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Extracting a specific substring from a specific hyper-reference using Python

I'm new to Python, and for my second attempt at a project, I wanted to extract a substring – specifically, an identifying number – from a hyper-reference on a url.
For example, this url is the result of my search query, giving the hyper-reference http://www.chessgames.com/perl/chessgame?gid=1012809. From this I want to extract the identifying number "1012809" and append it to navigate to the url http://www.chessgames.com/perl/chessgame?gid=1012809, after which I plan to download the file at the url http://www.chessgames.com/pgn/alekhine_naegeli_1932.pgn?gid=1012809 . But I am currently stuck a few steps behind this because I can't figure out a way to extract the identifier.
Here is my MWE:
from bs4 import BeautifulSoup
url = 'http://www.chessgames.com/perl/chess.pl?yearcomp=exactly&year=1932&playercomp=white&pid=&player=Alekhine&pid2=&player2=Naegeli&movescomp=exactly&moves=&opening=&eco=&result=1%2F2-1%2F2'
page = urllib2.urlopen(url)
soup = BeautifulSoup(page, 'html.parser')
import re
y = str(soup)
x = re.findall("gid=[0-9]+",y)
print x
z = re.sub("gid=", "", x(1)) #At this point, things have completely broken down...

As Albin Paul commented, re.findall return a list, you need to extract elements from it. By the way, you don't need BeautifulSoup here, use urllib2.urlopen(url).read() to get the string of the content, and the re.sub is also not needed here, one regex pattern (?:gid=)([0-9]+) is enough.
import re
import urllib2
url = 'http://www.chessgames.com/perl/chess.pl?yearcomp=exactly&year=1932&playercomp=white&pid=&player=Alekhine&pid2=&player2=Naegeli&movescomp=exactly&moves=&opening=&eco=&result=1%2F2-1%2F2'
page = urllib2.urlopen(url).read()
result = re.findall(r"(?:gid=)([0-9]+)",page)
print(result[0])
#'1012809'

You don't need regex here at all. Css selector along with string manipulation will lead you to the right direction. Try the below script:
import requests
from bs4 import BeautifulSoup
page_link = 'http://www.chessgames.com/perl/chess.pl?yearcomp=exactly&year=1932&playercomp=white&pid=&player=Alekhine&pid2=&player2=Naegeli&movescomp=exactly&moves=&opening=&eco=&result=1%2F2-1%2F2'
soup = BeautifulSoup(requests.get(page_link).text, 'lxml')
item_num = soup.select_one("[href*='gid=']")['href'].split("gid=")[1]
print(item_num)
Output:
1012809

Beautiful Soup Nested Tag Search

I am trying to write a python program that will count the words on a web page. I use Beautiful Soup 4 to scrape the page but I have difficulties accessing nested HTML tags (for example: <p class="hello"> inside <div>).
Every time I try finding such tag using page.findAll() (page is Beautiful Soup object containing the whole page) method it simply doesn't find any, although there are. Is there any simple method or another way to do it?

Maybe I'm guessing what you are trying to do is first looking in a specific div tag and the search all p tags in it and count them or do whatever you want. For example:
soup = bs4.BeautifulSoup(content, 'html.parser')
# This will get the div
div_container = soup.find('div', class_='some_class')
# Then search in that div_container for all p tags with class "hello"
for ptag in div_container.find_all('p', class_='hello'):
# prints the p tag content
print(ptag.text)
Hope that helps

Try this one :
data = []
for nested_soup in soup.find_all('xyz'):
data = data + nested_soup.find_all('abc')
Maybe you can turn in into lambda and make it cool, but this works. Thanks.

