I'm working on a project to measure and visualize image similarity. The images in my dataset come from photographs of images in books, some of which have very high or low exposure rates. For example, the images below come from two different books; the one on the top is an over-exposed reprint of the one on the bottom, wherein the exposure looks good:
I'd like to normalize each image's exposure in Python. I thought I could do so with the following naive approach, which attempts to center each pixel value between 0 and 255:
from scipy.ndimage import imread
import sys
def normalize(img):
'''
Normalize the exposure of an image.
#args:
{numpy.ndarray} img: an array of image pixels with shape:
(height, width)
#returns:
{numpy.ndarray} an image with shape of `img` wherein
all values are normalized such that the min=0 and max=255
'''
_min = img.min()
_max = img.max()
return img - _min * 255 / (_max - _min)
img = imread(sys.argv[1])
normalized = normalize(img)
Only after running this did I realize that this normalization will only help images whose lightest value is less than 255 or whose darkest value is greater than 0.
Is there a straightforward way to normalize the exposure of an image such as the top image above? I'd be grateful for any thoughts others can offer on this question.
Histogram equalisation works surprisingly well for this kind of thing. It's usually better for photographic images, but it's helpful even on line art, as long as there are some non-black/white pixels.
It works well for colour images too: split the bands up, equalize each one separately, and recombine.
I tried on your sample image:
Using libvips:
$ vips hist_equal sample.jpg x.jpg
Or from Python with pyvips:
x = pyvips.Image.new_from_file("sample.jpg")
x = x.hist_equal()
x.write_to_file("x.jpg")
It's very hard to say if it will work for you without seeing a larger sample of your images, but you may find an "auto-gamma" useful. There is one built into ImageMagick and the description - so that you can calculate it yourself - is:
Automagically adjust gamma level of image.
This calculates the mean values of an image, then applies a calculated
-gamma adjustment so that the mean color in the image will get a value of 50%.
This means that any solid 'gray' image becomes 50% gray.
This works well for real-life images with little or no extreme dark
and light areas, but tend to fail for images with large amounts of
bright sky or dark shadows. It also does not work well for diagrams or
cartoon like images.
You can try it out yourself on the command line very simply before you go and spend a lot of time coding something that may not work:
convert Tribunal.jpg -auto-gamma result.png
You can do -auto-level as per your own code beforehand, and a thousand other things too:
convert Tribunal.jpg -auto-level -auto-gamma result.png
I ended up using a numpy implementation of the histogram normalization method #user894763 pointed out. Just save the below as normalize.py then you can call:
python normalize.py cats.jpg
Script:
import numpy as np
from scipy.misc import imsave
from scipy.ndimage import imread
import sys
def get_histogram(img):
'''
calculate the normalized histogram of an image
'''
height, width = img.shape
hist = [0.0] * 256
for i in range(height):
for j in range(width):
hist[img[i, j]]+=1
return np.array(hist)/(height*width)
def get_cumulative_sums(hist):
'''
find the cumulative sum of a numpy array
'''
return [sum(hist[:i+1]) for i in range(len(hist))]
def normalize_histogram(img):
# calculate the image histogram
hist = get_histogram(img)
# get the cumulative distribution function
cdf = np.array(get_cumulative_sums(hist))
# determine the normalization values for each unit of the cdf
sk = np.uint8(255 * cdf)
# normalize the normalization values
height, width = img.shape
Y = np.zeros_like(img)
for i in range(0, height):
for j in range(0, width):
Y[i, j] = sk[img[i, j]]
# optionally, get the new histogram for comparison
new_hist = get_histogram(Y)
# return the transformed image
return Y
img = imread(sys.argv[1])
normalized = normalize_histogram(img)
imsave(sys.argv[1] + '-normalized.jpg', normalized)
Output:
Related
Say I have an image, and I want to have it fade out to greyscale over a distance.
I already know that to entirely convert an image to greyscale with Numpy, I'd do something like
import numpy as np
import cv2
myImage = cv2.imread("myImage.jpg")
grey = np.dot(an_image[...,:3], [0.2989, 0.5870, 0.1140])
This is not what I'm looking for. I already can get that to work.
