Question

I’m trying to open multiple pages following a certain format using mechanize. I want to start with a certain page, and have mechanize follow all the links that have a certain class or piece of text in a link. For example, the root url would be something like

http://hansard.millbanksystems.com/offices/prime-minister

and I want to follow every link on the page that has a format such as

<li class='office-holder'><a href="http://hansard.millbanksystems.com/people/mr-tony-blair">Mr Tony Blair</a> May  2, 1997 - June 27, 2007</li>

In other words, I want to follow every link that has the class ‘office-holder’ or that has /people/ in the URL. I’ve tried the following code, but it hasn’t worked.

import mechanize

br = mechanize.Browser()
response = br.open("http://hansard.millbanksystems.com/offices/prime-minister")
links = br.links(url_regex="/people/")

print links

I’m trying to print the links so I can make sure that I’m getting the right links/information before writing any more code. The error(?) I get from this is:

<generator object _filter_links at 0x10121e6e0>

Any pointers or tips are appreciated.

Answer 1

That’s not an error – it means that Browser.links() returns an generator object rather than a list.

An iterator is an object that acts “like a list”, meaning that you can do things like

for link in links:
    print link

and so on. But you can only access things in whatever order it defines; you can’t necessarily do link[5], and once you’ve gone through the iterator, it’s used up.

A generator is, for most purposes, just an iterator that doesn’t necessarily know all its results in advance. This is very useful in generator expressions, and you can actually write very simple functions that return generators with the yield keyword:

def odds():
    x = 1
    while True:
        yield x
        x += 2

 os = odds()
 os.next() # returns 1
 os.next() # returns 3

This is a Good Thing because it means that you don’t have to store all of your data in memory at once (which for odds() would be impossible…), and if you only need the first few elements of the result you don’t have to bother computing the rest. The itertools module has a bunch of handy functions for dealing with iterators.

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Anyway, if you just want to print out the contents of links, you can turn it into a list with the list() function (which takes an iterable and returns a list of its elements):

 print list(links)

or make a list of strings with a list comprehension:

 print [l.url for l in list(links)]

or walk over its elements and print them out:

 for l in links:
      print l.url

But note that after you do this, links will be “exhausted” – so if you want to actually do anything with it, you’ll need to get it again.

Maybe the simplest option is to immediately turn it into a list and not worry about it being an iterator at all:

links = list(br.links(url_regex="/people/"))

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Also, you’re obviously not yet getting links that have the class you want. There might be some mechanize trick to do an “or” here, but a nifty way to do it using sets and generator expressions would be something like this:

 links = set(l.url for l in br.links(url_regex='/people/'))
 links.update(l.url for l in br.get_links_with_class('office-holder'))

Obviously replace get_links_with_class with the real way to get those links. Then you’ll end up with a set of all the link URLs that have /people/ in their URL and/or have the class office-holder, with no duplicates. (Note that you can’t put the Link objects in the set directly because they’re not hashable.)