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• Installing PygraphViz on Windows, Python 2.6

  - by jbochi

  Anybody out there has successfully installed PygraphViz on Windows? Since there is not an official release for Windows, I'm trying to build it myself, but it fails to compile. I'm not the first one to face this issue, but I could not find an answer. This is the console output: C:\Python26\Lib\site-packages\pygraphviz-0.99.1>c:\python26\python.exe setup.py install library_path=C:/Program Files/Graphviz2.26.3/lib/debug/dll include_path=C:/Program Files/Graphviz2.26.3/include/graphviz running install running build running build_py running build_ext building 'pygraphviz._graphviz' extension C:\Program Files\Microsoft Visual Studio 9.0\VC\BIN\cl.exe /c /nologo /Ox /MD /W 3 /GS- /DNDEBUG "-IC:/Program Files/Graphviz2.26.3/include/graphviz" -Ic:\python 26\include -Ic:\python26\PC /Tcpygraphviz/graphviz_wrap.c /Fobuild\temp.win32-2. 6\Release\pygraphviz/graphviz_wrap.obj graphviz_wrap.c warning: I don't know what to do with 'runtime_library_dirs': ['C:/Program Files /Graphviz2.26.3/lib/debug/dll'] error: don't know how to set runtime library search path for MSVC++ Any help would be appreciated!

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• Bitbucket API authentication with Python's HTTPBasicAuthHandler

  - by jbochi

  I'm trying to get the list of issues on a private repository using bitbucket's API. I have confirmed that HTTP Basic authentication works with hurl, but I am unable to authenticate in Python. Adapting the code from this tutorial, I have written the following script. import cookielib import urllib2 class API(): api_url = 'http://api.bitbucket.org/1.0/' def __init__(self, username, password): self._opener = self._create_opener(username, password) def _create_opener(self, username, password): cj = cookielib.LWPCookieJar() cookie_handler = urllib2.HTTPCookieProcessor(cj) password_manager = urllib2.HTTPPasswordMgrWithDefaultRealm() password_manager.add_password(None, self.api_url, username, password) auth_handler = urllib2.HTTPBasicAuthHandler(password_manager) opener = urllib2.build_opener(cookie_handler, auth_handler) return opener def get_issues(self, username, repository): query_url = self.api_url + 'repositories/%s/%s/issues/' % (username, repository) try: handler = self._opener.open(query_url) except urllib2.HTTPError, e: print e.headers raise e return handler.read() api = API(username='my_username', password='XXXXXXXX') api.get_issues('my_username', 'my_repository') results in: >>> Server: nginx/0.7.62 Date: Mon, 19 Apr 2010 16:15:06 GMT Content-Type: text/plain Connection: close Vary: Authorization,Cookie Content-Length: 9 Traceback (most recent call last): File "C:/USERS/personal/bitbucket-burndown/bitbucket-api.py", line 29, in <module> print api.get_issues('my_username', 'my_repository') File "C:/USERS/personal/bitbucket-burndown/bitbucket-api.py", line 25, in get_issues raise e HTTPError: HTTP Error 401: UNAUTHORIZED api.get_issues('jespern', 'bitbucket') works like a charm. What's wrong with my code?

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• Fastest way to list all primes below N in python

  - by jbochi

  This is the best algorithm I could come up with after struggling with a couple of Project Euler's questions. def get_primes(n): numbers = set(range(n, 1, -1)) primes = [] while numbers: p = numbers.pop() primes.append(p) numbers.difference_update(set(range(p*2, n+1, p))) return primes >>> timeit.Timer(stmt='get_primes.get_primes(1000000)', setup='import get_primes').timeit(1) 1.1499958793645562 Can it be made even faster? EDIT: This code has a flaw: Since numbers is an unordered set, there is no guarantee that numbers.pop() will remove the lowest number from the set. Nevertheless, it works (at least for me) for some input numbers: >>> sum(get_primes(2000000)) 142913828922L #That's the correct sum of all numbers below 2 million >>> 529 in get_primes(1000) False >>> 529 in get_primes(530) True EDIT: The rank so far (pure python, no external sources, all primes below 1 million): Sundaram's Sieve implementation by myself: 327ms Daniel's Sieve: 435ms Alex's recipe from Cookbok: 710ms EDIT: ~unutbu is leading the race.

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• Is there a django template filter to display percentages?

  - by jbochi

  I would like something similar to the string formatting from the standard library. '%' Percentage. Multiplies the number by 100 and displays in fixed ('f') format, followed by a percent sign.

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