Multi otsu(multi-thresholding) with openCV - opencv

Question

I am trying to carry out multi-thresholding with otsu. The method I am using currently is actually via maximising the between class variance, I have managed to get the same threshold value given as that by the OpenCV library. However, that is just via running otsu method once.
Documentation on how to do multi-level thresholding or rather recursive thresholding is rather limited. Where do I do after obtaining the original otsu's value? Would appreciate some hints, I been playing around with the code, adding one external for loop, but the next value calculated is always 254 for any given image:(
My code if need be:
//compute histogram first
cv::Mat imageh; //image edited to grayscale for histogram purpose
//imageh=image; //to delete and uncomment below;
cv::cvtColor(image, imageh, CV_BGR2GRAY);
int histSize[1] = {256}; // number of bins
float hranges[2] = {0.0, 256.0}; // min andax pixel value
const float* ranges[1] = {hranges};
int channels[1] = {0}; // only 1 channel used
cv::MatND hist;
// Compute histogram
calcHist(&imageh, 1, channels, cv::Mat(), hist, 1, histSize, ranges);
IplImage* im = new IplImage(imageh);//assign the image to an IplImage pointer
IplImage* finalIm = cvCreateImage(cvSize(im->width, im->height), IPL_DEPTH_8U, 1);
double otsuThreshold= cvThreshold(im, finalIm, 0, 255, cv::THRESH_BINARY | cv::THRESH_OTSU );
cout<<"opencv otsu gives "<<otsuThreshold<<endl;
int totalNumberOfPixels= imageh.total();
cout<<"total number of Pixels is " <<totalNumberOfPixels<< endl;
float sum = 0;
for (int t=0 ; t<256 ; t++)
{
sum += t * hist.at<float>(t);
}
cout<<"sum is "<<sum<<endl;
float sumB = 0; //sum of background
int wB = 0; // weight of background
int wF = 0; //weight of foreground
float varMax = 0;
int threshold = 0;
//run an iteration to find the maximum value of the between class variance(as between class variance shld be maximise)
for (int t=0 ; t<256 ; t++)
{
wB += hist.at<float>(t); // Weight Background
if (wB == 0) continue;
wF = totalNumberOfPixels - wB; // Weight Foreground
if (wF == 0) break;
sumB += (float) (t * hist.at<float>(t));
float mB = sumB / wB; // Mean Background
float mF = (sum - sumB) / wF; // Mean Foreground
// Calculate Between Class Variance
float varBetween = (float)wB * (float)wF * (mB - mF) * (mB - mF);
// Check if new maximum found
if (varBetween > varMax) {
varMax = varBetween;
threshold = t;
}
}
cout<<"threshold value is: "<<threshold;

Answer 1

To extend Otsu's thresholding method to multi-level thresholding the between class variance equation becomes:
Please check out Deng-Yuan Huang, Ta-Wei Lin, Wu-Chih Hu, Automatic
Multilevel Thresholding Based on Two-Stage Otsu's Method with Cluster
Determination by Valley Estimation, Int. Journal of Innovative
Computing, 2011, 7:5631-5644 for more information.
http://www.ijicic.org/ijicic-10-05033.pdf
Here is my C# implementation of Otsu Multi for 2 thresholds:
/* Otsu (1979) - multi */
Tuple < int, int > otsuMulti(object sender, EventArgs e) {
//image histogram
int[] histogram = new int[256];
//total number of pixels
int N = 0;
//accumulate image histogram and total number of pixels
foreach(int intensity in image.Data) {
if (intensity != 0) {
histogram[intensity] += 1;
N++;
}
}
double W0K, W1K, W2K, M0, M1, M2, currVarB, optimalThresh1, optimalThresh2, maxBetweenVar, M0K, M1K, M2K, MT;
optimalThresh1 = 0;
optimalThresh2 = 0;
W0K = 0;
W1K = 0;
M0K = 0;
M1K = 0;
MT = 0;
maxBetweenVar = 0;
for (int k = 0; k <= 255; k++) {
MT += k * (histogram[k] / (double) N);
}
for (int t1 = 0; t1 <= 255; t1++) {
W0K += histogram[t1] / (double) N; //Pi
M0K += t1 * (histogram[t1] / (double) N); //i * Pi
M0 = M0K / W0K; //(i * Pi)/Pi
W1K = 0;
M1K = 0;
for (int t2 = t1 + 1; t2 <= 255; t2++) {
W1K += histogram[t2] / (double) N; //Pi
M1K += t2 * (histogram[t2] / (double) N); //i * Pi
M1 = M1K / W1K; //(i * Pi)/Pi
W2K = 1 - (W0K + W1K);
M2K = MT - (M0K + M1K);
if (W2K <= 0) break;
M2 = M2K / W2K;
currVarB = W0K * (M0 - MT) * (M0 - MT) + W1K * (M1 - MT) * (M1 - MT) + W2K * (M2 - MT) * (M2 - MT);
if (maxBetweenVar < currVarB) {
maxBetweenVar = currVarB;
optimalThresh1 = t1;
optimalThresh2 = t2;
}
}
}
return new Tuple(optimalThresh1, optimalThresh2);
}
And this is the result I got by thresholding an image scan of soil with the above code:
(T1 = 110, T2 = 147).
Otsu's original paper: "Nobuyuki Otsu, A Threshold Selection Method
from Gray-Level Histogram, IEEE Transactions on Systems, Man, and
Cybernetics, 1979, 9:62-66" also briefly mentions the extension to
Multithresholding.
https://engineering.purdue.edu/kak/computervision/ECE661.08/OTSU_paper.pdf
Hope this helps.

Answer 2

Here is a simple general approach for 'n' thresholds in python (>3.0) :
# developed by- SUJOY KUMAR GOSWAMI
# source paper- https://people.ece.cornell.edu/acharya/papers/mlt_thr_img.pdf
import cv2
import numpy as np
import math
img = cv2.imread('path-to-image')
img = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
a = 0
b = 255
n = 6 # number of thresholds (better choose even value)
k = 0.7 # free variable to take any positive value
T = [] # list which will contain 'n' thresholds
def sujoy(img, a, b):
if a>b:
s=-1
m=-1
return m,s
img = np.array(img)
t1 = (img>=a)
t2 = (img<=b)
X = np.multiply(t1,t2)
Y = np.multiply(img,X)
s = np.sum(X)
m = np.sum(Y)/s
return m,s
for i in range(int(n/2-1)):
img = np.array(img)
t1 = (img>=a)
t2 = (img<=b)
X = np.multiply(t1,t2)
Y = np.multiply(img,X)
mu = np.sum(Y)/np.sum(X)
Z = Y - mu
Z = np.multiply(Z,X)
W = np.multiply(Z,Z)
sigma = math.sqrt(np.sum(W)/np.sum(X))
T1 = mu - k*sigma
T2 = mu + k*sigma
x, y = sujoy(img, a, T1)
w, z = sujoy(img, T2, b)
T.append(x)
T.append(w)
a = T1+1
b = T2-1
k = k*(i+1)
T1 = mu
T2 = mu+1
x, y = sujoy(img, a, T1)
w, z = sujoy(img, T2, b)
T.append(x)
T.append(w)
T.sort()
print(T)
For full paper and more informations visit this link.

Answer 3

I've written an example on how otsu thresholding work in python before. You can see the source code here: https://github.com/subokita/Sandbox/blob/master/otsu.py
In the example there's 2 variants, otsu2() which is the optimised version, as seen on Wikipedia page, and otsu() which is more naive implementation based on the algorithm description itself.
If you are okay in reading python codes (in this case, they are pretty simple, almost pseudo code like), you might want to look at otsu() in the example and modify it. Porting it to C++ code is not hard either.

