Modify URL components in Python 2

Question

Is there a cleaner way to modify some parts of a URL in Python 2?

For example

http://foo/bar -> http://foo/yah

At present, I'm doing this:

import urlparse

url = 'http://foo/bar'

# Modify path component of URL from 'bar' to 'yah'
# Use nasty convert-to-list hack due to urlparse.ParseResult being immutable
parts = list(urlparse.urlparse(url))
parts[2] = 'yah'

url = urlparse.urlunparse(parts)

Is there a cleaner solution?

Answer 1

Unfortunately, the documentation is out of date; the results produced by urlparse.urlparse() (and urlparse.urlsplit()) use a collections.namedtuple()-produced class as a base.

Don't turn this namedtuple into a list, but make use of the utility method provided for just this task:

parts = urlparse.urlparse(url)
parts = parts._replace(path='yah')

url = parts.geturl()

The namedtuple._replace() method lets you create a new copy with specific elements replaced. The ParseResult.geturl() method then re-joins the parts into a url for you.

Demo:

>>> import urlparse
>>> url = 'http://foo/bar'
>>> parts = urlparse.urlparse(url)
>>> parts = parts._replace(path='yah')
>>> parts.geturl()
'http://foo/yah'

mgilson filed a bug report (with patch) to address the documentation issue.

Answer 2

I guess the proper way to do it is this way.

As using _replace private methods or variables is not suggested.

from urlparse import urlparse, urlunparse

res = urlparse('http://www.goog.com:80/this/is/path/;param=paramval?q=val&foo=bar#hash')
l_res = list(res)
# this willhave ['http', 'www.goog.com:80', '/this/is/path/', 'param=paramval', 'q=val&foo=bar', 'hash']
l_res[2] = '/new/path'
urlunparse(l_res)
# outputs 'http://www.goog.com:80/new/path;param=paramval?q=val&foo=bar#hash'