使用python3.4。在这里,我想使用singledispatch在__mul__
方法中调度不同的类型。像这样的代码:
class Vector(object):
## some code not paste
@functools.singledispatch
def __mul__(self, other):
raise NotImplementedError("can't mul these type")
@__mul__.register(int)
@__mul__.register(object) # Becasue can't use Vector , I have to use object
def _(self, other):
result = Vector(len(self)) # start with vector of zeros
for j in range(len(self)):
result[j] = self[j]*other
return result
@__mul__.register(Vector) # how can I use the self't type
@__mul__.register(object) #
def _(self, other):
pass # need impl
正如您可以看到代码,我想要支持Vector * Vertor,这有名称错误
Traceback (most recent call last):
File "p_algorithms\vector.py", line 6, in <module>
class Vector(object):
File "p_algorithms\vector.py", line 84, in Vector
@__mul__.register(Vector) # how can I use the self't type
NameError: name 'Vector' is not defined
问题可能是如何在班级方法中使用班级名称类型?我知道c ++有字体类声明。 python如何解决我的问题?看到result = Vector(len(self))
可以在方法体中使用Vector
,这很奇怪。
更新。看看http://lukasz.langa.pl/8/single-dispatch-generic-functions/之后
我可以选择这种方式来实现:
import unittest
from functools import singledispatch
class Vector(object):
"""Represent a vector in a multidimensional space."""
def __init__(self, d):
self._coords = [0 for i in range(0, d)]
self.__init__mul__()
def __init__mul__(self):
__mul__registry = self.__mul__.registry
self.__mul__ = singledispatch(__mul__registry[object])
self.__mul__.register(int, self.mul_int)
self.__mul__.register(Vector, self.mul_Vector)
def __setitem__(self, key, value):
self._coords[key] = value
def __getitem__(self, item):
return self._coords[item]
def __len__(self):
return len(self._coords)
def __str__(self):
return str(self._coords)
@singledispatch
def __mul__(self, other):
print ("error type is ", type(other))
print (type(other))
raise NotImplementedError("can't mul these type")
def mul_int(self,other):
print ("other type is ", type(other))
result = Vector(len(self)) # start with vector of zeros
for j in range(len(self)):
result[j] = self[j]*other
return result
def mul_Vector(self, other):
print ("other type is ", type(other))
#result = Vector(len(self)) # start with vector of zeros
sum = 0
for i in range(0,len(self)):
sum += self._coords[i] * other._coords[i]
return sum
class TestCase(unittest.TestCase):
def test_singledispatch(self):
# the following demonstrates usage of a few methods
v = Vector(5) # construct five-dimensional <0, 0, 0, 0, 0>
for i in range(1,6):
v[i-1] = i
print(v.__mul__(3))
print(v.__mul__(v))
print(v*3)
if __name__ == "__main__":
unittest.main()
ans很奇怪:
other type is <class 'int'> [3, 6, 9, 12, 15] other type is <class '__main__.Vector'> 55 error type is <class 'int'> Traceback (most recent call last): File "p_algorithms\vector.py", line 164, in <module> print(v*3) File "C:\Python34\lib\functools.py", line 710, in wrapper return dispatch(args[0].__class__)(*args, **kw) File "p_algorithms\vector.py", line 111, in __mul__ raise NotImplementedError("can't mul these type")
v.__mul__(3)
可行,但v*3
无法正常工作。这很奇怪我的选项v*3
与v.__mul__(3)
相同。
在@Martijn Pieters的评论之后更新,我仍然希望在课堂上实现v * 3。所以我试试这个
import unittest
from functools import singledispatch
class Vector(object):
@staticmethod
def static_mul_int(self,other):
print ("other type is ", type(other))
result = Vector(len(self)) # start with vector of zeros
for j in range(len(self)):
result[j] = self[j]*other
return result
@singledispatch
@staticmethod
def __static_mul__(cls, other):
print ("error type is ", type(other))
print (type(other))
raise NotImplementedError("can't mul these type")
__mul__registry2 = __static_mul__.registry
__mul__ = singledispatch(__mul__registry2[object])
__mul__.register(int, static_mul_int)
def __init__(self, d):
self._coords = [0 for i in range(0, d)]
self.__init__mul__()
