如何加载给定完整路径的Python模块?
请注意,文件可以位于文件系统中用户具有访问权限的任何位置。
另请参阅:如何导入以字符串形式命名的模块?
如何加载给定完整路径的Python模块?
请注意,文件可以位于文件系统中用户具有访问权限的任何位置。
另请参阅:如何导入以字符串形式命名的模块?
当前回答
在运行时导入包模块(Python配方)
http://code.activestate.com/recipes/223972/
###################
## #
## classloader.py #
## #
###################
import sys, types
def _get_mod(modulePath):
try:
aMod = sys.modules[modulePath]
if not isinstance(aMod, types.ModuleType):
raise KeyError
except KeyError:
# The last [''] is very important!
aMod = __import__(modulePath, globals(), locals(), [''])
sys.modules[modulePath] = aMod
return aMod
def _get_func(fullFuncName):
"""Retrieve a function object from a full dotted-package name."""
# Parse out the path, module, and function
lastDot = fullFuncName.rfind(u".")
funcName = fullFuncName[lastDot + 1:]
modPath = fullFuncName[:lastDot]
aMod = _get_mod(modPath)
aFunc = getattr(aMod, funcName)
# Assert that the function is a *callable* attribute.
assert callable(aFunc), u"%s is not callable." % fullFuncName
# Return a reference to the function itself,
# not the results of the function.
return aFunc
def _get_class(fullClassName, parentClass=None):
"""Load a module and retrieve a class (NOT an instance).
If the parentClass is supplied, className must be of parentClass
or a subclass of parentClass (or None is returned).
"""
aClass = _get_func(fullClassName)
# Assert that the class is a subclass of parentClass.
if parentClass is not None:
if not issubclass(aClass, parentClass):
raise TypeError(u"%s is not a subclass of %s" %
(fullClassName, parentClass))
# Return a reference to the class itself, not an instantiated object.
return aClass
######################
## Usage ##
######################
class StorageManager: pass
class StorageManagerMySQL(StorageManager): pass
def storage_object(aFullClassName, allOptions={}):
aStoreClass = _get_class(aFullClassName, StorageManager)
return aStoreClass(allOptions)
其他回答
听起来您不想专门导入配置文件(这会带来很多副作用和额外的复杂性)。您只需要运行它,并能够访问生成的命名空间。标准库以runpy.run_path的形式专门提供了一个API:
from runpy import run_path
settings = run_path("/path/to/file.py")
该接口在Python 2.7和Python 3.2+中可用。
一种非常简单的方法:假设您希望导入具有相对路径../..的文件/MyLibs/pyfunc.py
libPath = '../../MyLibs'
import sys
if not libPath in sys.path: sys.path.append(libPath)
import pyfunc as pf
但如果你在没有守卫的情况下成功,你最终会走上一条很长的路。
您可以使用pydoc中的importfile
from pydoc import importfile
module = importfile('/full/path/to/module/module.py')
name = module.myclass() # myclass is a class inside your python file
特殊的是使用Exec()导入具有绝对路径的模块:(exec采用代码字符串或代码对象。而eval采用表达式。)
PYMODULE = 'C:\maXbox\mX47464\maxbox4\examples\histogram15.py';
Execstring(LoadStringJ(PYMODULE));
然后使用eval()获取值或对象:
println('get module data: '+evalStr('pyplot.hist(x)'));
使用exec加载模块就像使用通配符命名空间导入:
Execstring('sys.path.append(r'+'"'+PYMODULEPATH+'")');
Execstring('from histogram import *');
要从给定文件名导入模块,可以临时扩展路径,并在finally块引用中恢复系统路径:
filename = "directory/module.py"
directory, module_name = os.path.split(filename)
module_name = os.path.splitext(module_name)[0]
path = list(sys.path)
sys.path.insert(0, directory)
try:
module = __import__(module_name)
finally:
sys.path[:] = path # restore