UPDATE: I noticed that text does not always return the expected result, at the same time, I realized there was a built-in way to get the text, sure enough reading the docs
we read that there is a method called get_text(), use it as:
from bs4 import BeautifulSoup
fd = open('index.html', 'r')
website= fd.read()
fd.close()
soup = BeautifulSoup(website)
contents= soup.get_text(separator=" ")
print "number of words %d" %len(contents.split(" "))
INCORRECT, please read above.Supposing that you have your html file locally in index.html you can:
from bs4 import BeautifulSoup
import re
BLACKLIST = ["html", "head", "title", "script"] # tags to be ignored
fd = open('index.html', 'r')
website= fd.read()
soup = BeautifulSoup(website)
tags=soup.find_all(True) # find everything
print "there are %d" %len(tags)
count= 0
matcher= re.compile("(\s|\n|<br>)+")
for tag in tags:
if tag.name.lower() in BLACKLIST:
continue
temp = matcher.split(tag.text) # Split using tokens such as \s and \n
temp = filter(None, temp) # remove empty elements in the list
count +=len(temp)
print "number of words in the document %d" %count
fd.close()
Please note that it may not be accurate, maybe because of errors in formatting, false positives(it detects any word, even if it is code), text that is shown dynamically using javascript or css, or other reason

You can find all <p> tags using regular expressions (re module).
Note that r.content is a string which contains the whole html of the site.
for eg:
r = requests.get(url,headers=headers)
p_tags = re.findall(r'<p>.*?</p>',r.content)
this should get you all the <p> tags irrespective of whether they are nested or not. And if you want the a tags specifically inside the tags you can add that whole tag as a string in the second argument instead of r.content.
Alternatively if you just want just the text you can try this:
from readability import Document #pip install readability-lxml
import requests
r = requests.get(url,headers=headers)
doc = Document(r.content)
simplified_html = doc.summary()
this will get you a more bare bones form of the html from the site, and now proceed with the parsing.

Trying to use bs4 to skip an attribute if there is no data

I'm trying to write a program that will pull data from a URL and format it so that I can copy into another program. I've got everything working except I can't get it to skip an item if there is no img src in the imagelink tag.
import requests, sys, webbrowser, bs4
res = requests.get('http://hzws.selco.info/prototype.php?type=new-arrivals&lib=nor&collect=Bnewnf,Bnewmys,Bnewf,Bnewsf&days=14&key=7a8adfa9aydfa999997af')
res.raise_for_status()
soup = bs4.BeautifulSoup(res.text, "lxml")
img = soup.select('imagelink') #why won't this pull anything?!?!?!?!
link = soup.select('cataloglink')
length = min([14, len(img)])
for i in range(length):
img1 = img[i].getText()
link1 = link[i].getText()
print('<div>' + link1 + img1 + '</a></div>')
Right now this prints all of the URLs regardless of whether or not there is an imagelink attached to it. I've tried numerous different things to get it to skip an item if there is no img src. Any ideas?

Looking at the BS4 docs, It looks like "lxml" is actually a HTML parser. You should replace it with "lxml-xml", since you're trying to scrape an XML page. That should be working.

how to extract specific csv from web page html containing multiple csv file links

I need to extract csv file from html page see below and once I get that I can do stuff with it. below is code to extract that particular line of html code from a previous assignment. The url is 'https://vincentarelbundock.github.io/Rdatasets/datasets.html'
that is test code so it breaks temporarly when it finds that line.
part of the line with my csv is href is csv/datasets/co2.csv ( unicode I think as type)
how to open the co2.csv?
sorry about any formatting issues with the question. The code has been sliced and diced by the editor.
import urllib
url = 'https://vincentarelbundock.github.io/Rdatasets/datasets.html'
from BeautifulSoup import *
def scrapper(url,k):
c=0
html = urllib.urlopen(url).read()
soup = BeautifulSoup(html)
#. Retrieve all of the anchor tags
tags = soup('a')
for tag in tags:
y= (tag.get('href', None))
#print ((y))
if y == 'csv/datasets/co2.csv':
print y
break
c= c+ 1
if c is k:
return y
print(type(y))
for w in range(29):
print(scrapper(url,w))

