dmz*_*rsk 3 python performance immutability python-3.x
我的应用程序中需要很多小对象.它们必须是不可变的,并在设置新属性时返回一个新实例.
我找到了许多禁用obj.prop = newValue行为的方法,现在我需要这个:
newObj = obj.setTitle(title)
newObj = obj.setDirection(x, y)
newObj = obj.incrementCount()
newObj = obj.swap()
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目前我这样做:
class Info(object):
__slots__ = ['_x', '_y', ...]
def setDirection(self, x, y):
newObj = copy.copy(self) # shallow copy is OK
newObj._x = x
newObj._y = y
return newObj
def swap(self):
newObj = copy.copy(self)
# methods can do different things
newObj._x, newObj._y = self._y, self._x
return newObj
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在性能方面这是好的吗?是否有更快的方法返回对象的克隆,并更改了一些属性?
我用__slots__.我的对象有预定义的属性.我没有通用的.set(prop, value)方法
(欢迎使用Python 3.5+)
为了获得真正的不变性,我宁愿继承collections.namedtuple并使用该方法_replace():
import collections as co
# this will create a class with five attributes
class Info(co.namedtuple('BaseInfo', 'x y a b c')):
__slots__ = ()
def setDirection(self, x, y):
return self._replace(x=x, y=y)
def swap(self):
return self._replace(x=self.y, y=self.x)
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我已经swap()在两个类中对该方法的性能进行了基准测试,并且派生的类namedtuple在python 3中的速度提高了3-4倍.以下是基准代码:
import copy
import collections as co
class Info(object):
__slots__ = ['x', 'y', 'a', 'b', 'c']
def swap(self):
newObj = copy.copy(self)
newObj.x, newObj.y = self.y, self.x
return newObj
# for the sake of convenience
def __init__(self, x, y, a, b, c):
self.x = x
self.y = y
class TupleInfo(co.namedtuple('BaseInfo', 'x y a b c')):
__slots__ = ()
def swap(self):
return self._replace(x=self.y, y=self.x)
if __name__ == "__main__":
from timeit import timeit
i1 = Info(1, 2, 0, 0, 0)
i2 = TupleInfo(1, 2, 0, 0, 0)
print("Built from scratch")
print(timeit("z = i1.swap()", "from __main__ import i1", number=100000))
print("Derived from namedtuple")
print(timeit("z = i2.swap()", "from __main__ import i2", number=100000))
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结果:
Built from scratch
1.8578372709998803
Derived from namedtuple
0.520611657999325
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