At the moment I\'m looking into doing some PDF merging with pyPdf, but sometimes the inputs are not in the right order, so I\'m looking into scraping each page for its page
The other answers use PyPDF/PyPDF2 which seems to read the entire file. This takes a long time for large files.
In the meantime I wrote something quick and dirty which doesn't take nearly as long to run. It does a shell call but I wasn't aware of any other way to do it. It can get the number of pages for pdfs that are ~5000 pages very quickly.
It works by just calling the "pdfinfo" shell command, so it probably only works in linux. I've only tested it on ubuntu so far.
One strange behavior I've seen is that surrounding this in a try/except block doesn't catch errors, you have to except subprocess.CalledProcessError.
from subprocess import check_output
def get_num_pages(pdf_path):
output = check_output(["pdfinfo", pdf_path]).decode()
pages_line = [line for line in output.splitlines() if "Pages:" in line][0]
num_pages = int(pages_line.split(":")[1])
return num_pages
The following worked for me:
from PyPDF2 import PdfFileReader
pdf = PdfFileReader(open('path/to/file.pdf','rb'))
pdf.getNumPages()
For full documentation, see Adobe's 978-page PDF Reference. :-)
More specifically, the PDF file contains metadata that indicates how the PDF's physical pages are mapped to logical page numbers and how page numbers should be formatted. This is where you go for canonical results. Example 2 of this page shows how this looks in the PDF markup. You'll have to fish that out, parse it, and perform a mapping yourself.
In PyPDF, to get at this information, try, as a starting point:
pdf.trailer["/Root"]["/PageLabels"]["/Nums"]
By the way, when you see an IndirectObject
instance, you can call its getObject()
method to retrieve the actual object being pointed to.
Your alternative is, as you say, to check the text objects and try to figure out which is the page number. You could use extractText()
of the page object for this, but you'll get one string back and have to try to fish out the page number from that. (And of course the page number might be Roman or alphabetic instead of numeric, and some pages may not be numbered.) Instead, have a look at how extractText()
actually does its job—PyPDF is written in Python, after all—and use it as a basis of a routine that checks each text object on the page individually to see if it's like a page number. Be wary of TOC/index pages that have lots of page numbers on them!
Another Option is pymupdf: https://pymupdf.readthedocs.io/en/latest/tutorial.html
import fitz
doc = fitz.open('Path To File')
doc.pageCount
pip install pymupdf
For large documents I was getting a recursion error when using pypdf2 so this was another quick and simple way.
The answer by kindall is very good. However, since a working code sample was requested later (by dreamer) and since I had the same problem today, I would like to add some notes.
pdf structure is not uniform; there are rather few things you can rely on, hence any working code sample is very unlikely to work for everyone. A very good explanation can be found in this answer.
As explained by kindall, you will most likely need to explore what pdf you are dealing with.
Like so:
import sys
import PyPDF2 as pyPdf
"""Open your pdf"""
pdf = pyPdf.PdfFileReader(open(sys.argv[1], "rb"))
"""Explore the /PageLabels (if it exists)"""
try:
page_label_type = pdf.trailer["/Root"]["/PageLabels"]
print(page_label_type)
except:
print("No /PageLabel object")
"""Select the item that is most likely to contain the information you desire; e.g.
{'/Nums': [0, IndirectObject(42, 0)]}
here, we only have "/Num". """
try:
page_label_type = pdf.trailer["/Root"]["/PageLabels"]["/Nums"]
print(page_label_type)
except:
print("No /PageLabel object")
"""If you see a list, like
[0, IndirectObject(42, 0)]
get the correct item from it"""
try:
page_label_type = pdf.trailer["/Root"]["/PageLabels"]["/Nums"][1]
print(page_label_type)
except:
print("No /PageLabel object")
"""If you then have an indirect object, like
IndirectObject(42, 0)
use getObject()"""
try:
page_label_type = pdf.trailer["/Root"]["/PageLabels"]["/Nums"][1].getObject()
print(page_label_type)
except:
print("No /PageLabel object")
"""Now we have e.g.
{'/S': '/r', '/St': 21}
meaning roman numerals, starting with page 21, i.e. xxi. We can now also obtain the two variables directly."""
try:
page_label_type = pdf.trailer["/Root"]["/PageLabels"]["/Nums"][1].getObject()["/S"]
print(page_label_type)
start_page = pdf.trailer["/Root"]["/PageLabels"]["/Nums"][1].getObject()["/St"]
print(start_page)
except:
print("No /PageLabel object")
Script:
import sys
import PyPDF2 as pyPdf
def arabic_to_roman(arabic):
roman = ''
while arabic >= 1000:
roman += 'm'
arabic -= 1000
diffs = [900, 500, 400, 300, 200, 100, 90, 50, 40, 30, 20, 10, 9, 5, 4, 3, 2, 1]
digits = ['cm', 'd', 'cd', 'ccc', 'cc', 'c', 'xc', 'l', 'xl', 'xxx', 'xx', 'x', 'ix', 'v', 'iv', 'iii', 'ii', 'i']
for i in range(len(diffs)):
if arabic >= diffs[i]:
roman += digits[i]
arabic -= diffs[i]
return(roman)
def get_page_labels(pdf):
try:
page_label_type = pdf.trailer["/Root"]["/PageLabels"]["/Nums"][1].getObject()["/S"]
except:
page_label_type = "/D"
try:
page_start = pdf.trailer["/Root"]["/PageLabels"]["/Nums"][1].getObject()["/St"]
except:
page_start = 1
page_count = pdf.getNumPages()
##or, if you feel fancy, do:
#page_count = pdf.trailer["/Root"]["/Pages"]["/Count"]
page_stop = page_start + page_count
if page_label_type == "/D":
page_numbers = list(range(page_start, page_stop))
for i in range(len(page_numbers)):
page_numbers[i] = str(page_numbers[i])
elif page_label_type == '/r':
page_numbers_arabic = range(page_start, page_stop)
page_numbers = []
for i in range(len(page_numbers_arabic)):
page_numbers.append(arabic_to_roman(page_numbers_arabic[i]))
print(page_label_type)
print(page_start)
print(page_count)
print(page_numbers)
pdf = pyPdf.PdfFileReader(open(sys.argv[1], "rb"))
get_page_labels(pdf)