问题
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 number to determine the order it should go in (e.g. if someone split up a book into 20 10-page PDFs and I want to put them back together).
I have two questions - 1.) I know that sometimes the page number is stored in the document data somewhere, as I've seen PDFs that render on Adobe as something like [1243] (10 of 150), but I've read documents of this sort into pyPDF and I can't find any information indicating the page number - where is this stored?
2.) If avenue #1 isn't available, I think I could iterate through the objects on a given page to try to find a page number - likely it would be its own object that has a single number in it. However, I can't seem to find any clear way to determine the contents of objects. If I run:
pdf.getPage(0).getContents()
This usually either returns:
{'/Filter': '/FlateDecode'}
or it returns a list of IndirectObject(num, num) objects. I don't really know what to do with either of these and there's no real documentation on it as far as I can tell. Is anyone familiar with this kind of thing that could point me in the right direction?
回答1:
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!
回答2:
The following worked for me:
from PyPDF2 import PdfFileReader
pdf = PdfFileReader(open('path/to/file.pdf','rb'))
pdf.getNumPages()
回答3:
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")
- As you can see from the ISO pdf 1.7 specification (relevant section here) there are lots of possibilities of how to label pages. As a simple working example consider this script that will at least deal with decimal (arabic) and with roman numerals:
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)
回答4:
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
来源：https://stackoverflow.com/questions/12360999/retrieve-page-numbers-from-document-with-pypdf