Question

我正在尝试用otsu进行多阈值处理。我目前使用的方法实际上是通过最大化类间差异，我已经设法获得与OpenCV库相同的阈值。但是，这只是通过运行otsu方法一次。

关于如何进行多级阈值处理或者更确切地说是递归阈值处理的文档相当有限。获得原始otsu的价值后我该怎么办？我会欣赏一些提示，我一直在玩代码，添加一个外部for循环，但对于任何给定的图像，下一个计算的值总是254 :(

我的代码如果需要：

//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

为了将Otsu的阈值方法扩展到多级阈值化，类间方差方程变为：

请查看黄登元，林大伟，胡志虎，自动   基于聚类的两阶段Otsu方法的多级阈值分割   通过Valley Estimation，Int。确定。创新学报   计算，2011,7：5631-5644获取更多信息。

http://www.ijicic.org/ijicic-10-05033.pdf

这是我的Otsu Multi的C＃实现2阈值：

/* 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);
}

这是我用以上代码对土壤进行阈值处理的结果：

（T1 = 110，T2 = 147）。

Otsu的原始论文：“Nobuyuki Otsu，门限选择方法   从灰度直方图，IEEE系统，人和   Cybernetics，1979,9：62-66“也简要提到了扩展到   Multithresholding。

https://engineering.purdue.edu/kak/computervision/ECE661.08/OTSU_paper.pdf

希望这有帮助。

Answer 2

我之前写过一个关于otsu阈值处理如何在python中工作的例子。您可以在此处查看源代码：https://github.com/subokita/Sandbox/blob/master/otsu.py

在示例中有2个变体，otsu2（）是优化版本，如维基百科页面所示，otsu（）是基于算法描述本身的更天真的实现。

如果您可以阅读python代码（在这种情况下，它们非常简单，几乎就像伪代码一样），您可能需要查看示例中的otsu（）并进行修改。将它移植到C ++代码也不难。

Answer 3

@ Antoni4在我看来给出了最好的答案，并且非常直接地增加了关卡数量。

这是针对三级阈值处理：

#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 4

@Guilherme Silva

您的代码有BUG

你必须更换：

    W3K = 0;
    M3K = 0;

与

    W2K = 0;
    M2K = 0;

和

        W3K = 1 - (W1K + W2K);
        M3K = MT - (M1K + M2K);

与

        W3K = 1 - (W0K + W1K + W2K);
        M3K = MT - (M0K + M1K + M2K);

;-) 此致

编辑（1）：[Toby Speight] 我发现这个错误是通过在不同的重塑（尺寸）上将效果应用于相同的图片，并看到输出结果彼此差异很大（甚至更改分辨率）

W3K和M3K必须是总数减去前一个WK和MK。（我认为这与代码相似性与少一级的相似）目前由于我缺乏英语，我无法解释更好的方法和原因

说实话，我仍然不能100％确定这种方式是正确的，即使从我的输出中我也可以看出它可以提供更好的结果。（即使有1级以上（5级灰色））你可以试试自己;-) 遗憾

我的输出：

3 Thresholds 4 Thresholds

Answer 5

我在此线程中找到了一段有用的代码。我一直在寻找用于双/浮动图像的多级Otsu实现。因此，我尝试使用double / float矩阵作为输入来归纳N级示例。在下面的代码中，我将armadillo库用作依赖项。但是此代码可以轻松地适应标准C ++数组，只需将vec，uvec对象替换为一维双精度和整数数组，将mat和umat替换为二维即可。这里使用的犰狳的其他两个功能是：vectorise和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;
}

Answer 6

这是 python （> 3.0）中'n'个阈值的简单通用方法：

# 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)

有关全文和更多信息，请访问this link。

使用openCV进行多otsu（多阈值处理）

6 个答案: