将字符串转换为Python类对象？

Question

给定一个字符串作为Python函数的用户输入,如果在当前定义的命名空间中有一个具有该名称的类,我想从中获取一个类对象.从本质上讲,我想要一个能产生这种结果的函数的实现:

class Foo:
    pass

str_to_class("Foo")
==> <class __main__.Foo at 0x69ba0>

这是可能吗？

Answer 1

这可能有效:

import sys

def str_to_class(classname):
    return getattr(sys.modules[__name__], classname)

Answer 2

你可以这样做:

globals()[class_name]

Answer 3

警告:eval()可用于执行任意Python代码.你永远eval()不应该使用不受信任的字符串.(请参阅Python的eval()对不受信任的字符串的安全性？)

这似乎最简单.

>>> class Foo(object):
...     pass
... 
>>> eval("Foo")
<class '__main__.Foo'>

Answer 4

你想要这个类Baz,它存在于模块中foo.bar.使用Python 2.7,您可以使用importlib.import_module(),因为这将使转换到Python 3更容易:

import importlib

def class_for_name(module_name, class_name):
    # load the module, will raise ImportError if module cannot be loaded
    m = importlib.import_module(module_name)
    # get the class, will raise AttributeError if class cannot be found
    c = getattr(m, class_name)
    return c

使用Python <2.7:

def class_for_name(module_name, class_name):
    # load the module, will raise ImportError if module cannot be loaded
    m = __import__(module_name, globals(), locals(), class_name)
    # get the class, will raise AttributeError if class cannot be found
    c = getattr(m, class_name)
    return c

使用:

loaded_class = class_for_name('foo.bar', 'Baz')

Answer 5

import sys
import types

def str_to_class(field):
    try:
        identifier = getattr(sys.modules[__name__], field)
    except AttributeError:
        raise NameError("%s doesn't exist." % field)
    if isinstance(identifier, (types.ClassType, types.TypeType)):
        return identifier
    raise TypeError("%s is not a class." % field)

这准确地处理了旧式和新式类.

Answer 6

我看过django如何处理这个问题

django.utils.module_loading有这个

def import_string(dotted_path):
    """
    Import a dotted module path and return the attribute/class designated by the
    last name in the path. Raise ImportError if the import failed.
    """
    try:
        module_path, class_name = dotted_path.rsplit('.', 1)
    except ValueError:
        msg = "%s doesn't look like a module path" % dotted_path
        six.reraise(ImportError, ImportError(msg), sys.exc_info()[2])

    module = import_module(module_path)

    try:
        return getattr(module, class_name)
    except AttributeError:
        msg = 'Module "%s" does not define a "%s" attribute/class' % (
            module_path, class_name)
        six.reraise(ImportError, ImportError(msg), sys.exc_info()[2])

你可以像使用它一样 import_string("module_path.to.all.the.way.to.your_class")

Answer 7

是的，你可以这样做。假设你的类存在于全局命名空间中，类似这样的事情就可以做到：

import types

class Foo:
    pass

def str_to_class(s):
    if s in globals() and isinstance(globals()[s], types.ClassType):
            return globals()[s]
    return None

str_to_class('Foo')

==> <class __main__.Foo at 0x340808cc>

Answer 8

如果你真的想检索你用字符串创建的类，你应该将它们存储（或正确措辞，引用）在字典中。毕竟，这也将允许在更高级别命名您的类并避免暴露不需要的类。

例如，来自一个在 Python 中定义演员类的游戏，并且您希望避免用户输入访问其他通用类。

另一种方法（如下例所示）将创建一个全新的类，它包含dict上述内容。这个会：

• 允许创建多个类持有者以便于组织（例如，一个用于演员类，另一个用于声音类型）；
• 使对持有者和所持有的类的修改更容易；
• 您可以使用类方法将类添加到字典中。（虽然下面的抽象并不是真正必要的，它只是为了...... “插图”）。

例子：

class ClassHolder:
    def __init__(self):
        self.classes = {}

    def add_class(self, c):
        self.classes[c.__name__] = c

    def __getitem__(self, n):
        return self.classes[n]

class Foo:
    def __init__(self):
        self.a = 0

    def bar(self):
        return self.a + 1

class Spam(Foo):
    def __init__(self):
        self.a = 2

    def bar(self):
        return self.a + 4

class SomethingDifferent:
    def __init__(self):
        self.a = "Hello"

    def add_world(self):
        self.a += " World"

    def add_word(self, w):
        self.a += " " + w

    def finish(self):
        self.a += "!"
        return self.a

aclasses = ClassHolder()
dclasses = ClassHolder()
aclasses.add_class(Foo)
aclasses.add_class(Spam)
dclasses.add_class(SomethingDifferent)

print aclasses
print dclasses

print "======="
print "o"
print aclasses["Foo"]
print aclasses["Spam"]
print "o"
print dclasses["SomethingDifferent"]

print "======="
g = dclasses["SomethingDifferent"]()
g.add_world()
print g.finish()

print "======="
s = []
s.append(aclasses["Foo"]())
s.append(aclasses["Spam"]())

for a in s:
    print a.a
    print a.bar()
    print "--"

print "Done experiment!"

这让我返回：

<__main__.ClassHolder object at 0x02D9EEF0>
<__main__.ClassHolder object at 0x02D9EF30>
=======
o
<class '__main__.Foo'>
<class '__main__.Spam'>
o
<class '__main__.SomethingDifferent'>
=======
Hello World!
=======
0
1
--
2
6
--
Done experiment!

另一个有趣的实验是添加一个腌制方法，ClassHolder这样你就不会丢失你所做的所有类:^)

更新：也可以使用装饰器作为速记。

class ClassHolder:
    def __init__(self):
        self.classes = {}

    def add_class(self, c):
        self.classes[c.__name__] = c

    # -- the decorator
    def held(self, c):
        self.add_class(c)

        # Decorators have to return the function/class passed (or a modified variant thereof), however I'd rather do this separately than retroactively change add_class, so.
        # "held" is more succint, anyway.
        return c 

    def __getitem__(self, n):
        return self.classes[n]

food_types = ClassHolder()

@food_types.held
class bacon:
    taste = "salty"

@food_types.held
class chocolate:
    taste = "sweet"

@food_types.held
class tee:
    taste = "bitter" # coffee, ftw ;)

@food_types.held
class lemon:
    taste = "sour"

print(food_types['bacon'].taste) # No manual add_class needed! :D