Getting the nth element using BeautifulSoup

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 ~]$

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

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]