Question

我找到了一种使用vtk显示dcm图像的方法。但是vtk太多了我想要的东西，我只想显示一个dcm图像。 dcmtk将为我处理dcm图像。那么我有一个简单的方法来显示dcm图像吗？提前谢谢。

Answer 1

最小的学习曲线和代码要求可能是使用Grass Roots DICOM。（http://gdcm.sourceforge.net/wiki/index.php/Main_Page）此库将链接到Qt并为您提供加载图像的快速方法。唯一要记住的是DICOM图像文件不包含Qt（或其他任何东西）可以直接显示的图像。您必须加载DICOM数据并转换图像才能显示它。

这些是要添加到项目文件的行。请注意，路径必须与您的机器和版本匹配，而不是我的;

    #INCLUDEPATH += /usr/local/include/gdcm-2.4/
    #LIBS += -L"/usr/local/lib/" -lgdcmCommon -lgdcmDICT -lgdcmDSED -lgdcmIOD -lgdcmMEXD -lgdcmMSFF -lgdcmjpeg12 -lgdcmjpeg16 -lgdcmopenjpeg -lgdcmjpeg8
    #LIBS += -L"/usr/local/lib/" -lgdcmcharls -lexpat -lgdcmzlib

这是一个示例转换器，一旦你加载了dicom图像，它就会将它转换为Qt QImage。

bool imageConverters::convertToFormat_RGB888(gdcm::Image const & gimage, char *buffer, QImage* &imageQt)
{
    unsigned int dimX;
    unsigned int dimY;
    int         photoInterp;

  const unsigned int* dimension = gimage.GetDimensions();

  if (dimension == 0)
  {
      dimX = 800;
      dimY = 600;
  }
    else
    {
        dimX = dimension[0];
        dimY = dimension[1];
   }
    gimage.GetBuffer(buffer);
    photoInterp = gimage.GetPhotometricInterpretation();

   qDebug() << "photo interp = " << photoInterp;
   qDebug() << "pixel format = " << gimage.GetPixelFormat();
    // Let's start with the easy case:

    if( photoInterp == gdcm::PhotometricInterpretation::RGB )
    {
        if( gimage.GetPixelFormat() != gdcm::PixelFormat::UINT8 )
        {
            return false;
        }
        unsigned char *ubuffer = (unsigned char*)buffer;
        // QImage::Format_RGB888  13  The image is stored using a 24-bit RGB format (8-8-8).Format_RGB888 Format_ARGB32
        imageQt = new QImage((unsigned char *)ubuffer, dimX, dimY, 3*dimX, QImage::Format_RGB888);
        //imageQt = &imageQt->rgbSwapped();
    }
    else
    if( photoInterp == gdcm::PhotometricInterpretation::MONOCHROME2 ||
            photoInterp == gdcm::PhotometricInterpretation::MONOCHROME1
            )
    {
        if( gimage.GetPixelFormat() == gdcm::PixelFormat::UINT8 || gimage.GetPixelFormat() == gdcm::PixelFormat::INT8
                || gimage.GetPixelFormat() == gdcm::PixelFormat::UINT16)
        {
            // We need to copy each individual 8bits into R / G and B:
            unsigned char *ubuffer = new unsigned char[dimX*dimY*3];
            unsigned char *pubuffer = ubuffer;
            for(unsigned int i = 0; i < dimX*dimY; i++)
            {
                *pubuffer++ = *buffer;
                *pubuffer++ = *buffer;
                *pubuffer++ = *buffer++;
            }

        imageQt = new QImage(ubuffer, dimX, dimY, QImage::Format_RGB888);
    }
    else
    if( gimage.GetPixelFormat() == gdcm::PixelFormat::INT16 )
    {
      // We need to copy each individual 16bits into R / G and B (truncate value)
      short *buffer16 = (short*)buffer;
      unsigned char *ubuffer = new unsigned char[dimX*dimY*3];
      unsigned char *pubuffer = ubuffer;
      for(unsigned int i = 0; i < dimX*dimY; i++)
        {
        // Scalar Range of gdcmData/012345.002.050.dcm is [0,192], we could simply do:
        // *pubuffer++ = *buffer16;
        // *pubuffer++ = *buffer16;
        // *pubuffer++ = *buffer16;
        // instead do it right:
        *pubuffer++ = (unsigned char)std::min(255, (32768 + *buffer16) / 255);
        *pubuffer++ = (unsigned char)std::min(255, (32768 + *buffer16) / 255);
        *pubuffer++ = (unsigned char)std::min(255, (32768 + *buffer16) / 255);
        buffer16++;
        }

      imageQt = new QImage(ubuffer, dimX, dimY, QImage::Format_RGB888);
      }
    else
      {
      std::cerr << "Pixel Format is: " << gimage.GetPixelFormat() << std::endl;
      return false;
      }
    }
  else
    {
    std::cerr << "Unhandled PhotometricInterpretation: " << gimage.GetPhotometricInterpretation() << std::endl;
    return false;
    }

  return true;
}

Answer 2

@john elemans
如果我使用你给我的代码，那似乎就是故事。我的图像像素格式为UINT16，因此程序将执行以下句子。

        unsigned char *ubuffer = new unsigned char[dimX*dimY*3];
        unsigned char *pubuffer = ubuffer;
        for(unsigned int i = 0; i < dimX*dimY; i++)
        {
            *pubuffer++ = *buffer;
            *pubuffer++ = *buffer;
            *pubuffer++ = *buffer++;
        }

        imageQt = new QImage(ubuffer, dimX, dimY, QImage::Format_RGB888);

但转换后，结果不对。
原始图像是这样的：before
这是已转换的图像：after
结果证明代码不对。我也尝试了其他方法。我试过的其中一个代码是：

    short *buffer16 = (short*)buffer;
    unsigned char *ubuffer = new unsigned char[dimX*dimY*3];
    unsigned char *pubuffer = ubuffer;
    for (unsigned int i = 0; i < dimX*dimY; i++)
    {
         *pubuffer++ = *buffer16;
         *pubuffer++ = *buffer16;
         *pubuffer++ = *buffer16;
          buffer16++;
    }
    imageQt = new QImage(ubuffer, dimX, dimY, QImage::Format_RGB888);

在我使用此代码转换图像后，我几乎相信我已经成功了。但结果是这样的：对不起，我的帖子超过2个链接我没有足够的声誉。
我只想将DICOM图像转换为位图，但我不知道如何。
最后，谢谢你的帮助。

如何编写qt应用程序来显示dcm图像？

2 个答案: