Histogram matching in digital image processing
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- Опубліковано 27 сер 2024
- In this video, we talk about Histogram matching in digital image processing which includes equalization and mapping of histograms.
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Image processing exam saved because of you. Many thanks!
Hey ur a life saver!! Hats off great explanation
Thank you😊
What happend to remaining 10 pixels, and also there is a chance that final histogram have more pixel than the image.
SAVE MY DAY BEFORE PAPER . THIS VIDEO HELPS ALOT.
Thankyou so much from the land of mountains, Nepal
Last minute semester savior amazing and really helpful content. 👌
vit?
@@hardikarora7681 no
short and sweet explanation.... thankyou
Thank you so much . Beautiful represention.
Thank you😊
After studying this video. great presentation. I feel like the output should be 0 0 0 80 0 100 180 30. I might be wrong
Thank you for uploading this kind of video its help me for preparations of exam.🙂
🇵🇰🇵🇰🇵🇰
Your most welcome! :)
Hi faaizuddin
Hi
ig this is wrong because the no of pixels values of the original image and final image are not same.
A simple unified way to treat both histogram equalization and histogram specification is the following: represent each pixel as a triple: (row index , column index, gray-level value). Now, sort all the pixels in the input image based on their grey-level in increasing order and store all the pixels of the input image in the triple format-- (row, column, value) - in an array whose size is equal to the number of pixels in the input image. This can be done efficiently as the number of gray-levels is limited, with number of operations proportional to the number of pixels (not N logN, but only N). If the entries of the given histogram to be matched are h0, h1, h2, ..., hk, with k+1 grey levels, then the output image having this histogram is computed easily as follows. As the pixels are sorted in increasing order, the monotonic transformation of gray-values is naturally satisfied. The first h0 pixels in the sorted list of pixels is assigned value 0, the next h1 pixels in the sorted list of pixels is assigned value 1, then the next h2 pixels are assigned value 2, and so on until the last hk pixels are assigned the value k. This assignment is done easily because the sorted list contains not only the gray level value of each pixel, but it also contains the associated position of the pixel in the image specified by row index and the column index, because the pixels are stored as triplets (row, column, value) in the sorted list. One efficient algorithm for sorting the pixels in time proportional to the number of pixels is this: use an array of k linked lists, one list corresponding to each grey level, and inserting each pixel into the list corresponding to its gray value, by scanning the input image for each pixel, and inserting all the pixels. After this, merge the k lists by simply appending the second list to the end of the first, the third list to the end of the second, the fourth list to the end of the third, etc. This merging is similar to that found in the radix-sort algorithm, but with only one iteration. There is an easier way to sort pixels using the cumulative histogram and an array instead of linked lists. That is left as an exercise for the reader! There are variations/modifications of this algorithm that avoid the sorted list but use only the cumulative histograms, and those that map pixels with same values in the input image to a new but same values in the output image, etc. But those are details that can be studied by a more serious student of digital image processing. Good luck! If you have questions, post your email address here in reply. I will answer the first 10 questions. I have the source code in C++.
Looks not quite simple bro
After the mapping do you still have to normalize? Because the new histogram has 380 pixels, which is 10 less than the original
bro calculate it is not 380
390 aa raha hai ek baar firr kar le calculate
i think the teacher is wrong, 是不是博主讲错了……
@@UpendraYadav-gc3bd
bsd wale calculate krke hi bola tha
Your voice is so sweet
Nice explaination !!
Mapping should be done from the input image to the specified image.
The video did it reversely.
exactly, thats what i was thinking too!!
Many many thanks for the video. U helped me to understand this subject :)
@5:45 why does Histogram 2 need to be greater than Histogram 1?
same question? Also one more thing that we are taking same pixel value of 1 for 2 and you are saying it should be greater?
Clear explanation!
Thank you for your video!
thankyou
very nice explanation !!!
Great explanation U explained the topic better than my cllge sir😂😂thank u so much🫶🏽👏🏾
I have a question, the total number of pixels in the final histogram is not matching with the first.
I have the same question.
@@shakyaabeytunge4459 Tell me also if you found solution
I got the answer
@@franckrooseveltngongtiomo8184 tell the answer then
@@shivamgoyal3888 While equalizing histogram (ii) for the gray level 4 we get (sk *7) = 2.506 so histogram equilization level is 3 so mapping it we get 100 and in final we get same pixel values
thank you.helped a lot
what a explaination......
Eazy pezy!
Cause of the video😇
tysm mam saved my a**
Mam, I have a doubt.
The last 2 columns for histogram 1 are Histogram equalization level and Nk. that means Nk is the last column.
But for histogram 2 the last column is Histogram equalization and not Nk.
Why you did not calculated Nk for histogram 2.
thanks it was great
god bless you
nice implementation thanks, how can I express the histogram of the product of 2 images as their own histograms of those images individually?
Great video , just a little bit sound issue .
Thanks, will take care of that next time!
pls also make videos on this topics:-
Discrete Fourier transform : Importance of fourier spectrum and Phase angle, Steps of Frequency
domain filtering algorithm, Ideal Lowpass filter, Ideal Highpass filter.
Color image models: RGB and CMY.
Image Restoration: Image Degradation model, Gaussian Noise Model, Spatial domain restoration
in presence of noise: Mean filter, Order statistic filters-Max, Min, Median.
Image Compression: Coding Inter-pixel and Psycho visual redundancy, Image Compression
model, Metrics: Compression Ratio, Entropy and Mean Square Error, Error free compression:
Huffman, Runlength coding, Lossy Compression: Introduction to still Image Compression standardJPEG.
Image Segmentation: Detection of discontinuities- Laplacian operator, Gradient operator,
Thresholding, Region Growing.
Image Representation: Using chain code Method and Image description using statistical moments.
Image Recognition: Pattern, Pattern Classes, Matching by Correlation.
Define the terms: Digital Watermarking, Image Morphing, Image registration, Image Spoofing and
Digital image forensics
I have tried to include most of these in the series. But will try to work on the remaining if possible
Thank tou
I'm still confused, why are we matching histogram 2 with histogram 1? Aren't they both two different histograms?
actually my proffesor said that we take step values not rounded off values
Mam make a video on haar function 😭😭 pls
Apex College kathmandu bata ko ko xan. Eta attendance dini hai.
mam can you please provide notes of ip that you tought
At 2.44, histogram equalization should be 2 not 1, because, 1.9 approximate is 2 not 1
it's 1.4, not 1.9
Hey, is your name Greata???
sound very less
Super 150 mai ho kya?
gla mei ho kya
imp:solve ques urself
you done little bit wrong.....
Can u share ig id or account?