You're re-downloading and reparsing the full html page for all of the 30 iterations of your loop, just to get the next csv file and see if that is the one you want. That is very inefficient, and not very polite to the server. Just read the html page once, and use the loop over the tags you already had to check if the tag is the one you want! If so, do something with it, and stop looping to avoid needless further processing because you said you only needed one particular file.
The other issue related to your question is that in the html file the csv hrefs are relative urls. So you have to join them on the base url of the document they're in. urlparse.urljoin() does just that.
Not related to the question directly, but you should also try to clean up your code;
fix your indentation (the comment on line 9)
choose better variable names; y/c/k/w are meaningless.
Resulting in something like:
import urllib
import urlparse
url = 'https://vincentarelbundock.github.io/Rdatasets/datasets.html'
from BeautifulSoup import *
def scraper(url):
html = urllib.urlopen(url).read()
soup = BeautifulSoup(html)
# Retrieve all of the anchor tags
tags = soup('a')
for tag in tags:
href = (tag.get('href', None))
if href.endswith("/co2.csv"):
csv_url = urlparse.urljoin(url, href)
# ... do something with the csv file....
contents = urllib.urlopen(csv_url).read()
print "csv file size=", len(contents)
break # we only needed this one file, so we end the loop.
scraper(url)

Python BS4 crawler indexerror

I am trying to create a simple crawler that pulls meta data from websites and saves the information into a csv. So far I am stuck here, I have followed some guides but am now stuck with the error:
IndexError: list of index out of range.
from urllib import urlopen
from BeautifulSoup import BeautifulSoup
import re
# Copy all of the content from the provided web page
webpage = urlopen('http://www.tidyawaytoday.co.uk/').read()
# Grab everything that lies between the title tags using a REGEX
patFinderTitle = re.compile('<title>(.*)</title>')
# Grab the link to the original article using a REGEX
patFinderLink = re.compile('<link rel.*href="(.*)" />')
# Store all of the titles and links found in 2 lists
findPatTitle = re.findall(patFinderTitle,webpage)
findPatLink = re.findall(patFinderLink,webpage)
# Create an iterator that will cycle through the first 16 articles and skip a few
listIterator = []
listIterator[:] = range(2,16)
# Print out the results to screen
for i in listIterator:
print findPatTitle[i] # The title
print findPatLink[i] # The link to the original article
articlePage = urlopen(findPatLink[i]).read() # Grab all of the content from original article
divBegin = articlePage.find('<div>') # Locate the div provided
article = articlePage[divBegin:(divBegin+1000)] # Copy the first 1000 characters after the div
# Pass the article to the Beautiful Soup Module
soup = BeautifulSoup(article)
# Tell Beautiful Soup to locate all of the p tags and store them in a list
paragList = soup.findAll('p')
# Print all of the paragraphs to screen
for i in paragList:
print i
print '\n'
# Here I retrieve and print to screen the titles and links with just Beautiful Soup
soup2 = BeautifulSoup(webpage)
print soup2.findAll('title')
print soup2.findAll('link')
titleSoup = soup2.findAll('title')
linkSoup = soup2.findAll('link')
for i in listIterator:
print titleSoup[i]
print linkSoup[i]
print '\n'
Any help would be greatly appreciated.
The error I get is
File "C:\Users......", line 24, in (module)
print findPatTitle[i] # the title
IndexError:list of index out of range
Thank you.

It seems that you are not using all the power that bs4 can give you.
You are getting this error because the lenght of patFinderTitle is just one, since all html has usually only one title element per document.
A simple way to grab the title of a HTML, is using bs4 itself:
from bs4 import BeautifulSoup
from urllib import urlopen
webpage = urlopen('http://www.tidyawaytoday.co.uk/').read()
soup = BeautifulSoup(webpage)
# get the content of title
title = soup.title.text
You will probably get the same error if you try to iterate over your findPatLink in the currently way, since it has length 6. For me, it is not clear enough if you want to get all the link elements or all the anchor elements, but stickying with the first idea, you can improve your code using bs4 again:
link_href_list = [link['href'] for link in soup.find_all("link")]
And finally, since you don't want some urls, you can slice link_href_list in the way that you want. An improved version of the last expression which excludes the first and the second result could be:
link_href_list = [link['href'] for link in soup.find_all("link")[2:]]