是否有一种方法可以方便地在Python中定义类c结构?我厌倦了写这样的东西:

class MyStruct():
    def __init__(self, field1, field2, field3):
        self.field1 = field1
        self.field2 = field2
        self.field3 = field3

当前回答

还可以按位置将初始化参数传递给实例变量

# Abstract struct class       
class Struct:
    def __init__ (self, *argv, **argd):
        if len(argd):
            # Update by dictionary
            self.__dict__.update (argd)
        else:
            # Update by position
            attrs = filter (lambda x: x[0:2] != "__", dir(self))
            for n in range(len(argv)):
                setattr(self, attrs[n], argv[n])

# Specific class
class Point3dStruct (Struct):
    x = 0
    y = 0
    z = 0

pt1 = Point3dStruct()
pt1.x = 10

print pt1.x
print "-"*10

pt2 = Point3dStruct(5, 6)

print pt2.x, pt2.y
print "-"*10

pt3 = Point3dStruct (x=1, y=2, z=3)
print pt3.x, pt3.y, pt3.z
print "-"*10

其他回答

一本字典怎么样?

就像这样:

myStruct = {'field1': 'some val', 'field2': 'some val'}

然后你可以使用这个来操作值:

print myStruct['field1']
myStruct['field2'] = 'some other values'

值不一定是字符串。它们可以是其他任何物体。

https://stackoverflow.com/a/32448434/159695在Python3中不起作用。

https://stackoverflow.com/a/35993/159695在Python3中工作。

然后我扩展它来添加默认值。

class myStruct:
    def __init__(self, **kwds):
        self.x=0
        self.__dict__.update(kwds) # Must be last to accept assigned member variable.
    def __repr__(self):
        args = ['%s=%s' % (k, repr(v)) for (k,v) in vars(self).items()]
        return '%s(%s)' % ( self.__class__.__qualname__, ', '.join(args) )

a=myStruct()
b=myStruct(x=3,y='test')
c=myStruct(x='str')

>>> a
myStruct(x=0)
>>> b
myStruct(x=3, y='test')
>>> c
myStruct(x='str')

使用命名元组,该元组被添加到Python 2.6标准库中的collections模块中。如果你需要支持Python 2.4,也可以使用Raymond Hettinger的命名元组配方。

它适用于基本示例,但也适用于稍后可能遇到的一些边缘情况。你上面的片段可以写成:

from collections import namedtuple
MyStruct = namedtuple("MyStruct", "field1 field2 field3")

新创建的类型可以这样使用:

m = MyStruct("foo", "bar", "baz")

你也可以使用命名参数:

m = MyStruct(field1="foo", field2="bar", field3="baz")

还可以按位置将初始化参数传递给实例变量

# Abstract struct class       
class Struct:
    def __init__ (self, *argv, **argd):
        if len(argd):
            # Update by dictionary
            self.__dict__.update (argd)
        else:
            # Update by position
            attrs = filter (lambda x: x[0:2] != "__", dir(self))
            for n in range(len(argv)):
                setattr(self, attrs[n], argv[n])

# Specific class
class Point3dStruct (Struct):
    x = 0
    y = 0
    z = 0

pt1 = Point3dStruct()
pt1.x = 10

print pt1.x
print "-"*10

pt2 = Point3dStruct(5, 6)

print pt2.x, pt2.y
print "-"*10

pt3 = Point3dStruct (x=1, y=2, z=3)
print pt3.x, pt3.y, pt3.z
print "-"*10

下面结构的解决方案是受namedtuple实现和前面一些答案的启发。然而,与namedtuple不同的是,它的值是可变的,但就像c风格的结构体在名称/属性中是不可变的,而普通的类或dict不是。

_class_template = """\
class {typename}:
def __init__(self, *args, **kwargs):
    fields = {field_names!r}

    for x in fields:
        setattr(self, x, None)            

    for name, value in zip(fields, args):
        setattr(self, name, value)

    for name, value in kwargs.items():
        setattr(self, name, value)            

def __repr__(self):
    return str(vars(self))

def __setattr__(self, name, value):
    if name not in {field_names!r}:
        raise KeyError("invalid name: %s" % name)
    object.__setattr__(self, name, value)            
"""

def struct(typename, field_names):

    class_definition = _class_template.format(
        typename = typename,
        field_names = field_names)

    namespace = dict(__name__='struct_%s' % typename)
    exec(class_definition, namespace)
    result = namespace[typename]
    result._source = class_definition

    return result

用法:

Person = struct('Person', ['firstname','lastname'])
generic = Person()
michael = Person('Michael')
jones = Person(lastname = 'Jones')


In [168]: michael.middlename = 'ben'
Traceback (most recent call last):

  File "<ipython-input-168-b31c393c0d67>", line 1, in <module>
michael.middlename = 'ben'

  File "<string>", line 19, in __setattr__

KeyError: 'invalid name: middlename'