List all pixel coordinates in ROI

[Charlie MAO]

Falks

Does anyone know how to obtain all pixel (x, y) coordinates from selected
ROI from either macro or plugin?

Thanks

Charlie Mao

[Michael Cammer]

One problem is that the getCoordinates command does not return all
coordinates.  When there is a horizontal or vertical, it only returns
vertices.  And for polygon tool it always returns only vertices.  Thus the
following:



//  Macro by Michael Cammer [hidden email]
// This routine gets all the pixels along a ROI, not just vertices.
//  It only works with the magic wand tool or a complete freehand ROI tool.
//  It does not work with the polygon tool!
      getSelectionCoordinates(x_, y_);
      x = newArray(x_.length);
      y = newArray(x_.length);
      for (i=0; i<x_.length; i){
         x[i] = x_[i];
         y[i] = y_[i];
         }
       x[x_.length] = x_[0];
       y[y_.length] = y_[0];

      //print("*************************************************************");
      //for (i=0; i<x.length; i)
          //print(x[i]" "[i]);
      //print("------------");

      newx = newArray(65535);
      newy = newArray(65535);
      newindex = 0;
      for (i=0; i<x.length; i){
          if (i < (x.length-1)){
            if (x[i] == x[i]){
              y0 = y[i];
              y1 = y[i];
              dy = y1 - y0;
              if (dy > 0) step = 1; else step = -1;
              for (k=y0; k!=y1; k=k×) {
                newx[newindex] = x[i];
                newy[newindex] = k;
                newindex;
              } // for k
           } //  if (x[i] == x[i])

            if (y[i] == y[i]){
              x0 = x[i];
              x1 = x[i];
              dx = x1 - x0;
              if (dx > 0) step = 1; else step = -1;
              for (k=x0; k!=x1; k=k×) {
                newx[newindex] = k;
                newy[newindex] = y[i];
                newindex;
              } // for k
           } //  if (y[i] == y[i])

          if ( (x[i] != x[i]) && (y[i]!=y[i]) ){
            newx[newindex] = x[i];
            newy[newindex] = y[i];
            newindex;      }

      } // if  (i < (x.length-1))
      } // for i

// This is the end of the routine that gets the coordinates along the edge.
// The coordinates are stored in the arrays x and y.




At 03:21 AM 07/28/05 -0400, you wrote:
>Falks
>
>Does anyone know how to obtain all pixel (x, y) coordinates from selected
>ROI from either macro or plugin?
>
>Thanks
>
>Charlie Mao

____________________________________________________________________________
Michael Cammer   Analytical Imaging Facility   Albert Einstein Coll. of Med.
Jack & Pearl Resnick Campus      1300 Morris Park Ave.     Bronx, NY  10461
[phone]       Fax:  [phone]      URL:  http://www.aecom.yu.edu/aif/
   **This electronic transmission contains information that is privileged.**



____________________________________________________________________________
Michael Cammer   Analytical Imaging Facility   Albert Einstein Coll. of Med.
Jack & Pearl Resnick Campus      1300 Morris Park Ave.     Bronx, NY  10461
[phone]       Fax:  [phone]      URL:  http://www.aecom.yu.edu/aif/
   **This electronic transmission contains information that is privileged.**

[Daniel Pensold]

Hello,
Iam new at this forum/mailing list and found this interesting macro here, but unfortunately I didn’t get it work :(
maybe there is already another function or macro, which gets all x,y-coordinates of a free-hand ROI?
I have not much experience with macro editing and Iam working either with ImageJ or Fiji
As I went  step by step through this macro I found some troubles which I couldn’t solve or explain
*******
//  Macro by Michael Cammer [hidden email]
// This routine gets all the pixels along a ROI, not just vertices.
//  It only works with the magic wand tool or a complete freehand ROI tool.
//  It does not work with the polygon tool!
      getSelectionCoordinates(x_, y_);
      x = newArray(x_.length);
      y = newArray(x_.length);
      for (i=0; i<x_.length; i){    --> should it be here ; i++ so the index increase per round?
         x[i] = x_[i];
         y[i] = y_[i];
         }
       x[x_.length] = x_[0];  --> I don’t get, what this should do... I get an “out of range” error, as the new array x is always x_.length-1 as it starts from zero. So should this function add a new value at the end (corresponding to the first value of the old x_) or should it overwrite the last value of the new array? If it should add a new value it would be easy to create the x-array in the beginning with x_.length+1
       y[y_.length] = y_[0];  --> same as comment before

