rep_movsl的Clobber列表(Clobber list for rep_movsl)

 大卫布拉班特让你已经朝着正确的方向发展（如果你想与DICOM合作，你一定要阅读并珍惜dclunie的医学影像常见问题 ）。 让我们看看我是否可以详细阐述它，并使其更容易实施。 
 
 我假设你有一个工具/库来从DICOM文件中提取标签（Offis'DCMTK？）。 为了举例说明，我将参考CT扫描（许多切片，即许多图像）和一个侦察图像，您要在其上显示定位线。 每个DICOM图像（包括您的CT片和您的侦察器）都包含有关它们在太空中的位置的完整信息，在这两个标签中： 
 
 
 
Group,Elem  VR Value                           Name of the tag   
---------------------------------------------------------------------
(0020,0032) DS [-249.51172\-417.51172\-821]  # ImagePositionPatient
                X0         Y0         Z0

(0020,0037) DS [1\0\0\0\1\0]                 # ImageOrientationPatient
                A B C D E F
 
 
 ImagePositionPatient以（x，y，z）表示的传输的第一个像素（左上角像素，清晰）的全局坐标为mm。 我将它们标记为X0，Y0，Z0。 ImageOrientationPatient包含两个矢量，它们都是三个分量，用于指定第一行像素的方向余弦和图像的第一列像素。 理解方向余弦不会伤害（参见http://mathworld.wolfram.com/DirectionCosine.html ），但dclunie建议的方法直接与它们一起工作，所以现在让我们说，他们给你的空间方向图像平面。 我标记他们自动对焦，使公式更容易。 
 
 现在，在dclunie给出的代码中（我相信它的目的是C，但它非常简单，它应该和Java，C＃，awk，Vala，Octave等一样好），约定如下： 
 
 
 
 scr_ * =指的是soruce图像，即CT切片 
 
 
 dst_ * =指目标图像，即侦察员 
 
 
 * _pos_x，* _pos_y，* _pos_z =上面的X0，Y0，Z0 
 
 
 * _row_dircos_x，* _row_dircos_y，* _row_dircos_z =上面的A，B，C 
 
 
 * _col_dircos_x，* _col_dircos_y，* _col_dircos_z =上面的D，E，F 
 
 
 设置正确的值后，只需应用这些值： 
 
dst_nrm_dircos_x = dst_row_dircos_y * dst_col_dircos_z 
                   - dst_row_dircos_z * dst_col_dircos_y; 
dst_nrm_dircos_y = dst_row_dircos_z * dst_col_dircos_x 
                   - dst_row_dircos_x * dst_col_dircos_z; 
dst_nrm_dircos_z = dst_row_dircos_x * dst_col_dircos_y 
                   - dst_row_dircos_y * dst_col_dircos_x; 

src_pos_x -= dst_pos_x;
src_pos_y -= dst_pos_y;
src_pos_z -= dst_pos_z;

dst_pos_x = dst_row_dircos_x * src_pos_x
          + dst_row_dircos_y * src_pos_y
          + dst_row_dircos_z * src_pos_z;

dst_pos_y = dst_col_dircos_x * src_pos_x
          + dst_col_dircos_y * src_pos_y
          + dst_col_dircos_z * src_pos_z;

dst_pos_z = dst_nrm_dircos_x * src_pos_x
          + dst_nrm_dircos_y * src_pos_y
          + dst_nrm_dircos_z * src_pos_z;
 
 或者，如果你有一些奇特的矩阵类，你可以建立这个矩阵并将它乘以你的点坐标。 
 
 
 
    [ dst_row_dircos_x  dst_row_dircos_y  dst_row_dircos_z  -dst_pos_x ] 
M = [ dst_col_dircos_x  dst_col_dircos_y  dst_col_dircos_z  -dst_pos_y ]
    [ dst_nrm_dircos_x  dst_nrm_dircos_y  dst_nrm_dircos_z  -dst_pos_z ]
    [ 0                 0                 0                 1          ]
 
 
 这将是这样的： 
 
 
 
Scout_Point(x,y,z,1) = M * CT_Point(x,y,z,1)
 
 
 说这一切，我们应该转换CT的哪些点在侦察员创建一条线？ 此外，这个dclunie已经提出了一个通用的解决方案： 
 
 “ 我的方法是投影源图像边界框的正方形（即连接切片的TLHC，TRHC，BRHC和BLHC的线）。 ” 
 
 如果您投影CT切片的四个角点，您将有一条垂直于侦察线的CT切片线，以及非垂直切片时的梯形线。 现在，如果您的CT切片与坐标轴对齐（即ImageOrientationPatient = [1 \ 0 \ 0 \ 0 \ 1 \ 0]），则这四个点很平常。 您可以使用行数/列数和沿x / y方向的像素距离计算图像的宽度/高度，并将其适当地相加。 如果你想实现一般情况，那么你需要一个小三角...或者可能不是。 如果你还没有阅读方向余弦的定义，那也许是时候了。 
 
 我会尽力让你走上正轨。 例如，在TRHC上工作时，您知道像素在图像平面中的位置： 
 
 
 
# Pixel location of the TRHC
x_pixel = number_of_columns-1 # Counting from 0
y_pixel = 0
z_pixel = 0 # We're on a plane!
 
 
 DICOM中的像素距离值被称为图像平面，因此您可以简单地将x和y乘以这些值，使其位置以毫米为单位，而z为0（像素和毫米）。 我正在谈论这些价值观： 
 
 
 
(0028,0011) US 512                       #   2, 1 Columns
(0028,0010) US 512                       #   2, 1 Rows
(0028,0030) DS [0.9765625\0.9765625]     #  20, 2 PixelSpacing
 
 
 上面的矩阵M是从全局到图像坐标的一般变换，具有可用的方向余弦。 你现在需要的是反作业（图像到全局）和源图像（CT切片）。 我会让你去挖掘几何图书，但我认为它应该是这样的（旋转部分被转置，翻译没有符号变化，当然我们使用src_ *值）： 
 
 
 
     [src_row_dircos_x  src_col_dircos_x  src_nrm_dircos_x  src_pos_x ]
M2 = [src_row_dircos_y  src_col_dircos_y  src_nrm_dircos_y  src_pos_y ]
     [src_row_dircos_z  src_col_dircos_z  src_nrm_dircos_z  src_pos_z ]
     [0                 0                 0                 1         ]
 
 
 将CT切片中的点（例如四个角）转换为毫米，然后应用M2将它们放在全局坐标中。 然后，您可以将它们提供给dclunie报告的过程。 在使用它之前交叉检查我的数学，例如用于患者诊断！ ;-) 
 
 希望这有助于理解更好的dclunie的方法。 干杯 

David Brabant put you already in the right direction (if you want to work with DICOM you should definitely read and treasure dclunie's medical image FAQ). Let's see if I can elaborate on it and make it easier for you to implement.
 
I assume you have a tool/library to extract tags from a DICOM file (Offis' DCMTK?). For the sake of exemplification I'll refer to a CT scan (many slices, i.e. many images) and a scout image, onto which you want to display localizer lines. Each DICOM image, including your CT slices and your scout, contain full information about their location in space, in these two tags:
 
 
 
Group,Elem  VR Value                           Name of the tag   
---------------------------------------------------------------------
(0020,0032) DS [-249.51172\-417.51172\-821]  # ImagePositionPatient
                X0         Y0         Z0

(0020,0037) DS [1\0\0\0\1\0]                 # ImageOrientationPatient
                A B C D E F
 
 
ImagePositionPatient has the global coordinates in mm of the first pixel transmitted (the top left-hand corner pixel, to be clear) expressed as (x,y,z). I marked them X0, Y0, Z0. ImageOrientationPatient contains two vectors, both of three components, specifying the direction cosines of the first row of pixels and first column of pixels of the image. Understanding direction cosines doesn't hurt (see e.g. http://mathworld.wolfram.com/DirectionCosine.html), but the method suggested by dclunie works directly with them, so for now let's just say they give you the orientation in space of the image plane. I marked them A-F to make formulas easier.
 
Now, in the code given by dclunie (I believe it's intended to be C, but it's so simple it should work as well as Java, C#, awk, Vala, Octave, etc.) the conventions are the following:
 
 
 
scr_* = refers to the soruce image, i.e. the CT slice 
 
 
dst_* = refers to the destination image, i.e. the scout
 
 
*_pos_x, *_pos_y, *_pos_z = the X0, Y0, Z0 above
 
 
*_row_dircos_x, *_row_dircos_y, *_row_dircos_z = the A, B, C above
 
 
*_col_dircos_x, *_col_dircos_y, *_col_dircos_z = the D, E, F above
 
 
After setting the right values just apply these:
 
dst_nrm_dircos_x = dst_row_dircos_y * dst_col_dircos_z 
                   - dst_row_dircos_z * dst_col_dircos_y; 
dst_nrm_dircos_y = dst_row_dircos_z * dst_col_dircos_x 
                   - dst_row_dircos_x * dst_col_dircos_z; 
dst_nrm_dircos_z = dst_row_dircos_x * dst_col_dircos_y 
                   - dst_row_dircos_y * dst_col_dircos_x; 

src_pos_x -= dst_pos_x;
src_pos_y -= dst_pos_y;
src_pos_z -= dst_pos_z;

dst_pos_x = dst_row_dircos_x * src_pos_x
          + dst_row_dircos_y * src_pos_y
          + dst_row_dircos_z * src_pos_z;

dst_pos_y = dst_col_dircos_x * src_pos_x
          + dst_col_dircos_y * src_pos_y
          + dst_col_dircos_z * src_pos_z;

dst_pos_z = dst_nrm_dircos_x * src_pos_x
          + dst_nrm_dircos_y * src_pos_y
          + dst_nrm_dircos_z * src_pos_z;
 
Or, if you have some fancy matrix class, you can build this matrix and multiply it with your point coordinates.
 
 
 
    [ dst_row_dircos_x  dst_row_dircos_y  dst_row_dircos_z  -dst_pos_x ] 
M = [ dst_col_dircos_x  dst_col_dircos_y  dst_col_dircos_z  -dst_pos_y ]
    [ dst_nrm_dircos_x  dst_nrm_dircos_y  dst_nrm_dircos_z  -dst_pos_z ]
    [ 0                 0                 0                 1          ]
 
 
That would be like this:
 
 
 
Scout_Point(x,y,z,1) = M * CT_Point(x,y,z,1)
 
 
Said all that, which points of the CT should we convert to create a line on the scout? Also for this dclunie already suggests a general solution:
 
"My approach is to project the square that is the bounding box of the source image (i.e. lines joining the TLHC, TRHC,BRHC and BLHC of the slice)."
 
If you project the four corner points of the CT slice, you'll have a line for CT slices perpendicular to the scout, and a trapezoid in case of non perpendicular slices. Now, if your CT slice is aligned with the coordinate axes (i.e. ImageOrientationPatient = [1\0\0\0\1\0]), the four points are trivial. You compute the width/height of the image in mm using the number of rows/columns and the pixel distance along x/y direction and sum things up appropriately. If you want to implement the generic case, then you need a little trigonometry... or maybe not. It's maybe time you read the definition of the direction cosines if you haven't yet.
 
I'll try to put you on track. E.g. working on the TRHC, you know where the voxel is in the image plane:
 
 
 
# Pixel location of the TRHC
x_pixel = number_of_columns-1 # Counting from 0
y_pixel = 0
z_pixel = 0 # We're on a plane!
 
 
The pixel distance values in DICOM are referred to the image plane, so you can simply multiply x and y by those values to have their position in mm, while z is 0 (both pixels and mm). I am talking about these values:
 
 
 
(0028,0011) US 512                       #   2, 1 Columns
(0028,0010) US 512                       #   2, 1 Rows
(0028,0030) DS [0.9765625\0.9765625]     #  20, 2 PixelSpacing
 
 
The matrix M above, is a generic transformation from global to image coordinates, having the direction cosines available. What you need now is something that does the inverse job (image to global) and on the source images (the CT slices). I'll let you go and dig in the geometry books to be sure, but I think it should be something like this (the rotation part is transposed, translation has no sign change and of course we use the src_* values):
 
 
 
     [src_row_dircos_x  src_col_dircos_x  src_nrm_dircos_x  src_pos_x ]
M2 = [src_row_dircos_y  src_col_dircos_y  src_nrm_dircos_y  src_pos_y ]
     [src_row_dircos_z  src_col_dircos_z  src_nrm_dircos_z  src_pos_z ]
     [0                 0                 0                 1         ]
 
 
Convert points in the CT slice (e.g. the four corners) to millimeters and then apply M2 to have them in global coordinates. Then you can feed them to the procedure reported by dclunie. Cross-check my maths before using it e.g. for patient diagnostics! ;-) 
 
Hope this helps understanding better dclunie's method. Cheers