Is possible to extract the exact position from an XObject?

[pau1996]

Hello, I want to extract the bbox from an XObject so that I can add a border in the original PDF document, I can extract the XObject properly as seen in this example: https://github.com/pymupdf/PyMuPDF-Utilities/tree/master/examples/extract-xobj, but when I try to extract the exact position I'm always getting something like: BBox [ 0 0 531 650 ] so it is always starting in (0,0) also tried applying the transformation matrix which is something like: Matrix [ .75 0 0 .75 0 0 ], but it doesn't affect the position. Is this something it can be done right now? Can you help me with that? Thanks

[JorjMcKie]

Transferring this to "Discussions" because all information you need is already present.

[JorjMcKie]

You can use page.get_xobjects() which is a recursive list of Form XObjects directly or indirectly referred to by the page. For example:

pprint(page.get_xobjects())
[(13, 'fzFrm0', 0, (100.0, 642.0, 200.0, 742.0)),
 (12, 'fullpage', 13, (0.0, 0.0, 100.0, 100.0))]

As you see, objects at xref 12 and xref 13 are interconnected:
The page's displayed fzFrm0 in turn invokes fullpage at xref 12. To get the correct final bbox, you would have to follow this stack of xobjects backwards, i.e. look for an item with 12 at item position 2, etc.
If not found, you have reached the start of the chain.

Each of the xobject in this chain has BBox and Matrix (assume identity if omitted) entries:

print(doc.xref_object(12))
<<
  /Type /XObject
  /Subtype /Form
  /BBox [ 0 0 100 100 ]
  /Matrix [ 1 0 0 1 0 0 ]
  /Resources 5 0 R
  /Length 122
  /Filter /FlateDecode
>>
print(doc.xref_object(13))
<<
  /Type /XObject
  /Subtype /Form
  /BBox [ 0 0 100 100 ]
  /Matrix [ 1 0 0 1 100 642 ]
  /Resources <<
    /XObject <<
      /fullpage 12 0 R
    >>
  >>
  /Length 20
  /Filter /FlateDecode
>>

Starting with the endpoint of this chain, multiply all matrices and finally multiply fzFrm0's BBox with the result.
You can use PyMuPDF's mupdf submodule to help with this. You may want to look at this snippet:

from pymupdf import mupdf
pdf = pymupdf._as_pdf_document(doc)  # access underlying PDF document of "doc"
xobj13 = mupdf.pdf_load_object(pdf, 13)  # load the object of fzFrm0

bbox13 = xobj13.pdf_dict_get_rect(pymupdf.PDF_NAME("BBox"))  # extract "BBox"
bbox13
(x0=0 y0=0 x1=100 y1=100)

matrix13 =xobj13.pdf_dict_get_matrix(pymupdf.PDF_NAME("Matrix"))  # extract "Matrix"
matrix13
(a=1 b=0 c=0 d=1 e=100 f=642)

# same for xobject 12:
xobj12 = mupdf.pdf_load_object(pdf, 12)  # load the object of "fullpage"
matrix12 =xobj13.pdf_dict_get_matrix(pymupdf.PDF_NAME("Matrix"))  # extract "Matrix"

# Now compute the resulting bbox. Elegant algebra is available in PyMuPDF only, so:
bbox = pymupdf.Rect(bbox13) * pymupdf.Matrix(matrix13) * pymupdf.Matrix(matrix12)
bbox
Rect(100.0, 642.0, 200.0, 742.0)

# This is in PDF coordinates! So multiply with the page transformation matrix:
bbox * page.transformation_matrix
Rect(100.0, 100.0, 200.0, 200.0)

This should work.