      //print("*************************************************************");
      //for (i=0; i<x.length; i)       --> should it be here ; i++ so the index increase per round?
          //print(x[i]" "[i]);
      //print("------------");
 
--> From here the script results in an endless calculation or is doing nothing

      newx = newArray(65535);
      newy = newArray(65535);
      newindex = 0;
      for (i=0; i<x.length; i){    --> should it be here ; i++ so the index increase per round?
          if (i < (x.length-1)){
            if (x[i] == x[i]){
              y0 = y[i];
              y1 = y[i];  --> should this be really the same as the line before? In my opinion the next if function is therefore always wrong, as dy = 0? Maybe y1 = y[i+1]?
              dy = y1 - y0;
              if (dy > 0) step = 1; else step = -1;
              for (k=y0; k!=y1; k=k×) {   --> what means k×, I only find this term by google as kmeans and don’t understand its function here
                newx[newindex] = x[i];
                newy[newindex] = k;
                newindex;
              } // for k
           } //  if (x[i] == x[i])

            if (y[i] == y[i]){
              x0 = x[i];
              x1 = x[i];  --> same as 2 comments above
              dx = x1 - x0;
              if (dx > 0) step = 1; else step = -1;
              for (k=x0; k!=x1; k=k×) {
                newx[newindex] = k;
                newy[newindex] = y[i];
                newindex;
              } // for k
           } //  if (y[i] == y[i])

          if ( (x[i] != x[i]) && (y[i]!=y[i]) ){
            newx[newindex] = x[i];
            newy[newindex] = y[i];
            newindex;      }

      } // if  (i < (x.length-1))
      } // for i

// This is the end of the routine that gets the coordinates along the edge.
// The coordinates are stored in the arrays x and y.
******
When I change the script as stated in the comments, then it runs without error, but the newindex stays 0 after running, so I think there happened nothing?

Thanks
Daniel

[Rasband, Wayne (NIH/NIMH) [E]]

On May 5, 2014, at 1:06 PM, Daniel Pensold wrote:

> Hello,
> Iam new at this forum/mailing list and found this interesting macro here,
> but unfortunately I didn’t get it work :(
> maybe there is already another function or macro, which gets all
> x,y-coordinates of a free-hand ROI?

You can do this with two lines of macro code:

  run("Interpolate", "interval=1"); // Edit>Selection>Interpolate
  getSelectionCoordinates(xpoints, ypoints);