I have a NxMx3 matrix (where N and M are the dimensions of the image), and this matrix is a dimension with the red transform, green transform, and blue transform.
So, for a given origin and radius of "keep this colored", I have
greyscaleWeights = np.array([0.2989, 0.5870, 0.1140])
# We flip this so we can weight down the transformation
greyscaleWeightOffsets = np.ones(3) - greyscaleWeights
from scipy.spatial.distance import cdist as getDistances
transformWeighter = list()
for rowNumber in np.arange(rowCount, dtype= 'int'):
# Create a row of tuples containing the coordinate we are at in the picture
row = [(x, rowNumber) for x in np.arange(columnCount, dtype= 'int')]
# Transform this into a row of distances from our in-color center
rowDistances = getDistances(row, [self.focusOrigin]).T[0]
# Get the transformation weights: inside of the focus radius we have no transform,
# outside of the pixelDistanceToFullTransform we have a weight of 1, and an even
# gradation in-between
rowWeights = [np.clip((x - self.focusRadius) / pixelDistanceToFullTransform, 0, 1) for x in rowDistances]
transformWeighter.append(rowWeights)
# Convert this into an numpy array
transformWeighter = np.array(transformWeighter)
# Change this 1-D set of weights into 3-D weights (for each color channel)
transformRGB = np.repeat(transformWeighter[:, :, None],3, axis=1).reshape(self.image.shape)
# Change the weight offsets back into greyscale weights
greyscaleTransform = 1 - greyscaleWeightOffsets * transformRGB
greyscaleishImage = self.image * greyscaleTransform
I do get the fade behaviour I was hoping for, but it just fades into the green channel while nuking the red and blue, so far as I can tell.
So, for example:
transforms into
which is the correct transformation behaviour, but fading to green instead of greyscale...
Well, the answer was both easy and hard.
The premise of my question was fundamentally flawed. To quote this answer on answers.opencv.org:
First, you must understand that a MxNx3 in greyscale doesn't exist. I mean, the concept of greyscale is that you have one channel describing the intensity on a gradual scale between black and white. So, it is not clear why would you need a 3 channels greyscale image, but if you do, I suggest that you take the value of each pixel of your 1 channel greyscale image and that you copy it three times, one on each channel of a BGR image. When a BGR image has the same value on each channel, it appears to be grey.
The correct answer then was to change the color space then desaturate the image, so
imageHSV = cv2.cvtColor(self.image, cv2.COLOR_RGB2HSV)
newSaturationChannel = saturationWeighter * imageHSV[:,:,1]
imageHSV[:,:,1] = newSaturationChannel
greyscaleishImage = cv2.cvtColor(imageHSV, cv2.COLOR_HSV2RGB)
I have 10 greyscale brain MRI scans from BrainWeb. They are stored as a 4d numpy array, brains, with shape (10, 181, 217, 181). Each of the 10 brains is made up of 181 slices along the z-plane (going through the top of the head to the neck) where each slice is 181 pixels by 217 pixels in the x (ear to ear) and y (eyes to back of head) planes respectively.
All of the brains are type dtype('float64'). The maximum pixel intensity across all brains is ~1328 and the minimum is ~0. For example, for the first brain, I calculate this by brains[0].max() giving 1328.338086605072 and brains[0].min() giving 0.0003886114541273855. Below is a plot of a slice of a brain[0]:
I want to binarize all these brain images by rescaling the pixel intensities from [0, 1328] to {0, 1}. Is my method correct?
I do this by first normalising the pixel intensities to [0, 1]:
normalized_brains = brains/1328
And then by using the binomial distribution to binarize each pixel:
binarized_brains = np.random.binomial(1, (normalized_brains))
The plotted result looks correct:
A 0 pixel intensity represents black (background) and 1 pixel intensity represents white (brain).
I experimented by implementing another method to normalise an image from this post but it gave me just a black image. This is because np.finfo(np.float64) is 1.7976931348623157e+308, so the normalization step
normalized_brains = brains/1.7976931348623157e+308
just returned an array of zeros which in the binarizition step also led to an array of zeros.
Am I binarising my images using a correct method?