Answer 4

#Antoni4 gives the best answer in my opinion and it's very straight forward to increase the number of levels.
This is for three-level thresholding:
#include "Shadow01-1.cuh"
void multiThresh(double &optimalThresh1, double &optimalThresh2, double &optimalThresh3, cv::Mat &imgHist, cv::Mat &src)
{
double W0K, W1K, W2K, W3K, M0, M1, M2, M3, currVarB, maxBetweenVar, M0K, M1K, M2K, M3K, MT;
unsigned char *histogram = (unsigned char*)(imgHist.data);
int N = src.rows*src.cols;
W0K = 0;
W1K = 0;
M0K = 0;
M1K = 0;
MT = 0;
maxBetweenVar = 0;
for (int k = 0; k <= 255; k++) {
MT += k * (histogram[k] / (double) N);
}
for (int t1 = 0; t1 <= 255; t1++)
{
W0K += histogram[t1] / (double) N; //Pi
M0K += t1 * (histogram[t1] / (double) N); //i * Pi
M0 = M0K / W0K; //(i * Pi)/Pi
W1K = 0;
M1K = 0;
for (int t2 = t1 + 1; t2 <= 255; t2++)
{
W1K += histogram[t2] / (double) N; //Pi
M1K += t2 * (histogram[t2] / (double) N); //i * Pi
M1 = M1K / W1K; //(i * Pi)/Pi
W2K = 1 - (W0K + W1K);
M2K = MT - (M0K + M1K);
if (W2K <= 0) break;
M2 = M2K / W2K;
W3K = 0;
M3K = 0;
for (int t3 = t2 + 1; t3 <= 255; t3++)
{
W2K += histogram[t3] / (double) N; //Pi
M2K += t3 * (histogram[t3] / (double) N); // i*Pi
M2 = M2K / W2K; //(i*Pi)/Pi
W3K = 1 - (W1K + W2K);
M3K = MT - (M1K + M2K);
M3 = M3K / W3K;
currVarB = W0K * (M0 - MT) * (M0 - MT) + W1K * (M1 - MT) * (M1 - MT) + W2K * (M2 - MT) * (M2 - MT) + W3K * (M3 - MT) * (M3 - MT);
if (maxBetweenVar < currVarB)
{
maxBetweenVar = currVarB;
optimalThresh1 = t1;
optimalThresh2 = t2;
optimalThresh3 = t3;
}
}
}
}
}

Answer 5

#Guilherme Silva
Your code has a BUG
You Must Replace:
W3K = 0;
M3K = 0;
with
W2K = 0;
M2K = 0;
and
W3K = 1 - (W1K + W2K);
M3K = MT - (M1K + M2K);
with
W3K = 1 - (W0K + W1K + W2K);
M3K = MT - (M0K + M1K + M2K);
;-)
Regards
EDIT(1): [Toby Speight]
I discovered this bug by applying the effect to the same picture at different resoultions(Sizes) and seeing that the output results were to much different from each others (Even changing resolution a little bit)
W3K and M3K must be the totals minus the Previous WKs and MKs.
(I thought about this for Code-similarity with the one with one level less)
At the moment due to my lacks of English I cannot explain Better How and Why
To be honest I'm still not 100% sure that this way is correct, even thought from my outputs I could tell that it gives better results. (Even with 1 Level more (5 shades of gray))
You could try yourself ;-)
Sorry
My Outputs:
3 Thresholds
4 Thresholds

