorithm is non-maximum suppression. This is where the continuous gradient is converted to a set of points. For each pixel, it checks to see if it is a local maximum in the direction of the gradient. Because this is done per pixel, a different set of locations are tested in the rotated image compared to the original. Nonetheless, it should detect points along the same ridges in both cases.
Next, a hysteresis threshold is applied. This is a two-threshold operation that keeps pixels above one threshold as long as at least one pixel above a second threshold is present in the same connected component. This is where the differences could occur between rotated and original image. Remember we're dealing with a set of pixels. We have samples the continuous gradient function at discrete points. There could be an edge that has one pixel above the second threshold in one version of the image, but not in the other. This would only occur for edges very close to the chosen threshold, of course.
Next comes a thinning. Because the non-maximum suppression can yield points along a thicker line, a thinning operation is applied that removes pixels from the set that are not needed to maintain connectivity of the lines. Which pixels are selected here will also differ between rotated and original images, but this does not change the geometry of the solution, so we still have the same set of points.
So, the answer is yes and no. :)
Note that the same logic applies to translation.
The image below has many circles. Click and zoom in to see the circles.
https://drive.google.com/open?id=1ox3kiRX5hf2tHDptWfgcbMTAHKCDizSI
What I want is counting the circles using any free language, such as python.
Is there a function or idea to do it?
Edit: I came up with a better solution, partially inspired by this answer below. I thought of this method originally (as noted in the OP comments) but I decided against it. The original image was just not good enough quality for it. However I improved that method and it works brilliantly for the better quality image. The original approach is first, and then the new approach at the bottom.
First approach
So here's a general approach that seems to work well, but definitely just gives estimates. This assumes that circles are roughly the same size.
First, the image is mostly blue---so it seems reasonable to just do the analysis on the blue channel. Thresholding the blue channel, in this case, using Otsu thresholding (which determines an optimal threshold value without input) seems to work very well. This isn't too much of a surprise since the distribution of color values is pretty much binary. Check the mask that results from it!
Then, do a connected component analysis on the mask to get the area of each component (component = white blob in the mask). The statistics returned from connectedComponentsWithStats() give (among other things) the area, which is exactly what we need. Then we can simply count the circles by estimating how many circles fit in a given component based on its area. Also note that I'm taking the statistics for every label except the first one: this is the background label 0, and not any of the white blobs.
Now, how large in area is a single circle? It would be best to let the data tell us. So you could compute a histogram of all the areas, and since there are more single circles than anything else, there will be a high concentration around 250-270 pixels or so for the area. Or you could just take an average of all the areas between something like 50 and 350 which should also get you in a similar ballpark.
Really in this histogram you can see the demarcations between single circles, double circles, triple, and so on quite easily. Only the larger components will give pretty rough estimates. And in fact, the area doesn't seem to scale exactly linearly. Blobs of two circles are slightly larger than two single circles, and blobs of three are larger still than three single circles, and so on, so this makes it a little difficult to estimate nicely, but rounding should still keep us close. If you want you could include a small multiplication parameter that increases as the area increases to account for that, but that would be hard to quantify without going through the histogram analytically...so, I didn't worry about this.
A single circle area divided by the average single circle area should be close to 1. And the area of a 5-circle group divided by the average circle area should be close to 5. And this also means that small insignificant components, that are 1 or 10 or even 100 pixels in area, will not count towards the total since round(50/avg_circle_size) < 1/2, so those will round down to a count of 0. Thus I should just be able to take all the component areas, divide them by the average circle size, round, and get to a decent estimate by summing them all up.
import cv2
import numpy as np
img = cv2.imread('circles.png')
mask = cv2.threshold(img[:, :, 0], 255, 255, cv2.THRESH_BINARY_INV + cv2.THRESH_OTSU)[1]
stats = cv2.connectedComponentsWithStats(mask, 8)[2]
label_area = stats[1:, cv2.CC_STAT_AREA]
min_area, max_area = 50, 350 # min/max for a single circle
singular_mask = (min_area < label_area) & (label_area <= max_area)
circle_area = np.mean(label_area[singular_mask])
n_circles = int(np.sum(np.round(label_area / circle_area)))
print('Total circles:', n_circles)
This code is simple and effective for rough counts.
However, there are definitely some assumptions here about the groups of circles compared to a normal circle size, and there are issues where circles that are at the boundaries will not be counted correctly (these aren't well defined---a two circle blob that is half cut off will look more like one circle---no clear way to count or not count these with this method). Further I just used automatic thresholding via Otsu here; you could get (probably better) results with more careful color filtering. Additionally in the mask generated by Otsu, some circles that are masked have a few pixels removed from their center. Morphology could add these pixels back in, which would give you a (slightly larger) more accurate area for the single circle components. Either way, I just wanted to give the general idea towards how you could easily estimate this with minimal code.
New approach
Before, the goal was to count circles. This new approach instead counts the centers of the circles. The general idea is you threshold and then flood fill from a background pixel to fill in the background (flood fill works like the paint bucket tool in photo editing apps), that way you only see the centers, as shown in this answer below.
