对数缩放算法由x而不是y [英] Logarithmic scaling algo by x, not y

问题描述

好吧，好像我变得愚蠢，因为我无法接受以下内容：



我有一组数据y（x）。基本上，x是离散时间（秒），y是一些数据值，取决于x。缩放y很简单：ScaledY = ln（y - min（y）+ 1）。如果我需要压缩数据，让我们说，按分钟，它也很容易：每分钟计算平均y，然后使用上面的公式计算ScaledY。现在，我试图解决的问题是 - 而不是Y，以对数方式缩放X.到目前为止，我唯一的想法就是：



- 将我的数据分为两部分。计算左边所有x的平均y;将右边部分分成两部分，并重复这个除法/计算平均逻辑直到所有处理完毕。



我不喜欢这种方法的原因是平均在左边部分，因为从数学的角度来看这是不正确的。



有没有其他方法或者可能是标准算法？

OK, seems like I got stupid, as I can't embrace myself on the following:

I have a set of data y(x). Basically, x is discrete time (seconds), y is some data values, dependent on that x. Scaling y is simple: ScaledY = ln(y - min(y) + 1). If I need to compress data, let's say, by minute, it's also easy: calc average y for each minute, then calc ScaledY using the above formula. Now, the problem I'm trying to solve is - instead of Y, scale X logarithmically. The only idea that I have so far is this:

- divide my data set in two parts. Calculate average y for all x on the left part; divide right part in two parts, and repeat this divide/calc average logic until all is processed.

The reason I don't like this approach is an "average" on the left part, because this is not correct from a math point of view.

Is there any other approach or may be a standard algo?

推荐答案

感谢CPallini，他给了我一个很好的提示，如果有人有兴趣，我想出了这个功能。 （这是第一个，粗略的代码，但它产生了我正在寻找的东西）：

Thanks to CPallini, who gave me a good hint, I came up with this function if anyone is interested. (This is first, rough code, but it produced exactly what I was looking for):

void scaleXLn(std::vector<double>& data, std::vector<double>& result)
{
   size_t oldSize = data.size();
   size_t newSize = log((double)oldSize);
   ++newSize; // because it is always truncated during conversion
   result.resize(newSize, 0.0);
   // now we have "newSize" number of "new" points.
   // Let's loop by them:
   size_t leftIdx = 0; 
   for (int i=newSize - 1; i>=0; --i)
   {
      // get the right index:
      size_t rightIdx = i ? oldSize - exp(i) : oldSize - 1;
      // now we have "old" interval "left" to "right".
      // Our w(x) is 1/log(x + 1)
      // Let's calculate weighted average:
      double wxSum = 0.0f;
      double avg = 0.0f;
      for (size_t j = leftIdx; j<=rightIdx; ++j)
      {
         double wx = 1.0f/log(oldSize - j + 1);
         wxSum += wx;
         avg += wx*data[j];
      }
      result[newSize - i - 1] = avg/wxSum;
      // next
      leftIdx = rightIdx + 1;
   }
}

这篇关于对数缩放算法由x而不是y的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！