the question with get_textpage_ocr DPI

[xiaolibuzai-ovo]

I used different DPIs, but it seems that I got the same coordinates. Is this reasonable? Doesn't DPI affect the coordinate positions? Or is there some kind of conversion applied to the results?

code:

docs=pymupdf.open("1.pdf")
page=docs[1]
gt = page.get_textpage_ocr(flags=0,full=True,dpi=150,language='eng')
blocks=page.get_text("dict",textpage=gt)["blocks"]
with open('blocks-test.json', 'w') as f:
    json.dump(blocks, f, indent=4)  

gt = page.get_textpage_ocr(flags=0,full=True,dpi=300,language='eng')
blocks=page.get_text("dict",textpage=gt)["blocks"]
with open('blocks1-test.json', 'w') as f:
    json.dump(blocks, f, indent=4) 

result:

thanks for u help!

[JorjMcKie]

Or is there some kind of conversion applied to the results?

Of course - this is a must! The positions of text on the page cannot depend on the precision by which it is detected. Detection precision is the only sense in life of DPI.

[xiaolibuzai-ovo]

May I ask what this conversion rule is? In other words, how can I obtain the original OCR coordinates? Or how do the coordinates obtained from pdmupdf get_text('dict') correspond to the OCR coordinates?

[JorjMcKie]

When making the pixmap from the original page, it will have a Pixmap.irect. Coordinate transformation from that IRect to the page rectangle is done by the matrix created by method torect.

[xiaolibuzai-ovo]

So, is there a good way to convert the coordinates from PyMuPDF to the original OCR coordinates, assuming the DPI used is 1600?

[JorjMcKie]

I think I answered this?

• An A4 page has the rectangle Rect(0.0, 0.0, 595.0, 842.0). The page's Pixmap in default resolution is this Pixmap(DeviceRGB, (0, 0, 595, 842), 0), pix.xres, pix.yres = (96, 96).

• So the matrix transforming page coordinates to image coordinates is m = page.rect.torect(pix.irect) the identity matrix Matrix(1.0, 0.0, 0.0, 1.0, 0.0, 0.0).

• If we create the pixmap with some non-default DPI, things change like this:
  

• A page rectangle bbox = Rect(100, 100, 200, 200) (imagine a text boundary box here) will therefore have these coordinates on the image (actually more precisely the IRect version of it):
  

• And of course vice versa: given a bbox or a point on the image / pixmap, the corresponding page objects can be computed via bbox * ~m, point * ~m, where ~m is the inverse of above m.