Your method of converting the image to a binary image basically amounts to random dithering, which is a poor method of creating the illusion of grey values on a binary medium. Old-fashioned print is a binary medium, they have fine-tuned the methods to represent grey-value photographs in print over centuries. This process is called halftoning, and is shaped in part by properties of ink on paper, that we do not have to deal with in binary images.
So what methods have people come up with outside of print? Ordered dithering (mostly Bayer matrix), and error diffusion dithering. Read more about dithering on Wikipedia. I wrote a blog post showing how to implement all of these methods in MATLAB some years ago.
I would recommend you use error diffusion dithering for your particular application. Here is some code in MATLAB (taken from my blog post liked above) for the Floyd-Steinberg algorithm, I hope that you can translate this to Python:
img = imread('https://i.stack.imgur.com/d5E9i.png');
img = img(:,:,1);
out = double(img);
sz = size(out);
for ii=1:sz(1)
for jj=1:sz(2)
old = out(ii,jj);
%new = 255*(old >= 128); % Original Floyd-Steinberg
new = 255*(old >= 128+(rand-0.5)*100); % Simple improvement
out(ii,jj) = new;
err = new-old;
if jj<sz(2)
% right
out(ii ,jj+1) = out(ii ,jj+1)-err*(7/16);
end
if ii<sz(1)
if jj<sz(2)
% right-down
out(ii+1,jj+1) = out(ii+1,jj+1)-err*(1/16);
end
% down
out(ii+1,jj ) = out(ii+1,jj )-err*(5/16);
if jj>1
% left-down
out(ii+1,jj-1) = out(ii+1,jj-1)-err*(3/16);
end
end
end
end
imshow(out)
Resampling the image before applying the dithering greatly improves the results:
img = imresize(img,4);
% (repeat code above)
imshow(out)
NOTE that the above process expects the input to be in the range [0,255]. It is easy to adapt to a different range, say [0,1328] or [0,1], but it is also easy to scale your images to the [0,255] range.
Have you tried a threshold on the image?
This is a common way to binarize images, rather than trying to apply a random binomial distribution. You could try something like:
binarized_brains = (brains > threshold_value).astype(int)
which returns an array of 0s and 1s according to whether the image value was less than or greater than your chosen threshold value.
You will have to experiment with the threshold value to find the best one for your images, but it does not need to be normalized first.
If this doesn't work well, you can also experiment with the thresholding options available in the skimage filters package.
IT is easy in OpenCV. as mentioned a very common way is defining a threshold, But your result looks like you are allocating random values to your intensities instead of thresholding it.
import cv2
im = cv2.imread('brain.png', cv2.CV_LOAD_IMAGE_GRAYSCALE)
(th, brain_bw) = cv2.threshold(imy, 128, 255, cv2.THRESH_BINARY | cv2.THRESH_OTSU)
th = (DEFINE HERE)
im_bin = cv2.threshold(im, th, 255, cv
cv2.imwrite('binBrain.png', brain_bw)
brain
binBrain
I'm trying to stretch an image's histogram using a logarithmic transformation. Basically, I am applying a log operation to each pixel's intensity. When I'm trying to change image's value in each pixel, the new values are not saved but the histogram looks OK. Also, the maximum value is not correct. This is my code:
import cv2
import numpy as np
import math
from matplotlib import pyplot as plt
img = cv2.imread('messi.jpg',0)
img2 = img
for i in range(0,img2.shape[0]-1):
for j in range(0,img2.shape[1]-1):
if (math.log(1+img2[i,j],2)) < 0:
img2[i,j]=0
else:
img2[i,j] = np.int(math.log(1+img2[i,j],2))
print (np.int(math.log(1+img2[i,j],2)))
print (img2.ravel().max())
cv2.imshow('LSP',img2)
cv2.waitKey(0)
fig = plt.gcf()
fig.canvas.set_window_title('LSP histogram')
plt.hist(img2.ravel(),256,[0,256]); plt.show()
img3 = img2
B = np.int(img3.max())
A = np.int(img3.min())
print ("Maximum intensity = ", B)
print ("minimum intensity = ", A)
This is also the histogram I get:
However, the maximum intensity shows 186! This isn't applying the proper logarithmic operation at all.
Any ideas?
The code you wrote performs a logarithmic transformation applied to the image intensities. The reason why you are getting such a high spurious intensity as the maximum is because your for loops are wrong. Specifically, your range is incorrect. range is exclusive of the ending interval, which means that you must go up to img.shape[0] and img.shape[1] respectively, and not img.shape[0]-1 or img.shape[1]-1. Therefore, you are missing the last row and last column of the image, and these don't get touched by logarithmic operation. The maximum that is reported is from one of these pixels in the last row or column that you didn't touch.
Once you correct this, you don't get those bad intensities anymore:
for i in range(0,img2.shape[0]): # Change
for j in range(0,img2.shape[1]): # Change
if (math.log(1+img2[i,j],2)) < 0:
img2[i,j]=0
else:
img2[i,j] = np.int(math.log(1+img2[i,j],2))
Doing that now gives us:
('Maximum intensity = ', 7)
('minimum intensity = ', 0)
However, what you're going to get now is a very dark image. The histogram that you have shown us illustrates that all of the image pixels are in the dark range... roughly between [0-7]. Because of that, the majority of your image is going to be dark if you use uint8 as the data type for visualization. Take note that I searched for the Lionel Messi image that's part of the OpenCV tutorials, and this is the image I found:
Source: https://opencv-python-tutroals.readthedocs.org/en/latest/_images/roi.jpg
Your code is converting this to grayscale, and that's fine for the purpose of your question. Now, using the above image, if you actually show what the histogram count looks like as well as what the intensities are per bin in the histogram, this is what we get for img2:
In [41]: np.unique(img2)
Out[41]: array([0, 1, 2, 3, 4, 5, 6, 7], dtype=uint8)
In [42]: np.bincount(img2.ravel())
Out[42]: array([ 86, 88, 394, 3159, 14841, 29765, 58012, 19655])
As you can see, the bulk of the image pixels are hovering between the [0-7] range, which is why everything looks black. If you want to see this better, perhaps scale the image by roughly 255 / 7 = 36 or so we can see the image better:
img2 = 36*img2
cv2.imshow('LSP',img2)
cv2.waitKey(0)
We get this image:
I also get this histogram:
That personally looks very ugly... at least to me. As such, I would recommend that you choose a more meaningful image transformation if you want to stretch the histogram. In fact, the log operation compresses the dynamic range of the histogram. If you want to stretch the histogram, go the opposite way and try a power-law operation. Specifically, given an input intensity and the output is defined as:
out = c*in^(p)
in is the input intensity, p is a power and c is a constant to ensure that you scale the image so that the maximum intensity gets mapped to the same maximum intensity of the input when you're finished and not anything larger. That can be done by calculating c so that:
c = (img2.max()) / (img2.max()**p)
... where p is the power you want. In addition, the transformation via power-law can be explained with this nice diagram:
Source: http://www.nptel.ac.in/courses/117104069/chapter_8/8_14.html
Basically, powers that are less than 1 perform an intensity expansion where darker intensities get pushed towards the lighter side. Similarly, powers that are greater than 1 perform an intensity compression where lighter intensities get pushed to the darker side. In your case, you want to expand the histogram, and so you want the first option. Specifically, try making the intensities that are smaller go towards the larger range. This can be done by choosing a power that's smaller than 1... try 0.5 for example.
You'd modify your code so that it is like this:
img2 = img2.astype(np.float) # Cast to float
c = (img2.max()) / (img2.max()**(0.5))
for i in range(0,img2.shape[0]-1):
for j in range(0,img2.shape[1]-1):
img2[i,j] = np.int(c*img2[i,j]**(0.5))
# Cast back to uint8 for display
img2 = img2.astype(np.uint8)
Doing that, I get this image:
I also get this histogram:
Minor Note
If I can suggest something in terms of efficiency, I wouldn't recommend that you loop through the entire image and set each pixel individually... that's how numpy arrays were not supposed to be used. You can achieve what you want vectorized in a single line of code.