-wayne

> I have not much experience with macro editing and Iam working either with
> ImageJ or Fiji
> As I went  step by step through this macro I found some troubles which I
> couldn’t solve or explain
> *******
> //  Macro by Michael Cammer [hidden email]
> // This routine gets all the pixels along a ROI, not just vertices.
> //  It only works with the magic wand tool or a complete freehand ROI tool.
> //  It does not work with the polygon tool!
>      getSelectionCoordinates(x_, y_);
>      x = newArray(x_.length);
>      y = newArray(x_.length);
>      for (i=0; i<x_.length; i){    --> should it be here ; i++ so the index
> increase per round?
>         x[i] = x_[i];
>         y[i] = y_[i];
>         }
>       x[x_.length] = x_[0];  *--> I don’t get, what this should do... I get
> an “out of range” error, as the new array x is always x_.length-1 as it
> starts from zero. So should this function add a new value at the end
> (corresponding to the first value of the old x_) or should it overwrite the
> last value of the new array? If it should add a new value it would be easy
> to create the x-array in the beginning with x_.length+1*
>       y[y_.length] = y_[0];  *--> same as comment before*
>
>
> //print("*************************************************************");
>      //for (i=0; i<x.length; i)       --> should it be here ; i++ so the
> index increase per round?
>          //print(x[i]" "[i]);
>      //print("------------");
>
> *--> From here the script results in an endless calculation or is doing
> nothing *
>
>      newx = newArray(65535);
>      newy = newArray(65535);
>      newindex = 0;
>      for (i=0; i<x.length; i){    --> should it be here ; i++ so the index
> increase per round?
>          if (i < (x.length-1)){
>            if (x[i] == x[i]){
>              y0 = y[i];
>              y1 = y[i];  *--> should this be really the same as the line
> before? In my opinion the next if function is therefore always wrong, as dy
> = 0? Maybe y1 = y[i+1]?*
>              dy = y1 - y0;
>              if (dy > 0) step = 1; else step = -1;
>              for (k=y0; k!=y1; k=k×) {   *--> what means k×, I only find
> this term by google as kmeans and don’t understand its function here*
>                newx[newindex] = x[i];
>                newy[newindex] = k;
>                newindex;
>              } // for k
>           } //  if (x[i] == x[i])
>
>            if (y[i] == y[i]){
>              x0 = x[i];
>              x1 = x[i];  *--> same as 2 comments above*
>              dx = x1 - x0;
>              if (dx > 0) step = 1; else step = -1;
>              for (k=x0; k!=x1; k=k×) {
>                newx[newindex] = k;
>                newy[newindex] = y[i];
>                newindex;
>              } // for k
>           } //  if (y[i] == y[i])
>
>          if ( (x[i] != x[i]) && (y[i]!=y[i]) ){
>            newx[newindex] = x[i];
>            newy[newindex] = y[i];
>            newindex;      }
>
>      } // if  (i < (x.length-1))
>      } // for i
>
> // This is the end of the routine that gets the coordinates along the edge.
> // The coordinates are stored in the arrays x and y.
> ******
> When I change the script as stated in the comments, then it runs without
> error, but the newindex stays 0 after running, so I think there happened
> nothing?
>
> Thanks
> Daniel
>
>
>
>
> --
> View this message in context: http://imagej.1557.x6.nabble.com/List-all-pixel-coordinates-in-ROI-tp3705127p5007571.html
> Sent from the ImageJ mailing list archive at Nabble.com.
>
> --
> ImageJ mailing list: http://imagej.nih.gov/ij/list.html

--
ImageJ mailing list: http://imagej.nih.gov/ij/list.html

[Daniel Pensold]

Thx for your reply. Unfortunately this return only the border of the selection, but I need all x,y coordinates of the whole selection/ROI (also the inside coordinates).

best regards
daniel

Rasband, Wayne (NIH/NIMH) [E] wrote

On May 5, 2014, at 1:06 PM, Daniel Pensold wrote:

> Hello,
> Iam new at this forum/mailing list and found this interesting macro here,
> but unfortunately I didn’t get it work :(
> maybe there is already another function or macro, which gets all
> x,y-coordinates of a free-hand ROI?

You can do this with two lines of macro code:

  run("Interpolate", "interval=1"); // Edit>Selection>Interpolate
  getSelectionCoordinates(xpoints, ypoints);