def __init__mul__(self):
__mul__registry = self.__mul__.registry
print ("__mul__registry",__mul__registry,__mul__registry[object])
self.__mul__ = singledispatch(__mul__registry[object])
self.__mul__.register(int, self.mul_int)
print ("at last __mul__registry",self.__mul__.registry)
# @singledispatch
# def __mul__(self, other):
# print ("error type is ", type(other))
# print (type(other))
# raise NotImplementedError("can't mul these type")
def mul_int(self,other):
print ("other type is ", type(other))
result = Vector(len(self)) # start with vector of zeros
for j in range(len(self)):
result[j] = self[j]*other
return result
def __setitem__(self, key, value):
self._coords[key] = value
def __getitem__(self, item):
return self._coords[item]
def __len__(self):
return len(self._coords)
def __str__(self):
return str(self._coords)
class TestCase(unittest.TestCase):
def test_singledispatch(self):
# the following demonstrates usage of a few methods
v = Vector(5) # construct five-dimensional <0, 0, 0, 0, 0>
for i in range(1,6):
v[i-1] = i
print(v.__mul__(3))
print("type(v).__mul__'s registry:",type(v).__mul__.registry)
type(v).__mul__(v, 3)
print(v*3)
if __name__ == "__main__":
unittest.main()
这一次。我的实现就像我实现v.__mul__(3)
。但错误是
Traceback (most recent call last): File "test.py", line 73, in test_singledispatch type(v).__mul__(v, 3) File "/usr/lib/python3.4/functools.py", line 708, in wrapper return dispatch(args[0].__class__)(*args, **kw) TypeError: 'staticmethod' object is not callable
对我来说,静态methond应该像实例methond一样。
答案 0 :(得分:12)
你不能在方法 >>上使用functools.singledispatch
,至少不能作为装饰者使用。
Vector
此处尚未定义,这并不重要;任何方法的第一个参数总是self
,而你在这里使用单个调度用于第二个参数。
因为装饰器在创建类对象之前应用于函数对象,所以你也可以注册你的方法&#39;作为函数,而不是类主体的外部,因此您可以访问Vector
名称:
class Vector(object):
@functools.singledispatch
def __mul__(self, other):
return NotImplemented
@Vector.__mul__.register(int)
@Vector.__mul__.register(Vector)
def _(self, other):
result = Vector(len(self)) # start with vector of zeros
for j in range(len(self)):
result[j] = self[j]*other
return result
对于不受支持的类型,您需要返回NotImplemented
单例,而不是引发异常。这样Python也会尝试逆操作。
但是,由于此处的调度将关键错误的参数(self
),因此您必须提出自己的单一调度机制。< / p>
如果你真的想使用@functools.singledispatch
,你必须委托常规函数,参数 inversed :
@functools.singledispatch
def _vector_mul(other, self):
return NotImplemented
class Vector(object):
def __mul__(self, other):
return _vector_mul(other, self)
@_vector_mul.register(int)
def _vector_int_mul(other, self):
result = Vector(len(self))
for j in range(len(self)):
result[j] = self[j] * other
return result
对于使用__init__mul__
的更新:v * 3
未已翻译为v.__mul__(3)
。它转换为type(v).__mul__(v, 3)
,请参阅Python数据模型参考中的Special method lookup。此总是绕过直接在实例上设置的任何方法。
此处type(v)
为Vector
; Python查找函数,它不会在这里使用绑定方法。同样,因为functools.singledispatch
一直在第一个参数上调度,所以你不能直接在Vector
的方法上使用单一调度,因为第一个参数总是一个{ {1}}实例。
换句话说,Python将不使用您在Vector
中self
上设置的方法;特殊方法从不在实例上查找,请参阅datamodel文档中的Special method lookup。
答案 1 :(得分:1)
这有点难看,因为您需要推迟绑定Vector
/ Vector
乘法的实现,直到实际定义Vector
之后。但是这个想法是单调度函数需要第一个参数是任意类型的,所以Vector.__mul__
将调用该函数,self
作为第二个参数。
import functools
class Vector:
def __mul__(self, other):
# Python has already dispatched Vector() * object() here, so
# swap the arguments so that our single-dispatch works. Note
# that in general if a*b != b*a, then the _mul_by_other
# implementations need to compensate.
return Vector._mul_by_other(other, self)
@functools.singledispatch
def _mul_by_other(x, y):
raise NotImplementedError("Can't multiply vector by {}".format(type(x)))
@_mul_by_other.register(int)
def _(x, y):
print("Multiply vector by int")
@Vector._mul_by_other.register(Vector)
def _(x, y):
print("Multiply vector by another vector")
x = Vector()
y = Vector()
x * 3
x * y
try:
x * "foo"
except NotImplementedError:
print("Caught attempt to multiply by string")