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
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
你可以通过以下方式在python中访问C-Style struct。
class cstruct:
var_i = 0
var_f = 0.0
var_str = ""
如果你只想使用cstruct的对象
obj = cstruct()
obj.var_i = 50
obj.var_f = 50.00
obj.var_str = "fifty"
print "cstruct: obj i=%d f=%f s=%s" %(obj.var_i, obj.var_f, obj.var_str)
如果你想创建一个cstruct对象的数组
obj_array = [cstruct() for i in range(10)]
obj_array[0].var_i = 10
obj_array[0].var_f = 10.00
obj_array[0].var_str = "ten"
#go ahead and fill rest of array instaces of struct
#print all the value
for i in range(10):
print "cstruct: obj_array i=%d f=%f s=%s" %(obj_array[i].var_i, obj_array[i].var_f, obj_array[i].var_str)
注意: 请使用你的struct名称,而不是'cstruct'名称 请定义结构的成员变量,而不是var_i, var_f, var_str。
其他回答
每当我需要一个“行为像字典一样的即时数据对象”(我不会想到C结构体!),我就会想到这个可爱的hack:
class Map(dict):
def __init__(self, **kwargs):
super(Map, self).__init__(**kwargs)
self.__dict__ = self
现在你可以说:
struct = Map(field1='foo', field2='bar', field3=42)
self.assertEquals('bar', struct.field2)
self.assertEquals(42, struct['field3'])
当你需要一个“不是类的数据包”的时候,非常方便,当命名元组是不可理解的……
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')
更新:数据类
随着Python 3.7中数据类的引入,我们已经非常接近了。
下面的示例与下面的NamedTuple示例类似,但是生成的对象是可变的,并且允许使用默认值。
from dataclasses import dataclass
@dataclass
class Point:
x: float
y: float
z: float = 0.0
p = Point(1.5, 2.5)
print(p) # Point(x=1.5, y=2.5, z=0.0)
如果您想使用更特定的类型注释,这可以很好地配合新的类型模块。
我一直在绝望地等待这一刻!要我说,Data Classes和新的NamedTuple声明,再加上typing模块,简直就是天赐之物!
改进的NamedTuple声明
自从Python 3.6以来,它变得非常简单和美丽(恕我直言),只要你能忍受不变性。
引入了一种声明NamedTuples的新方法,它也允许类型注释:
from typing import NamedTuple
class User(NamedTuple):
name: str
class MyStruct(NamedTuple):
foo: str
bar: int
baz: list
qux: User
my_item = MyStruct('foo', 0, ['baz'], User('peter'))
print(my_item) # MyStruct(foo='foo', bar=0, baz=['baz'], qux=User(name='peter'))
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'
NamedTuple很舒服。但是没有人共享性能和存储。
from typing import NamedTuple
import guppy # pip install guppy
import timeit
class User:
def __init__(self, name: str, uid: int):
self.name = name
self.uid = uid
class UserSlot:
__slots__ = ('name', 'uid')
def __init__(self, name: str, uid: int):
self.name = name
self.uid = uid
class UserTuple(NamedTuple):
# __slots__ = () # AttributeError: Cannot overwrite NamedTuple attribute __slots__
name: str
uid: int
def get_fn(obj, attr_name: str):
def get():
getattr(obj, attr_name)
return get
if 'memory test':
obj = [User('Carson', 1) for _ in range(1000000)] # Cumulative: 189138883
obj_slot = [UserSlot('Carson', 1) for _ in range(1000000)] # 77718299 <-- winner
obj_namedtuple = [UserTuple('Carson', 1) for _ in range(1000000)] # 85718297
print(guppy.hpy().heap()) # Run this function individually.
"""
Index Count % Size % Cumulative % Kind (class / dict of class)
0 1000000 24 112000000 34 112000000 34 dict of __main__.User
1 1000000 24 64000000 19 176000000 53 __main__.UserTuple
2 1000000 24 56000000 17 232000000 70 __main__.User
3 1000000 24 56000000 17 288000000 87 __main__.UserSlot
...
"""
if 'performance test':
obj = User('Carson', 1)
obj_slot = UserSlot('Carson', 1)
obj_tuple = UserTuple('Carson', 1)
time_normal = min(timeit.repeat(get_fn(obj, 'name'), repeat=20))
print(time_normal) # 0.12550550000000005
time_slot = min(timeit.repeat(get_fn(obj_slot, 'name'), repeat=20))
print(time_slot) # 0.1368690000000008
time_tuple = min(timeit.repeat(get_fn(obj_tuple, 'name'), repeat=20))
print(time_tuple) # 0.16006120000000124
print(time_tuple/time_slot) # 1.1694481584580898 # The slot is almost 17% faster than NamedTuple on Windows. (Python 3.7.7)
如果你的__dict__没有被使用,请在__slots__(更高的性能和存储)和NamedTuple(清晰的阅读和使用)之间选择。
您可以查看此链接(插槽的使用 )来获取更多的__slots__信息。