用于序列化足压图的最佳无损压缩技术

时间：2016-08-05 21:13:44

标签： compression sparse-matrix image-compression run-length-encoding

我正在处理人脚的压力感应，我需要通过串行实时传输帧。

典型的框架如下所示，由平坦的背景和非平坦数据组成：

enter image description here

传输速度目前是由于Serial.send命令导致的微控制器开销导致的瓶颈，因此工程师正在使用Run Length Encoding来压缩图像，由于平坦，连续，这看起来很好背景，但我们想进一步压缩它。

我尝试过＆＃34;坐标列表＆＃34;编码格式（List<i, j, val>其中val＆gt; 0），但其大小与RLE相似，不会产生显着差异。

虽然研究了一下SO，人们说并没有重新发明轮子，但是对于任何类型的图像都有很多久经考验的压缩算法，所以我想知道是什么对于下面显示的图像类型，这将是最好的，考虑：

  1. 压缩性能（因为它将由微控制器执行）;
  2. 大小 - 因为它是通过串行发送的，这是目前的瓶颈（原文如此）。

    3. 其他方法是使用＆＃34;稀疏矩阵＆＃34;概念（而不是＆＃34;图像压缩＆＃34;概念），看起来有类似CRS或CSR的东西，我无法理解如何实现以及如何序列化正确，甚至更少与图像压缩技术相比。

    更新： 我用我用来创建图像的数据创建了一个Gist。这些是压缩方法的结果（每个条目一个字节）：

    • plain：（[n_rows, n_columns, *data]）： 2290 bytes;
    • 坐标列表：（[*(i, j, val)]）： 936 字节;
    • 游程编码：（[*(rowlength, rle-pairs)]）： 846 字节;
    • 列表清单： 690 字节;
    • 紧凑的列表清单:(参见要点） 498 字节;

2 个答案:

答案 0 :(得分：1)

建议的算法

下面是一个可能的算法，它只使用简单的操作 [1] ，内存占用少（没有双关语）。

它似乎工作得相当好，但当然，它应该在几个不同的数据集上进行测试，以便更准确地了解其效率。

  1. 将矩阵划分为13x11块4x4像素

  2. 对于每个块：

    • 如果该块为空，则发出位'0'
    • 如果块不为空：
      1. 发出位'1'
      2. 在此块中发出非零像素的16位位掩码
      3. 发出8位值，表示此块中的最小值（0除外）
      4. 如果只有一个非零像素，请在此处停止 [2]
      5. 发出3位值，表示编码此块中每个非零像素所需的位数：b = ceil（log 2 （max + 1 - min））
      6. 将非零像素数据发射为N x b位

    3. 基于以下观察：

    • 矩阵中的许多块都是空的
    • 足迹边界的非空区块通常有许多空单元格（传感器上的“压力”/“无压力”转换是突然的）

    [1] 特别是没有浮点运算。算法描述中使用的log2（）操作可以通过对1,2,4,8,16 ......最多256的简单比较轻松替换。

    [2] 这是一个不经常触发的次要优化。解码器必须通过计算来检测位掩码中只有一个位：(msk & -msk) == msk。

    块编码示例

    让我们考虑以下块：

     0,  0,  0,  0
12,  0,  0,  0
21, 20,  0,  0
28, 23,  0,  0

    非零像素的位掩码是：

     0,  0,  0,  0
 1,  0,  0,  0  =  0000100011001100
 1,  1,  0,  0
 1,  1,  0,  0

    最小值为12（00001100），编码每个非零像素所需的位数为5（101），log 2 （28） + 1 - 12）〜= 4.09。

    最后，让我们编码非零像素：

      [ 12, 21, 20, 28, 23 ]
- [ 12, 12, 12, 12, 12 ]
------------------------
= [  0,  9,  8, 16, 11 ] = [ 00000, 01001, 01000, 10000, 01011 ]

    所以，这个块的最终编码是：

    1 0000100011001100 00001100 101 00000 01001 01000 10000 01011

    ，长度为53位（与未压缩格式的16 * 8 = 128位相反）。

    但是，最大的增益来自空块，它们被编码为一个单独的位。矩阵中有许多空块的事实是该算法中的一个重要假设。

    演示

    以下是一些处理原始数据集的JS演示代码：

    var nEmpty, nFilled;

function compress(matrix) {
  var x, y, data = '';

  nEmpty = nFilled = 0;
  
  for(y = 0; y < 44; y += 4) {
    for(x = 0; x < 52; x += 4) {
      data += compressBlock(matrix, x, y);
    }
  }
  console.log("Empty blocks: " + nEmpty);
  console.log("Filled blocks: " + nFilled);
  console.log("Average bits per block: " + (data.length / (nEmpty + nFilled)).toFixed(2));
  console.log("Average bits per filled block: " + ((data.length - nEmpty) / nFilled).toFixed(2));
  console.log("Final packed size: " + data.length + " bits --> " + ((data.length + 7) >> 3) + " bytes");
}

function compressBlock(matrix, x, y) {
  var min = 0x100, max = 0, msk = 0, data = [],
      width, v, x0, y0;
  
  for(y0 = 0; y0 < 4; y0++) {
    for(x0 = 0; x0 < 4; x0++) {
      if(v = matrix[y + y0][x + x0]) {
        msk |= 1 << (15 - y0 * 4 - x0);
        data.push(v);
        min = Math.min(min, v);
        max = Math.max(max, v);
      }
    }
  }
  if(msk) {
    nFilled++;
    width = Math.ceil(Math.log(max + 1 - min) / Math.log(2));
    data = data.map(function(v) { return bin(v - min, width); }).join('');
    return '1' + bin(msk, 16) + bin(min, 8) + ((msk & -msk) == msk ? '' : bin(width, 3) + data);
  }
  nEmpty++;
  return '0';
}

function bin(n, sz) {
  var b = n.toString(2);
  return Array(sz + 1 - b.length).join('0') + b;
}

compress([
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  [ 0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0, 0 ],
  [ 0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0, 0 ],
  [ 0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0, 0 ],
  [ 0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0, 0 ],
  [ 0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0, 0 ],
  [ 0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0, 0 ]
]);

    最终输出 349字节长。

    Empty blocks: 102
Filled blocks: 41
Average bits per block: 19.50
Average bits per filled block: 65.51
Final packed size: 2788 bits --> 349 bytes

答案 1 :(得分：1)

我会测试JPEG-LS。它是一种非常快速的算法，可为许多类型的图像提供最先进的无损压缩结果。特别是，它的预测算法将为平坦区域提供与RLE相当的结果，并为脚区域提供更好的结果。

由于您正在传输多个帧，并且这些帧可能非常相似，因此您可能需要在应用JPEG-LS之前尝试从下一帧中减去一帧（您可能需要先将像素重新映射为正整数但是，使用JPEG-LS。

如果您不需要严格无损压缩（即，如果您可以容忍重建图像中的某些失真），您可以测试近无损模式，该模式限制在任何给定像素中引入的最大绝对误差。

您可以在https://jpeg.org/jpegls/software.html找到一个非常完善且完整的实施方案。

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