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

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')

其他回答

我认为Python结构字典适合这个需求。

d = dict{}
d[field1] = field1
d[field2] = field2
d[field2] = field3

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")

也许你正在寻找没有构造函数的struct:

class Sample:
  name = ''
  average = 0.0
  values = None # list cannot be initialized here!


s1 = Sample()
s1.name = "sample 1"
s1.values = []
s1.values.append(1)
s1.values.append(2)
s1.values.append(3)

s2 = Sample()
s2.name = "sample 2"
s2.values = []
s2.values.append(4)

for v in s1.values:   # prints 1,2,3 --> OK.
  print v
print "***"
for v in s2.values:   # prints 4 --> OK.
  print v

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。

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。