With your old code, use np.log2, not math.log with the base 2 with numpy arrays:
import cv2
import numpy as np
from matplotlib import pyplot as plt
# Your code
img = cv2.imread('messi.jpg',0)
# New code
img2 = np.log2(1 + img.astype(np.float)).astype(np.uint8)
# Back to your code
img2 = 36*img2 # Edit from before
cv2.imshow('LSP',img2)
cv2.waitKey(0)
fig = plt.gcf()
fig.canvas.set_window_title('LSP histogram')
plt.hist(img2.ravel(),256,[0,256]); plt.show()
img3 = img2
B = np.int(img3.max())
A = np.int(img3.min())
print ("Maximum intensity = ", B)
print ("minimum intensity = ", A)
cv2.destroyAllWindows() # Don't forget this
Similarly, if you want to apply a power-law transformation, it's very simply:
import cv2
import numpy as np
from matplotlib import pyplot as plt
# Your code
img = cv2.imread('messi.jpg',0)
# New code
c = (img2.max()) / (img2.max()**(0.5))
img2 = (c*img.astype(np.float)**(0.5)).astype(np.uint8)
#... rest of code as before
I can only ever find examples in C/C++ and they never seem to map well to the OpenCV API. I'm loading video frames (both from files and from a webcam) and want to reduce them to 16 color, but mapped to a 24-bit RGB color-space (this is what my output requires - a giant LED display).
I read the data like this:
ret, frame = self._vid.read()
image = cv2.cvtColor(frame, cv2.COLOR_RGB2BGRA)
I did find the below python example, but cannot figure out how to map that to the type of output data I need:
import numpy as np
import cv2
img = cv2.imread('home.jpg')
Z = img.reshape((-1,3))
# convert to np.float32
Z = np.float32(Z)
# define criteria, number of clusters(K) and apply kmeans()
criteria = (cv2.TERM_CRITERIA_EPS + cv2.TERM_CRITERIA_MAX_ITER, 10, 1.0)
K = 8
ret,label,center=cv2.kmeans(Z,K,None,criteria,10,cv2.KMEANS_RANDOM_CENTERS)
# Now convert back into uint8, and make original image
center = np.uint8(center)
res = center[label.flatten()]
res2 = res.reshape((img.shape))
cv2.imshow('res2',res2)
cv2.waitKey(0)
cv2.destroyAllWindows()
That obviously works for the OpenCV image viewer but trying to do the same errors on my output code since I need an RGB or RGBA format. My output works like this:
for y in range(self.height):
for x in range(self.width):
self._led.set(x,y,tuple(image[y,x][0:3]))
Each color is represented as an (r,g,b) tuple.
Any thoughts on how to make this work?
I think the following could be faster than kmeans, specially with a k = 16.
Convert the color image to gray
Contrast stretch this gray image to so that resulting image gray levels are between 0 and 255 (use normalize with NORM_MINMAX)
Calculate the histogram of this stretched gray image using 16 as the number of bins (calcHist)
Now you can modify these 16 values of the histogram. For example you can sort and assign ranks (say 0 to 15), or assign 16 uniformly distributed values between 0 and 255 (I think these could give you a consistent output for a video)
Backproject this histogram onto the stretched gray image (calcBackProject)
Apply a color-map to this backprojected image (you might want to scale the backprojected image befor applying a colormap using applyColorMap)
Tip for kmeans:
If you are using kmeans for video, you can use the cluster centers from the previous frame as the initial positions in kmeans for the current frame. That way, it'll take less time to converge, so kmeans in the subsequent frames will most probably run faster.
You can speed up your processing by applying the k-means on a downscaled version of your image. This will give you the cluster centroids. You can then quantify each pixel of the original image by picking the closest centroid.
I'd like some advice on performing a simple image analysis in python. I need to calculate a value for the "brightness" of an image. I know PIL is the goto library for doing something like this. There is a built-in histogram function.
What I need is a "perceived brightness" values I can decide if further adjustments to the image are necessary. So what are something of the basic techniques that will work in this situation? Should I just work with the RGB values, or will histogram give me something close enough?
One possible solution might be to combine the two, and generate average R,G,and B values using the histogram, then apply the "perceived brightness" formula.
Using the techniques mentioned in the question, I came up with a few different versions.
Each method returns a value close, but not exactly the same as the others. Also, all methods run about the same speed except for the last one, which is much slower depending on the image size.
