将光栅字节[]图像数据转换为C#中的列格式
Convert raster byte[] image data to column Format in C#
我正在尝试将字节数组从 Raster
格式的图像(从左到右读取)转换为 Column
格式(从上到下读取)。
问题看起来很简单,我们有二维位数组(width/height 图像)。在 Raster
格式中,您从左到右读取位,在 Column
格式中,您从上到下读取位。
我尝试这样做是为了支持ESC/POS
协议的Column
格式打印。我已经有了 Raster
格式的图像,现在我正在尝试将其转换为 Column
格式。
ESC/POS
Raster
打印文档:
ESC/POS
Column
打印文档:
- https://reference.epson-biz.com/modules/ref_escpos/index.php?content_id=106#gs_lparen_cl_fn68
- https://reference.epson-biz.com/modules/ref_escpos/index.php?content_id=90
目前,我通过直接使用 BitArray
处理位来进行转换。这个解决方案不是最优的,在我看来有必要直接在 Byte
中工作。
private byte[] ConvertRasterToColumnFormat(byte[] rasterData, int width, int height)
{
var finalHeight = height;
while (finalHeight % 8 != 0) finalHeight++;
var finalWidth = width;
while (finalWidth % 8 != 0) finalWidth++;
var rasterBitArray = new BitArray(rasterData);
var columnBitArray = new BitArray(finalHeight * finalWidth);
for (int y = 0; y < height; y++)
{
for (int x = 0; x < width; x++)
{
var rasterPosition = y * finalWidth;
var columnPosition = x * finalHeight;
rasterPosition += (x / 8) * 8;
columnPosition += (y / 8) * 8;
rasterPosition += 7 - x % 8;
columnPosition += 7 - y % 8;
var value = rasterBitArray[rasterPosition];
columnBitArray[columnPosition] = value;
}
}
var result = new byte[columnBitArray.Length / 8];
columnBitArray.CopyTo(result, 0);
return result;
}
.NET Fiddle 测试:https://dotnetfiddle.net/NBRBgt
有没有人有更好的解决方案?
一种可能的方法是获取一个 8x8 位块,使用 bitboard techniques 转置它,然后存储结果的字节。它不是很漂亮,但它避免了处理单个位,当然也避免了 BitArray
(即使在逐位方法中也很慢)。
以下代码通过了您的测试用例,但可能应该进行更好的测试..
private static byte[] ConvertRasterToColumnFormat(byte[] rasterData, int width, int height)
{
int h = height + 7 & -8;
int w = (width + 7) >> 3;
int hsmall = h >> 3;
var result = new byte[h * w];
for (int y = 0; y < height; y += 8)
{
for (int x = 0; x < w; x++)
{
// grab 8x8 block of bits
int i = x + w * y;
ulong block = rasterData[i];
if (i + w < rasterData.Length)
block |= (ulong)rasterData[i + w] << 8;
if (i + w * 2 < rasterData.Length)
block |= (ulong)rasterData[i + w * 2] << 16;
if (i + w * 3 < rasterData.Length)
block |= (ulong)rasterData[i + w * 3] << 24;
if (i + w * 4 < rasterData.Length)
block |= (ulong)rasterData[i + w * 4] << 32;
if (i + w * 5 < rasterData.Length)
block |= (ulong)rasterData[i + w * 5] << 40;
if (i + w * 6 < rasterData.Length)
block |= (ulong)rasterData[i + w * 6] << 48;
if (i + w * 7 < rasterData.Length)
block |= (ulong)rasterData[i + w * 7] << 56;
// transpose 8x8
// https://www.chessprogramming.org/Flipping_Mirroring_and_Rotating#Anti-Diagonal
ulong t;
const ulong k1 = 0xaa00aa00aa00aa00;
const ulong k2 = 0xcccc0000cccc0000;
const ulong k4 = 0xf0f0f0f00f0f0f0f;
t = block ^ (block << 36);
block ^= k4 & (t ^ (block >> 36));
t = k2 & (block ^ (block << 18));
block ^= t ^ (t >> 18);
t = k1 & (block ^ (block << 9));
block ^= t ^ (t >> 9);
// write block to columns
int j = (y >> 3) + h * x;
result[j] = (byte)block;
result[j + hsmall] = (byte)(block >> 8);
result[j + hsmall * 2] = (byte)(block >> 16);
result[j + hsmall * 3] = (byte)(block >> 24);
result[j + hsmall * 4] = (byte)(block >> 32);
result[j + hsmall * 5] = (byte)(block >> 40);
result[j + hsmall * 6] = (byte)(block >> 48);
result[j + hsmall * 7] = (byte)(block >> 56);
}
}
return result;
}
我在 256x256 阵列上的设置(Intel 4770K、x64、.NET Core 3.0)的简单基准测试给出了:
old: 1103 ticks
new: 56 ticks
非常明显的区别。
我正在尝试将字节数组从 Raster
格式的图像(从左到右读取)转换为 Column
格式(从上到下读取)。
问题看起来很简单,我们有二维位数组(width/height 图像)。在 Raster
格式中,您从左到右读取位,在 Column
格式中,您从上到下读取位。
我尝试这样做是为了支持ESC/POS
协议的Column
格式打印。我已经有了 Raster
格式的图像,现在我正在尝试将其转换为 Column
格式。
ESC/POS
Raster
打印文档:
ESC/POS
Column
打印文档:
- https://reference.epson-biz.com/modules/ref_escpos/index.php?content_id=106#gs_lparen_cl_fn68
- https://reference.epson-biz.com/modules/ref_escpos/index.php?content_id=90
目前,我通过直接使用 BitArray
处理位来进行转换。这个解决方案不是最优的,在我看来有必要直接在 Byte
中工作。
private byte[] ConvertRasterToColumnFormat(byte[] rasterData, int width, int height)
{
var finalHeight = height;
while (finalHeight % 8 != 0) finalHeight++;
var finalWidth = width;
while (finalWidth % 8 != 0) finalWidth++;
var rasterBitArray = new BitArray(rasterData);
var columnBitArray = new BitArray(finalHeight * finalWidth);
for (int y = 0; y < height; y++)
{
for (int x = 0; x < width; x++)
{
var rasterPosition = y * finalWidth;
var columnPosition = x * finalHeight;
rasterPosition += (x / 8) * 8;
columnPosition += (y / 8) * 8;
rasterPosition += 7 - x % 8;
columnPosition += 7 - y % 8;
var value = rasterBitArray[rasterPosition];
columnBitArray[columnPosition] = value;
}
}
var result = new byte[columnBitArray.Length / 8];
columnBitArray.CopyTo(result, 0);
return result;
}
.NET Fiddle 测试:https://dotnetfiddle.net/NBRBgt
有没有人有更好的解决方案?
一种可能的方法是获取一个 8x8 位块,使用 bitboard techniques 转置它,然后存储结果的字节。它不是很漂亮,但它避免了处理单个位,当然也避免了 BitArray
(即使在逐位方法中也很慢)。
以下代码通过了您的测试用例,但可能应该进行更好的测试..
private static byte[] ConvertRasterToColumnFormat(byte[] rasterData, int width, int height)
{
int h = height + 7 & -8;
int w = (width + 7) >> 3;
int hsmall = h >> 3;
var result = new byte[h * w];
for (int y = 0; y < height; y += 8)
{
for (int x = 0; x < w; x++)
{
// grab 8x8 block of bits
int i = x + w * y;
ulong block = rasterData[i];
if (i + w < rasterData.Length)
block |= (ulong)rasterData[i + w] << 8;
if (i + w * 2 < rasterData.Length)
block |= (ulong)rasterData[i + w * 2] << 16;
if (i + w * 3 < rasterData.Length)
block |= (ulong)rasterData[i + w * 3] << 24;
if (i + w * 4 < rasterData.Length)
block |= (ulong)rasterData[i + w * 4] << 32;
if (i + w * 5 < rasterData.Length)
block |= (ulong)rasterData[i + w * 5] << 40;
if (i + w * 6 < rasterData.Length)
block |= (ulong)rasterData[i + w * 6] << 48;
if (i + w * 7 < rasterData.Length)
block |= (ulong)rasterData[i + w * 7] << 56;
// transpose 8x8
// https://www.chessprogramming.org/Flipping_Mirroring_and_Rotating#Anti-Diagonal
ulong t;
const ulong k1 = 0xaa00aa00aa00aa00;
const ulong k2 = 0xcccc0000cccc0000;
const ulong k4 = 0xf0f0f0f00f0f0f0f;
t = block ^ (block << 36);
block ^= k4 & (t ^ (block >> 36));
t = k2 & (block ^ (block << 18));
block ^= t ^ (t >> 18);
t = k1 & (block ^ (block << 9));
block ^= t ^ (t >> 9);
// write block to columns
int j = (y >> 3) + h * x;
result[j] = (byte)block;
result[j + hsmall] = (byte)(block >> 8);
result[j + hsmall * 2] = (byte)(block >> 16);
result[j + hsmall * 3] = (byte)(block >> 24);
result[j + hsmall * 4] = (byte)(block >> 32);
result[j + hsmall * 5] = (byte)(block >> 40);
result[j + hsmall * 6] = (byte)(block >> 48);
result[j + hsmall * 7] = (byte)(block >> 56);
}
}
return result;
}
我在 256x256 阵列上的设置(Intel 4770K、x64、.NET Core 3.0)的简单基准测试给出了:
old: 1103 ticks
new: 56 ticks
非常明显的区别。