Getting the nth element using BeautifulSoup

Question

From a large table I want to read rows 5, 10, 15, 20 ... using BeautifulSoup. How do I do this? Is findNextSibling and an incrementing counter the way to go?

Answer 1

You could also use findAll to get all the rows in a list and after that just use the slice syntax to access the elements that you need:

rows = soup.findAll('tr')[4::5]

Answer 2

Here's how you could scrape every 5th distribution link on this Wikipedia page with gazpacho:

from gazpacho import Soup

url = "https://en.wikipedia.org/wiki/List_of_probability_distributions"
soup = Soup.get(url)

a_tags = soup.find("a", {"href": "distribution"})
links = ["https://en.wikipedia.org" + a.attrs["href"] for a in a_tags]

links[4::5] # start at 0,1,2,3,**4** and stride by 5

Answer 3

As a general solution, you can convert the table to a nested list and iterate...

import BeautifulSoup

def listify(table):
  """Convert an html table to a nested list""" 
  result = []
  rows = table.findAll('tr')
  for row in rows:
    result.append([])
    cols = row.findAll('td')
    for col in cols:
      strings = [_string.encode('utf8') for _string in col.findAll(text=True)]
      text = ''.join(strings)
      result[-1].append(text)
  return result

if __name__=="__main__":
    """Build a small table with one column and ten rows, then parse into a list"""
    htstring = """<table> <tr> <td>foo1</td> </tr> <tr> <td>foo2</td> </tr> <tr> <td>foo3</td> </tr> <tr> <td>foo4</td> </tr> <tr> <td>foo5</td> </tr>  <tr> <td>foo6</td> </tr>  <tr> <td>foo7</td> </tr>  <tr> <td>foo8</td> </tr>  <tr> <td>foo9</td> </tr>  <tr> <td>foo10</td> </tr></table>"""
    soup = BeautifulSoup.BeautifulSoup(htstring)
    for idx, ii in enumerate(listify(soup)):
        if ((idx+1)%5>0):
            continue
        print ii

Running that...

[mpenning@Bucksnort ~]$ python testme.py
['foo5']
['foo10']
[mpenning@Bucksnort ~]$

Answer 4

This can be easily done with select in beautiful soup if you know the row numbers to be selected. (Note : This is in bs4)

row = 5
while true
    element = soup.select('tr:nth-of-type('+ row +')')
    if len(element) > 0:
        # element is your desired row element, do what you want with it 
        row += 5
    else:
        break

Answer 5

Another option, if you prefer raw html...

"""Build a small table with one column and ten rows, then parse it into a list"""
htstring = """<table> <tr> <td>foo1</td> </tr> <tr> <td>foo2</td> </tr> <tr> <td>foo3</td> </tr> <tr> <td>foo4</td> </tr> <tr> <td>foo5</td> </tr>  <tr> <td>foo6</td> </tr>  <tr> <td>foo7</td> </tr>  <tr> <td>foo8</td> </tr>  <tr> <td>foo9</td> </tr>  <tr> <td>foo10</td> </tr></table>"""
result = [html_tr for idx, html_tr in enumerate(soup.findAll('tr')) \
     if (idx+1)%5==0]
print result

Running that...

[mpenning@Bucksnort ~]$ python testme.py
[<tr> <td>foo5</td> </tr>, <tr> <td>foo10</td> </tr>]
[mpenning@Bucksnort ~]$