Convert image to greyscale, return average pixel brightness.
def brightness( im_file ):
im = Image.open(im_file).convert('L')
stat = ImageStat.Stat(im)
return stat.mean[0]
Convert image to greyscale, return RMS pixel brightness.
def brightness( im_file ):
im = Image.open(im_file).convert('L')
stat = ImageStat.Stat(im)
return stat.rms[0]
Average pixels, then transform to "perceived brightness".
def brightness( im_file ):
im = Image.open(im_file)
stat = ImageStat.Stat(im)
r,g,b = stat.mean
return math.sqrt(0.241*(r**2) + 0.691*(g**2) + 0.068*(b**2))
RMS of pixels, then transform to "perceived brightness".
def brightness( im_file ):
im = Image.open(im_file)
stat = ImageStat.Stat(im)
r,g,b = stat.rms
return math.sqrt(0.241*(r**2) + 0.691*(g**2) + 0.068*(b**2))
Calculate "perceived brightness" of pixels, then return average.
def brightness( im_file ):
im = Image.open(im_file)
stat = ImageStat.Stat(im)
gs = (math.sqrt(0.241*(r**2) + 0.691*(g**2) + 0.068*(b**2))
for r,g,b in im.getdata())
return sum(gs)/stat.count[0]
Update Test Results
I ran a simulation against 200 images. I found that methods #2, #4 gave almost identical results. Also methods #3, #5 were also nearly identical. Method #1 closely followed #3, #5 (with a few exceptions).
Given that you're just looking for an average across the whole image, and not per-pixel brightness values, averaging PIL's histogram and applying the brightness function to the output seems like the best approach for that library.
If using ImageMagick (with the PythonMagick bindings), I would suggest using the identify command with the "verbose" option set. This will provide you with a mean value for each channel, saving you the need to sum and average a histogram — you can just multiply each channel directly.
I think your best result would come from converting the RGB to grayscale using your favorite formula, then taking the histogram of that result. I'm not sure if the mean or the median of the histogram would be more appropriate, but on most images they are probably similar.
I'm not sure how to do the conversion to grayscale in PIL using an arbitrary formula, but I'm guessing it's possible.
the code below will give you the brightness level of an image from 0-10
1- calculate the average brightness of the image after converting the image to HSV format using opencv.
2- find where this value lies in the list of brightness range.
import numpy as np
import cv2
import sys
from collections import namedtuple
#brange brightness range
#bval brightness value
BLevel = namedtuple("BLevel", ['brange', 'bval'])
#all possible levels
_blevels = [
BLevel(brange=range(0, 24), bval=0),
BLevel(brange=range(23, 47), bval=1),
BLevel(brange=range(46, 70), bval=2),
BLevel(brange=range(69, 93), bval=3),
BLevel(brange=range(92, 116), bval=4),
BLevel(brange=range(115, 140), bval=5),
BLevel(brange=range(139, 163), bval=6),
BLevel(brange=range(162, 186), bval=7),
BLevel(brange=range(185, 209), bval=8),
BLevel(brange=range(208, 232), bval=9),
BLevel(brange=range(231, 256), bval=10),
]
def detect_level(h_val):
h_val = int(h_val)
for blevel in _blevels:
if h_val in blevel.brange:
return blevel.bval
raise ValueError("Brightness Level Out of Range")
def get_img_avg_brightness():
if len(sys.argv) < 2:
print("USAGE: python3.7 brightness.py <image_path>")
sys.exit(1)
img = cv2.imread(sys.argv[1])
hsv = cv2.cvtColor(img, cv2.COLOR_BGR2HSV)
_, _, v = cv2.split(hsv)
return int(np.average(v.flatten()))
if __name__ == '__main__':
print("the image brightness level is:
{0}".format(detect_level(get_img_avg_brightness())))
This can be done by converting the BGR image from cv2 to grayscale and then finding the intensity - x and y are pixel coordinates. It's been explained well in this https://docs.opencv.org/3.4/d5/d98/tutorial_mat_operations.html document.
Scalar intensity = img.at<uchar>(y, x);
def calculate_brightness(image):
greyscale_image = image.convert('L')
histogram = greyscale_image.histogram()
pixels = sum(histogram)
brightness = scale = len(histogram)
for index in range(0, scale):
ratio = histogram[index] / pixels
brightness += ratio * (-scale + index)
return 1 if brightness == 255 else brightness / scale