However, this relies on global thresholding, which isn't robust to local lighting changes. This means that since some centers are brighter/darker than others, you won't always get good results with a single threshold.
Here I've created an animation to show looping through different threshold values; watch as some centers appear and disappear at different times, meaning you get different counts depending on the threshold you choose (this is just a small patch of the image, it happens everywhere):
Notice that the first blob to appear in the top left actually disappears as the threshold increases. However, if we actually OR each frame together, then each detected pixel persists:
But now every single speck appears, so we should clean up the mask each frame so that we remove single pixels as they come (otherwise they may build up and be hard to remove later). Simple morphological opening with a small kernel will remove them:
Applied over the whole image, this method works incredibly well and finds almost every single cell. There are only three false positives (detected blob that's not a center) and two misses I can spot, and the code is very simple. The final thing to do after the mask has been created is simply count the components, minus one for the background. The only user input required here is a single point to flood fill from that is in the background (seed_pt in the code).
img = cv2.imread('circles.png', 0)
seed_pt = (25, 25)
fill_color = 0
mask = np.zeros_like(img)
kernel = cv2.getStructuringElement(cv2.MORPH_RECT, (3, 3))
for th in range(60, 120):
prev_mask = mask.copy()
mask = cv2.threshold(img, th, 255, cv2.THRESH_BINARY)[1]
mask = cv2.floodFill(mask, None, seed_pt, fill_color)[1]
mask = cv2.bitwise_or(mask, prev_mask)
mask = cv2.morphologyEx(mask, cv2.MORPH_OPEN, kernel)
n_centers = cv2.connectedComponents(mask)[0] - 1
print('There are %d cells in the image.'%n_centers)
There are 874 cells in the image.
One possible solution would be to read the image using OpenCV, get its grayscale, then use Canny edge detection and perform countour finding in OpenCV. This will return a list of countours. It would look something like:
import cv2
image = cv2.imread('path-to-your-image')
gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)
# tweak the parameters of the GaussianBlur for best performance
blurred = cv2.GaussianBlur(gray, (7, 7), 0)
# again, try different values here
edged = cv2.Canny(blurred, 20, 140)
(_, contours, _) = cv2.findContours(edged.copy(), cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
print(len(contours))
If you have all images like this - consider thresholding it, not necessarily by auto threshold-seeking algorithm like Otsu, but rather using simplest threshold by a given threshold value. Yes, before thresholding you have to convert your color input to gray-scale, or take one of color channels. Then based on few experiments with channels and threshold values - determine threshold value to have circles with holes in monochrome thresholding result. Based on your png image I found value of 81 (intensity of gray varies from 0 to 255) to be great to threshold gray-scale version of your input to have such binary image with holes in place, as described above.
Then simply count those holes.
Holes can be determined by seed-filling white area, connected to image border. As result you will have white hole connected components on black background - so simply count them.
More details you can find here http://www.leptonica.com/filling.html and use leptonica primitives to do thresholding, hole counting an so on.
I am looking for a "very" simple way to check if an image bitmap is blur. I do not need accurate and complicate algorithm which involves fft, wavelet, etc. Just a very simple idea even if it is not accurate.
I've thought to compute the average euclidian distance between pixel (x,y) and pixel (x+1,y) considering their RGB components and then using a threshold but it works very bad. Any other idea?
Don't calculate the average differences between adjacent pixels.
Even when a photograph is perfectly in focus, it can still contain large areas of uniform colour, like the sky for example. These will push down the average difference and mask the details you're interested in. What you really want to find is the maximum difference value.
Also, to speed things up, I wouldn't bother checking every pixel in the image. You should get reasonable results by checking along a grid of horizontal and vertical lines spaced, say, 10 pixels apart.
Here are the results of some tests with PHP's GD graphics functions using an image from Wikimedia Commons (Bokeh_Ipomea.jpg). The Sharpness values are simply the maximum pixel difference values as a percentage of 255 (I only looked in the green channel; you should probably convert to greyscale first). The numbers underneath show how long it took to process the image.
If you want them, here are the source images I used:
original
slightly blurred
blurred
Update:
There's a problem with this algorithm in that it relies on the image having a fairly high level of contrast as well as sharp focused edges. It can be improved by finding the maximum pixel difference (maxdiff), and finding the overall range of pixel values in a small area centred on this location (range). The sharpness is then calculated as follows:
sharpness = (maxdiff / (offset + range)) * (1.0 + offset / 255) * 100%
where offset is a parameter that reduces the effects of very small edges so that background noise does not affect the results significantly. (I used a value of 15.)
This produces fairly good results. Anything with a sharpness of less than 40% is probably out of focus. Here's are some examples (the locations of the maximum pixel difference and the 9×9 local search areas are also shown for reference):
(source)
(source)
(source)
(source)
The results still aren't perfect, though. Subjects that are inherently blurry will always result in a low sharpness value:
(source)
Bokeh effects can produce sharp edges from point sources of light, even when they are completely out of focus:
(source)
You commented that you want to be able to reject user-submitted photos that are out of focus. Since this technique isn't perfect, I would suggest that you instead notify the user if an image appears blurry instead of rejecting it altogether.
