Using Regex with BeautifulSoup to parse a string in Python

Question

I have a series of strings that are all something like "Saturday, December 27th 2014" and I want to toss the "Saturday" and save the file with the name "141227" which is year + month + day. So far, everything is working except I can't get the regex for daypos or yearpos to work. They both give the same error:

Traceback (most recent call last): File "scrapewaybackblog.py", line 17, in daypos = byline.find(re.compile("[A-Z][a-z]*\s")) TypeError: expected a character buffer object

What's a character buffer object? Does that mean there's something wrong with my expression? Here's my script:

for i in xrange(3, 1, -1):
       page = urllib2.urlopen("http://web.archive.org/web/20090204221349/http://www.americansforprosperity.org/nationalblog?page={}".format(i))
       soup = BeautifulSoup(page.read())
       snippet = soup.find_all('div', attrs={'class': 'blog-box'})
       for div in snippet:
           byline =  div.find('div', attrs={'class': 'date'}).text.encode('utf-8')
           text = div.find('div', attrs={'class': 'right-box'}).text.encode('utf-8')

           monthpos = byline.find(",")
           daypos = byline.find(re.compile("[A-Z][a-z]*\s"))
           yearpos = byline.find(re.compile("[A-Z][a-z]*\D\d*\w*\s"))
           endpos = monthpos + len(byline)

           month = byline[monthpos+1:daypos]
           day = byline[daypos+0:yearpos]
           year = byline[yearpos+2:endpos]

           output_files_pathname = 'Data/'  # path where output will go
           new_filename = year + month + day + ".txt"
           outfile = open(output_files_pathname + new_filename,'w')
           outfile.write(date)
           outfile.write("\n")
           outfile.write(text)
           outfile.close()
       print "finished another url from page {}".format(i)

I also haven't figured out how to make December = 12 but that's for another time. Just please help me find the right positions.

score 5 · Accepted Answer · edited May 23 '17 at 12:23

Instead of parsing a date string with regex, parse it with dateutil:

from dateutil.parser import parse

for div in soup.select('div.blog-box'):
    byline = div.find('div', attrs={'class': 'date'}).text.encode('utf-8')
    text = div.find('div', attrs={'class': 'right-box'}).text.encode('utf-8')

    dt = parse(byline)
    new_filename = "{dt.year}{dt.month}{dt.day}.txt".format(dt=dt)
    ...

Or, you can parse the string with datetime.strptime(), but you need to take care of suffixes:

byline = re.sub(r"(?<=\d)(st|nd|rd|th)", "", byline)
dt = datetime.strptime(byline, '%A, %B %d %Y')

re.sub() here finds st or nd or rd or th string after a digit and replaces the suffixes with empty strings. After it a date string would match '%A, %B %d %Y' format, see:

strftime() and strptime() Behavior

Some additional notes:

you can pass the result of urlopen() directly to the BeautifulSoup constructor
instead of find_all() by class name, use a CSS Selector div.blog-box
to join system paths, use os.path.join()
use with context manager when dealing with files

Fixed version:

import os
import urllib2

from bs4 import BeautifulSoup
from dateutil.parser import parse


for i in xrange(3, 1, -1):
    page = urllib2.urlopen("http://web.archive.org/web/20090204221349/http://www.americansforprosperity.org/nationalblog?page={}".format(i))
    soup = BeautifulSoup(page)

    for div in soup.select('div.blog-box'):
        byline = div.find('div', attrs={'class': 'date'}).text.encode('utf-8')
        text = div.find('div', attrs={'class': 'right-box'}).text.encode('utf-8')

        dt = parse(byline)

        new_filename = "{dt.year}{dt.month}{dt.day}.txt".format(dt=dt)
        with open(os.path.join('Data', new_filename), 'w') as outfile:
            outfile.write(byline)
            outfile.write("\n")
            outfile.write(text)

    print "finished another url from page {}".format(i)

Using Regex with BeautifulSoup to parse a string in Python

1 Answers1