Answer 6

I found a useful piece of code in this thread. I was looking for a multi-level Otsu implementation for double/float images. So, I tried to generalize example for N-levels with double/float matrix as input. In my code below I am using armadillo library as dependency. But this code can be easily adapted for standard C++ arrays, just replace vec, uvec objects with single dimensional double and integer arrays, mat and umat with two-dimensional. Two other functions from armadillo used here are: vectorise and hist.
// Input parameters:
// map - input image (double matrix)
// mask - region of interest to be thresholded
// nBins - number of bins
// nLevels - number of Otsu thresholds
#include <armadillo>
#include <algorithm>
#include <vector>
mat OtsuFilterMulti(mat map, int nBins, int nLevels) {
mat mapr; // output thresholded image
mapr = zeros<mat>(map.n_rows, map.n_cols);
unsigned int numElem = 0;
vec threshold = zeros<vec>(nLevels);
vec q = zeros<vec>(nLevels + 1);
vec mu = zeros<vec>(nLevels + 1);
vec muk = zeros<vec>(nLevels + 1);
uvec binv = zeros<uvec>(nLevels);
if (nLevels <= 1) return mapr;
numElem = map.n_rows*map.n_cols;
uvec histogram = hist(vectorise(map), nBins);
double maxval = map.max();
double minval = map.min();
double odelta = (maxval - abs(minval)) / nBins; // distance between histogram bins
vec oval = zeros<vec>(nBins);
double mt = 0, variance = 0.0, bestVariance = 0.0;
for (int ii = 0; ii < nBins; ii++) {
oval(ii) = (double)odelta*ii + (double)odelta*0.5; // centers of histogram bins
mt += (double)ii*((double)histogram(ii)) / (double)numElem;
}
for (int ii = 0; ii < nLevels; ii++) {
binv(ii) = ii;
}
double sq, smuk;
int nComb;
nComb = nCombinations(nBins,nLevels);
std::vector<bool> v(nBins);
std::fill(v.begin(), v.begin() + nLevels, true);
umat ibin = zeros<umat>(nComb, nLevels); // indices from combinations will be stored here
int cc = 0;
int ci = 0;
do {
for (int i = 0; i < nBins; ++i) {
if(ci==nLevels) ci=0;
if (v[i]) {
ibin(cc,ci) = i;
ci++;
}
}
cc++;
} while (std::prev_permutation(v.begin(), v.end()));
uvec lastIndex = zeros<uvec>(nLevels);
// Perform operations on pre-calculated indices
for (int ii = 0; ii < nComb; ii++) {
for (int jj = 0; jj < nLevels; jj++) {
smuk = 0;
sq = 0;
if (lastIndex(jj) != ibin(ii, jj) || ii == 0) {
q(jj) += double(histogram(ibin(ii, jj))) / (double)numElem;
muk(jj) += ibin(ii, jj)*(double(histogram(ibin(ii, jj)))) / (double)numElem;
mu(jj) = muk(jj) / q(jj);
q(jj + 1) = 0.0;
muk(jj + 1) = 0.0;
if (jj>0) {
for (int kk = 0; kk <= jj; kk++) {
sq += q(kk);
smuk += muk(kk);
}
q(jj + 1) = 1 - sq;
muk(jj + 1) = mt - smuk;
mu(jj + 1) = muk(jj + 1) / q(jj + 1);
}
if (jj>0 && jj<(nLevels - 1)) {
q(jj + 1) = 0.0;
muk(jj + 1) = 0.0;
}
lastIndex(jj) = ibin(ii, jj);
}
}
variance = 0.0;
for (int jj = 0; jj <= nLevels; jj++) {
variance += q(jj)*(mu(jj) - mt)*(mu(jj) - mt);
}
if (variance > bestVariance) {
bestVariance = variance;
for (int jj = 0; jj<nLevels; jj++) {
threshold(jj) = oval(ibin(ii, jj));
}
}
}
cout << "Optimized thresholds: ";
for (int jj = 0; jj<nLevels; jj++) {
cout << threshold(jj) << " ";
}
cout << endl;
for (unsigned int jj = 0; jj<map.n_rows; jj++) {
for (unsigned int kk = 0; kk<map.n_cols; kk++) {
for (int ll = 0; ll<nLevels; ll++) {
if (map(jj, kk) >= threshold(ll)) {
mapr(jj, kk) = ll+1;
}
}
}
}
return mapr;
}
int nCombinations(int n, int r) {
if (r>n) return 0;
if (r*2 > n) r = n-r;
if (r == 0) return 1;
int ret = n;
for( int i = 2; i <= r; ++i ) {
ret *= (n-i+1);
ret /= i;
}
return ret;
}