I suppose that, philosophically speaking, all natural images are blurry...How blurry and to which amount, is something that depends upon your application. Broadly speaking, the blurriness or sharpness of images can be measured in various ways. As a first easy attempt I would check for the energy of the image, defined as the normalised summation of the squared pixel values:
1 2
E = --- Σ I, where I the image and N the number of pixels (defined for grayscale)
N
First you may apply a Laplacian of Gaussian (LoG) filter to detect the "energetic" areas of the image and then check the energy. The blurry image should show considerably lower energy.
See an example in MATLAB using a typical grayscale lena image:
This is the original image
This is the blurry image, blurred with gaussian noise
This is the LoG image of the original
And this is the LoG image of the blurry one
If you just compute the energy of the two LoG images you get:
E = 1265 E = 88
or bl
which is a huge amount of difference...
Then you just have to select a threshold to judge which amount of energy is good for your application...
calculate the average L1-distance of adjacent pixels:
N1=1/(2*N_pixel) * sum( abs(p(x,y)-p(x-1,y)) + abs(p(x,y)-p(x,y-1)) )
then the average L2 distance:
N2= 1/(2*N_pixel) * sum( (p(x,y)-p(x-1,y))^2 + (p(x,y)-p(x,y-1))^2 )
then the ratio N2 / (N1*N1) is a measure of blurriness. This is for grayscale images, for color you do this for each channel separately.
Given an image (Like the one given below) I need to convert it into a binary image (black and white pixels only). This sounds easy enough, and I have tried with two thresholding functions. The problem is I cant get the perfect edges using either of these functions. Any help would be greatly appreciated.
The filters I have tried are, the Euclidean distance in the RGB and HSV spaces.
Sample image:
Here it is after running an RGB threshold filter. (40% it more artefects after this)
Here it is after running an HSV threshold filter. (at 30% the paths become barely visible but clearly unusable because of the noise)
The code I am using is pretty straightforward. Change the input image to appropriate color spaces and check the Euclidean distance with the the black color.
sqrt(R*R + G*G + B*B)
since I am comparing with black (0, 0, 0)
Your problem appears to be the variation in lighting over the scanned image which suggests that a locally adaptive thresholding method would give you better results.
The Sauvola method calculates the value of a binarized pixel based on the mean and standard deviation of pixels in a window of the original image. This means that if an area of the image is generally darker (or lighter) the threshold will be adjusted for that area and (likely) give you fewer dark splotches or washed-out lines in the binarized image.
http://www.mediateam.oulu.fi/publications/pdf/24.p
I also found a method by Shafait et al. that implements the Sauvola method with greater time efficiency. The drawback is that you have to compute two integral images of the original, one at 8 bits per pixel and the other potentially at 64 bits per pixel, which might present a problem with memory constraints.
http://www.dfki.uni-kl.de/~shafait/papers/Shafait-efficient-binarization-SPIE08.pdf
I haven't tried either of these methods, but they do look promising. I found Java implementations of both with a cursory Google search.
Running an adaptive threshold over the V channel in the HSV color space should produce brilliant results. Best results would come with higher than 11x11 size window, don't forget to choose a negative value for the threshold.
Adaptive thresholding basically is:
if (Pixel value + constant > Average pixel value in the window around the pixel )
Pixel_Binary = 1;
else
Pixel_Binary = 0;
Due to the noise and the illumination variation you may need an adaptive local thresholding, thanks to Beaker for his answer too.
Therefore, I tried the following steps:
Convert it to grayscale.
Do the mean or the median local thresholding, I used 10 for the window size and 10 for the intercept constant and got this image (smaller values might also work):
Please refer to : http://homepages.inf.ed.ac.uk/rbf/HIPR2/adpthrsh.htm if you need more
information on this techniques.
To make sure the thresholding was working fine, I skeletonized it to see if there is a line break. This skeleton may be the one needed for further processing.
To get ride of the remaining noise you can just find the longest connected component in the skeletonized image.
Thank you.
You probably want to do this as a three-step operation.
use leveling, not just thresholding: Take the input and scale the intensities (gamma correct) with parameters that simply dull the mid tones, without removing the darks or the lights (your rgb threshold is too strong, for instance. you lost some of your lines).
edge-detect the resulting image using a small kernel convolution (5x5 for binary images should be more than enough). Use a simple [1 2 3 2 1 ; 2 3 4 3 2 ; 3 4 5 4 3 ; 2 3 4 3 2 ; 1 2 3 2 1] kernel (normalised)
threshold the resulting image. You should now have a much better binary image.
You could try a black top-hat transform. This involves substracting the Image from the closing of the Image. I used a structural element window size of 11 and a constant threshold of 0.1 (25.5 on for a 255 scale)
You should get something like:
Which you can then easily threshold:
Best of luck.