我从未听说过Otsus方法,但我理解一些基百页,因此我试图简化这一方法。

1 Count how many pixels are at each level of darkness.
2 "Guess" a threshold.
3 Calculate the variance of the counts of darkness less than the threshold
4 Calculate the variance of the counts of darkness greater than the threshold
5 If the variance of the darker side is greater, guess a darker threshold, 
  else guess a higher threshold.
  Do this like a binary search so that it ends.
6 Turn all pixels darker than threshold black, the rest white.

抽水法实际上是“尽量扩大跨层差异”,但我不理解这部分数学。

差异概念是“彼此的数值相距甚远”。 差异不大意味着一切都相似。 差异很大,意味着价值大相径庭。 雨果差异很大,有色。 st流的背景差异是零,因为它完全是白人,没有其他肤色。 差异的计算方式大致如下:

double variance(unsigned int* counts, int size, int threshold, bool above) {
    //this is a quick trick to turn the "upper" into lower, save myself code
    if (above) return variance(counts, size-threshold, size-threshold, false);
    //first we calculate the average value
    unsigned long long atotal=0;
    unsigned long long acount=0;
    for(int i=0; i<threshold; ++i) {
        atotal += counts[i]*i //number of px times value
        acount += counts[i];
    }
    //finish calculating average
    double average = double(atotal)/count;
    //next we calculate the variance
    double vtotal=0;
    for(int i=0; i<threshold; ++i) {
        //to do so we get each values s difference from the average
        double t = std::abs(i-average);
        //and square it (I hate mathmaticians)
        vtotal += counts[i]*t*t;
    }
    //and return the average of those squared values.
    return vtotal/count;
}

我将采取另一种做法处理这个问题:

• Compute the cumulative histogram of greyscaled values of the image.
• Use as threshold the pixel value in which this cumulative reaches half of the total pixels of the image.

算法如下:

  int bin [256];
  foreach pixel in image
       bin[pixelvalue]++;  
  endfor  // this computes the histogram of the image

  int thresholdCount = ImageWidth * ImageSize / 2;
  int count = 0;
  for int i = 0 to 255
    count = count + bin[i];
    if( count > thresholdCount)
         threshold = i;
         break; // we are done
    endif
  endfor

This algorithm does not compute the cumulative histogram itself but rather uses the image histogram to do what I said earlier.

如果你本人为白人背景做了适当的工作,但因黑人背景而失败,那么你就完全需要发现你有黑色背景,歪曲价值观。 如果你认为背景价值更为普遍,那么你就只能计算结果的1和0;如果0或更多,则不考虑结果。

Instead of using mean of min and max, you should use median of all points as threshold. In general kth percentile (k = what percentage of points you want as black) is more appropriate.

另一个解决办法是将数据分为两组。