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
我认为,最好的方法是从官方文件(29.1。imp-访问导入内部构件):
import imp
import sys
def __import__(name, globals=None, locals=None, fromlist=None):
# Fast path: see if the module has already been imported.
try:
return sys.modules[name]
except KeyError:
pass
# If any of the following calls raises an exception,
# there's a problem we can't handle -- let the caller handle it.
fp, pathname, description = imp.find_module(name)
try:
return imp.load_module(name, fp, pathname, description)
finally:
# Since we may exit via an exception, close fp explicitly.
if fp:
fp.close()
其他回答
您可以使用pkgutil模块(特别是walk_packages方法)获取当前目录中的包列表。从那里,使用importlib机制导入所需的模块很简单:
import pkgutil
import importlib
packages = pkgutil.walk_packages(path='.')
for importer, name, is_package in packages:
mod = importlib.import_module(name)
# do whatever you want with module now, it's been imported!
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)
这个答案是对Sebastian Rittau对评论的回答的补充:“但是如果你没有模块名怎么办?”这是一种快速而肮脏的方法,可以将可能的Python模块名指定为文件名——它只是沿着树向上移动,直到找到一个没有__init__.py文件的目录,然后将其转换回文件名。对于Python 3.4+(使用pathlib),这是有意义的,因为Python 2可以使用“imp”或其他方式进行相对导入:
import pathlib
def likely_python_module(filename):
'''
Given a filename or Path, return the "likely" python module name. That is, iterate
the parent directories until it doesn't contain an __init__.py file.
:rtype: str
'''
p = pathlib.Path(filename).resolve()
paths = []
if p.name != '__init__.py':
paths.append(p.stem)
while True:
p = p.parent
if not p:
break
if not p.is_dir():
break
inits = [f for f in p.iterdir() if f.name == '__init__.py']
if not inits:
break
paths.append(p.stem)
return '.'.join(reversed(paths))
当然有改进的可能性,可选的__init__.py文件可能需要进行其他更改,但如果您通常有__init__.pry,这就有了窍门。
可以使用__import__和chdir执行此操作:
def import_file(full_path_to_module):
try:
import os
module_dir, module_file = os.path.split(full_path_to_module)
module_name, module_ext = os.path.splitext(module_file)
save_cwd = os.getcwd()
os.chdir(module_dir)
module_obj = __import__(module_name)
module_obj.__file__ = full_path_to_module
globals()[module_name] = module_obj
os.chdir(save_cwd)
except Exception as e:
raise ImportError(e)
return module_obj
import_file('/home/somebody/somemodule.py')
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