You should take a look at using urlencode.

from urllib import urlencode

base_url = "http://en.wikipedia.org/w/api.php?"

arguments = dict(action="query",
                 meta="globaluserinfo",
                 guiuser="$cammer",
                 guiprop="groups|merged|unattached",
                 format="json")
url = base_url + urlencode(arguments)

If you don't need to build a complete url you can just use the quote function for a single string:

>>> import urllib
>>> urllib.quote("$cammer")
'%24cammer'

So you end up with:

r['guiprop'] = urllib.quote(u'groups|merged|unattached')
r['guiuser'] = urllib.quote(u'$cammer')

You can escape certain characters by specifying them explicitly as safe argument value

urllib.quote(str, safe='~()*!.\'')

More : https://docs.python.org/3.0/library/urllib.parse.html#urllib.parse.quote

Using urllib package (import urllib) :

Python 2.7

From official documentation :

urllib.unquote(string)

Replace %xx escapes by their single-character equivalent.

Example: unquote('/%7Econnolly/') yields '/~connolly/'.

Python 3

From official documentation :

urllib.parse.unquote(string, encoding='utf-8', errors='replace')

[…]

Example: unquote('/El%20Ni%C3%B1o/') yields '/El Niño/'.

From the Python 3 documentation:

urllib.parse.quote(string, safe='/', encoding=None, errors=None)

Replace special characters in string using the %xx escape. Letters, digits, and the characters '_.-~' are never quoted. By default, this function is intended for quoting the path section of a URL. The optional safe parameter specifies additional ASCII characters that should not be quoted — its default value is '/'.

That means passing '' for safe will solve your first issue:

>>> import urllib.parse
>>> urllib.parse.quote('/test')
'/test'
>>> urllib.parse.quote('/test', safe='')
'%2Ftest'

(The function quote was moved from urllib to urllib.parse in Python 3.)

By the way, have a look at urlencode.

About the second issue, there was a bug report about it and it was fixed in Python 3.

For Python 2, you can work around it by encoding as UTF-8 like this:

>>> query = urllib.quote(u"Müller".encode('utf8'))
>>> print urllib.unquote(query).decode('utf8')
Müller

You should have a look at urllib.quote - that should do the trick. Have a look at the docs for reference.

To expand on this answer: The problem is, that % (+ a hexadecimal number) is the escape sequence for special characters in URLs. If you want the server to interpret your % literaly, you need to escape it as well, which is done by replacing it with %25. The aforementioned qoute function does stuff like that for you.

Convert a mapping object or a sequence of two-element tuples to a “percent-encoded” string[...]

The urlencode() method is acting as expected. If you want to prevent the encoding then you can first encode the entire object and then replace the encoded characters with pipes.

>>> u = urlencode(params)
>>> u.replace('%7C', '|')
'p=1+2+3+4+5%266&l=ab|cd|ef'  

For Python 2.x, use urllib.quote

Replace special characters in string using the %xx escape. Letters, digits, and the characters '_.-' are never quoted. By default, this function is intended for quoting the path section of the URL. The optional safe parameter specifies additional characters that should not be quoted — its default value is '/'.

example:

In [1]: import urllib

In [2]: urllib.quote('%')
Out[2]: '%25'

EDIT:

In your case, in order to replace space by plus signs, you may use urllib.quote_plus

example:

In [4]: urllib.quote_plus('a b')
Out[4]: 'a+b'

For Python 3.x, use quote

>>> import urllib
>>> a = "asdas#@das"
>>> urllib.parse.quote(a)
'asdas%23%40das'

and for string with space use quote_plus

>>> import urllib
>>> a = "as da& s#@das"
>>> urllib.parse.quote_plus(a)
'as+da%26+s%23%40das'