Mercurial > repos > davidmurphy > codonlogo
view corebio/_future/__init__.py @ 5:b89bb51eba83
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author | davidmurphy |
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date | Mon, 16 Jan 2012 06:59:52 -0500 |
parents | c55bdc2fb9fa |
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""" Private compatability module for running under python version 2.3. Replacement for o string.Template -- introduced in python 2.4 o subprocess -- introduced in python 2.4 o resource_string -- introduced in pkg_resource of setuptools o resource_stream o resource_filename from string import Template -> from corebio._future import Template """ try : import pkg_resources except ImportError : pkg_resources = None try : from string import Template except ImportError : from _string import Template def resource_string( modulename, resource, basefilename = None): """Locate and return a resource as a string. >>> f = resource_string( __name, 'somedatafile', __file__) """ if pkg_resources : return pkg_resources.resource_string(modulename, resource) f = resource_stream( modulename, resource, basefilename) return f.read() def resource_stream( modulename, resource, basefilename = None): """Locate and return a resource as a stream. >>> f = resource_stream( __name__, 'somedatafile', __file__) """ if pkg_resources : return pkg_resources.resource_stream(modulename, resource) return open( resource_filename( modulename, resource, basefilename) ) def resource_filename( modulename, resource, basefilename = None): """Locate and return a resource filename. >>> f = resource_stream( __name__, 'somedatafile', __file__) A resource is a data file stored with the python code in a package. All three resource methods (resource_string, resource_stream, resource_filename) call the corresponding methods in the 'pkg_resources' module, if installed. Otherwise, we resort to locating the resource in the local filesystem. However, this does not work if the package is located inside a zip file. """ if pkg_resources : return pkg_resources.resource_filename(modulename, resource) if basefilename is None : raise NotImplementedError( "Require either basefilename or pkg_resources") import os return os.path.join(os.path.dirname(basefilename), resource)