-wayne

> I have not much experience with macro editing and Iam working either with
> ImageJ or Fiji
> As I went  step by step through this macro I found some troubles which I
> couldn’t solve or explain
> *******
> //  Macro by Michael Cammer [hidden email]
> // This routine gets all the pixels along a ROI, not just vertices.
> //  It only works with the magic wand tool or a complete freehand ROI tool.
> //  It does not work with the polygon tool!
>      getSelectionCoordinates(x_, y_);
>      x = newArray(x_.length);
>      y = newArray(x_.length);
>      for (i=0; i<x_.length; i){    --> should it be here ; i++ so the index
> increase per round?
>         x[i] = x_[i];
>         y[i] = y_[i];
>         }
>       x[x_.length] = x_[0];  *--> I don’t get, what this should do... I get
> an “out of range” error, as the new array x is always x_.length-1 as it
> starts from zero. So should this function add a new value at the end
> (corresponding to the first value of the old x_) or should it overwrite the
> last value of the new array? If it should add a new value it would be easy
> to create the x-array in the beginning with x_.length+1*
>       y[y_.length] = y_[0];  *--> same as comment before*
>
>
> //print("*************************************************************");
>      //for (i=0; i<x.length; i)       --> should it be here ; i++ so the
> index increase per round?
>          //print(x[i]" "[i]);
>      //print("------------");
>
> *--> From here the script results in an endless calculation or is doing
> nothing *
>
>      newx = newArray(65535);
>      newy = newArray(65535);
>      newindex = 0;
>      for (i=0; i<x.length; i){    --> should it be here ; i++ so the index
> increase per round?
>          if (i < (x.length-1)){
>            if (x[i] == x[i]){
>              y0 = y[i];
>              y1 = y[i];  *--> should this be really the same as the line
> before? In my opinion the next if function is therefore always wrong, as dy
> = 0? Maybe y1 = y[i+1]?*
>              dy = y1 - y0;
>              if (dy > 0) step = 1; else step = -1;
>              for (k=y0; k!=y1; k=k×) {   *--> what means k×, I only find
> this term by google as kmeans and don’t understand its function here*
>                newx[newindex] = x[i];
>                newy[newindex] = k;
>                newindex;
>              } // for k
>           } //  if (x[i] == x[i])
>
>            if (y[i] == y[i]){
>              x0 = x[i];
>              x1 = x[i];  *--> same as 2 comments above*
>              dx = x1 - x0;
>              if (dx > 0) step = 1; else step = -1;
>              for (k=x0; k!=x1; k=k×) {
>                newx[newindex] = k;
>                newy[newindex] = y[i];
>                newindex;
>              } // for k
>           } //  if (y[i] == y[i])
>
>          if ( (x[i] != x[i]) && (y[i]!=y[i]) ){
>            newx[newindex] = x[i];
>            newy[newindex] = y[i];
>            newindex;      }
>
>      } // if  (i < (x.length-1))
>      } // for i
>
> // This is the end of the routine that gets the coordinates along the edge.
> // The coordinates are stored in the arrays x and y.
> ******
> When I change the script as stated in the comments, then it runs without
> error, but the newindex stays 0 after running, so I think there happened
> nothing?
>
> Thanks
> Daniel
>
>
>
>
> --
> View this message in context: http://imagej.1557.x6.nabble.com/List-all-pixel-coordinates-in-ROI-tp3705127p5007571.html
> Sent from the ImageJ mailing list archive at Nabble.com.
>
> --
> ImageJ mailing list: http://imagej.nih.gov/ij/list.html

--
ImageJ mailing list: http://imagej.nih.gov/ij/list.html

[Michael Schmid]

Hi Daniel,

you can use selectionContains(x, y) to determine whether a point is inside a ROI.
To speed up the macro, don't do it for all points but only for these inside the bounding rectangle of the selection, like in this macro:

//needs an image with an area selection
getSelectionBounds(x0, y0, width, height);
for (y=y0; y<y0+height; y++)
  for (x=x0; x<x0+width; x++) {
    if (selectionContains(x, y))
      print(x," ",y);
  }

Michael
________________________________________________________________
On May 6, 2014, at 09:47, Daniel Pensold wrote:

> Thx for your reply. Unfortunately this return only the border of the
> selection, but I need all x,y coordinates of the whole selection/ROI (also
> the inside coordinates).
>
> best regards
> daniel
>
>
> Rasband, Wayne (NIH/NIMH) [E] wrote
>> On May 5, 2014, at 1:06 PM, Daniel Pensold wrote:
>>
>>> Hello,
>>> Iam new at this forum/mailing list and found this interesting macro here,
>>> but unfortunately I didn’t get it work :(
>>> maybe there is already another function or macro, which gets all
>>> x,y-coordinates of a free-hand ROI?
>>
>> You can do this with two lines of macro code:
>>
>>  run("Interpolate", "interval=1"); // Edit>Selection>Interpolate
>>  getSelectionCoordinates(xpoints, ypoints);
>>
>> -wayne
>>
>>> I have not much experience with macro editing and Iam working either with
>>> ImageJ or Fiji
>>> As I went  step by step through this macro I found some troubles which I
>>> couldn’t solve or explain
>>> *******
>>> //  Macro by Michael Cammer [hidden email]
>>> // This routine gets all the pixels along a ROI, not just vertices.
>>> //  It only works with the magic wand tool or a complete freehand ROI
>>> tool.
>>> //  It does not work with the polygon tool!
>>>     getSelectionCoordinates(x_, y_);
>>>     x = newArray(x_.length);
>>>     y = newArray(x_.length);
>>>     for (i=0; i&lt;x_.length; i){    --&gt; should it be here ; i++ so
>>> the index
>>> increase per round?
>>>        x[i] = x_[i];
>>>        y[i] = y_[i];
>>>        }
>>>      x[x_.length] = x_[0];  *--> I don’t get, what this should do... I
>>> get
>>> an “out of range” error, as the new array x is always x_.length-1 as it
>>> starts from zero. So should this function add a new value at the end
>>> (corresponding to the first value of the old x_) or should it overwrite
>>> the
>>> last value of the new array? If it should add a new value it would be
>>> easy
>>> to create the x-array in the beginning with x_.length+1*
>>>      y[y_.length] = y_[0];  *--> same as comment before*
>>>
>>>
>>> //print("*************************************************************");
>>>     //for (i=0; i&lt;x.length; i)       --&gt; should it be here ; i++
>>> so the
>>> index increase per round?
>>>         //print(x[i]" "[i]);
>>>     //print("------------");
>>>
>>> *--> From here the script results in an endless calculation or is doing
>>> nothing *
>>>
>>>     newx = newArray(65535);
>>>     newy = newArray(65535);
>>>     newindex = 0;
>>>     for (i=0; i&lt;x.length; i){    --&gt; should it be here ; i++ so
>>> the index
>>> increase per round?
>>>         if (i < (x.length-1)){
>>>           if (x[i] == x[i]){
>>>             y0 = y[i];
>>>             y1 = y[i];  *--> should this be really the same as the line
>>> before? In my opinion the next if function is therefore always wrong, as
>>> dy
>>> = 0? Maybe y1 = y[i+1]?*
>>>             dy = y1 - y0;
>>>             if (dy > 0) step = 1; else step = -1;
>>>             for (k=y0; k!=y1; k=k×) {   *--> what means k×, I only find
>>> this term by google as kmeans and don’t understand its function here*
>>>               newx[newindex] = x[i];
>>>               newy[newindex] = k;
>>>               newindex;
>>>             } // for k
>>>          } //  if (x[i] == x[i])
>>>
>>>           if (y[i] == y[i]){
>>>             x0 = x[i];
>>>             x1 = x[i];  *--> same as 2 comments above*
>>>             dx = x1 - x0;
>>>             if (dx > 0) step = 1; else step = -1;
>>>             for (k=x0; k!=x1; k=k×) {
>>>               newx[newindex] = k;
>>>               newy[newindex] = y[i];
>>>               newindex;
>>>             } // for k
>>>          } //  if (y[i] == y[i])
>>>
>>>         if ( (x[i] != x[i]) && (y[i]!=y[i]) ){
>>>           newx[newindex] = x[i];
>>>           newy[newindex] = y[i];
>>>           newindex;      }
>>>
>>>     } // if  (i < (x.length-1))
>>>     } // for i
>>>
>>> // This is the end of the routine that gets the coordinates along the
>>> edge.
>>> // The coordinates are stored in the arrays x and y.
>>> ******
>>> When I change the script as stated in the comments, then it runs without
>>> error, but the newindex stays 0 after running, so I think there happened
>>> nothing?
>>>
>>> Thanks
>>> Daniel
>>>
>>>
>>>
>>>
>>> --
>>> View this message in context:
>>> http://imagej.1557.x6.nabble.com/List-all-pixel-coordinates-in-ROI-tp3705127p5007571.html
>>> Sent from the ImageJ mailing list archive at Nabble.com.
>>>
>>> --
>>> ImageJ mailing list: http://imagej.nih.gov/ij/list.html
>>
>> --
>> ImageJ mailing list: http://imagej.nih.gov/ij/list.html
>
>
>
>
>
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