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
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……
其他回答
这里的一些答案非常详尽。我找到的最简单的选项是(from: http://norvig.com/python-iaq.html):)
class Struct:
"A structure that can have any fields defined."
def __init__(self, **entries): self.__dict__.update(entries)
初始化:
>>> options = Struct(answer=42, linelen=80, font='courier')
>>> options.answer
42
添加更多的:
>>> options.cat = "dog"
>>> options.cat
dog
编辑:对不起,没有看到这个例子已经进一步。
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
就我个人而言,我也喜欢这种变体。它扩展了@dF的答案。
class struct:
def __init__(self, *sequential, **named):
fields = dict(zip(sequential, [None]*len(sequential)), **named)
self.__dict__.update(fields)
def __repr__(self):
return str(self.__dict__)
它支持两种初始化模式(可以混合使用):
# Struct with field1, field2, field3 that are initialized to None.
mystruct1 = struct("field1", "field2", "field3")
# Struct with field1, field2, field3 that are initialized according to arguments.
mystruct2 = struct(field1=1, field2=2, field3=3)
而且,它打印得更好:
print(mystruct2)
# Prints: {'field3': 3, 'field1': 1, 'field2': 2}
你可以用元组来做很多你在C语言中使用结构体的事情(比如x,y坐标或RGB颜色)。
对于其他任何东西,你可以使用字典,或像这样的实用程序类:
>>> class Bunch:
... def __init__(self, **kwds):
... self.__dict__.update(kwds)
...
>>> mystruct = Bunch(field1=value1, field2=value2)
我认为“权威”的讨论在这里,在Python Cookbook的出版版本中。
我发现做到这一点的最好方法是使用自定义字典类,如本文所述:https://stackoverflow.com/a/14620633/8484485
如果需要iPython自动补全支持,只需像这样定义dir()函数:
class AttrDict(dict):
def __init__(self, *args, **kwargs):
super(AttrDict, self).__init__(*args, **kwargs)
self.__dict__ = self
def __dir__(self):
return self.keys()
然后像这样定义你的伪结构(这个是嵌套的)
my_struct=AttrDict ({
'com1':AttrDict ({
'inst':[0x05],
'numbytes':2,
'canpayload':False,
'payload':None
})
})
然后你可以像这样访问my_struct中的值:
打印(my_struct.com1.inst)
= > [5]