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

向sys.path添加路径(与使用imp相比)的优点是,当从单个包导入多个模块时,可以简化操作。例如:

import sys
# the mock-0.3.1 dir contains testcase.py, testutils.py & mock.py
sys.path.append('/foo/bar/mock-0.3.1')

from testcase import TestCase
from testutils import RunTests
from mock import Mock, sentinel, patch

其他回答

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)

我相信您可以使用imp.find_module()和imp.load_module)来加载指定的模块。您需要将模块名称从路径中分离出来,即,如果要加载/home/mypath/mymodule.py,则需要执行以下操作:

imp.find_module('mymodule', '/home/mypath/')

…但这应该能完成任务。

我发现这是一个简单的答案:

module = dict()

code = """
import json

def testhi() :
    return json.dumps({"key" : "value"}, indent = 4 )
"""

exec(code, module)
x = module['testhi']()
print(x)

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

在Linux中,可以在Python脚本所在的目录中添加符号链接。

即。:

ln -s /absolute/path/to/module/module.py /absolute/path/to/script/module.py

Python解释器将创建/aabsolute/path/to/script/module.pyc,如果您更改/aabsolute/path/to-module/module.py的内容,Python解释器将对其进行更新。

然后在文件mypythonscript.py中包含以下内容:

from module import *