Constructing objects in __init__
I've seen code that looks something like this:
class MyClass:
def __init__(self, someargs):
myObj = OtherClass()
myDict = {}
...code to setup myObj, myDict...
self.myObj = myObj
self.myDict = myDict
My first thought when I saw this was: Why not just use self.myObj and self.myDict in the beginning? It seems inefficient to construct local objects, then assign them to the members. The code to construct the objects do possibly throw exceptions, maybe they did this so it wouldn't leave a half-constructed object? Do you do this, or just construct the开发者_StackOverflow社区 members directly?
It's faster and more readable to construct the object and then attach it to self
.
class Test1(object):
def __init__(self):
d = {}
d['a'] = 1
d['b'] = 2
d['c'] = 3
self.d = d
class Test2(object):
def __init__(self):
self.d = {}
self.d['a'] = 1
self.d['b'] = 2
self.d['c'] = 3
import dis
print "Test1.__init__"
dis.dis(Test1.__init__)
print "Test2.__init__"
dis.dis(Test2.__init__)
disassemles to:
Test1.__init__
4 0 BUILD_MAP 0
3 STORE_FAST 1 (d)
5 6 LOAD_CONST 1 (1)
9 LOAD_FAST 1 (d)
12 LOAD_CONST 2 ('a')
15 STORE_SUBSCR
6 16 LOAD_CONST 3 (2)
19 LOAD_FAST 1 (d)
22 LOAD_CONST 4 ('b')
25 STORE_SUBSCR
7 26 LOAD_CONST 5 (3)
29 LOAD_FAST 1 (d)
32 LOAD_CONST 6 ('c')
35 STORE_SUBSCR
8 36 LOAD_FAST 1 (d)
39 LOAD_FAST 0 (self)
42 STORE_ATTR 0 (d)
45 LOAD_CONST 0 (None)
48 RETURN_VALUE
Test2.__init__
12 0 BUILD_MAP 0
3 LOAD_FAST 0 (self)
6 STORE_ATTR 0 (d)
13 9 LOAD_CONST 1 (1)
12 LOAD_FAST 0 (self)
15 LOAD_ATTR 0 (d)
18 LOAD_CONST 2 ('a')
21 STORE_SUBSCR
14 22 LOAD_CONST 3 (2)
25 LOAD_FAST 0 (self)
28 LOAD_ATTR 0 (d)
31 LOAD_CONST 4 ('b')
34 STORE_SUBSCR
15 35 LOAD_CONST 5 (3)
38 LOAD_FAST 0 (self)
41 LOAD_ATTR 0 (d)
44 LOAD_CONST 6 ('c')
47 STORE_SUBSCR
48 LOAD_CONST 0 (None)
51 RETURN_VALUE
You can see that STORE_ATTR
only gets called once doing it the first way at the end. Doing it the other way, STORE_ATTR
still gets called right at the beginning but now LOAD_ATTR
gets called for every access to the dictionary. The more assignments there are, the higher the cost. Every other instruction is the same. It's still a ridiculously small cost.
This trick can be exploited to make loops with many iterations run faster. It's not uncommon to see things like
foo = self.foo
factorial = math.factorial
for x in really_big_iterator:
foo(factorial(x))
another trick is to pass global functions as default arguments to a function that has a loop like that or gets called a whole bunch to save some attribute lookups: it's in the local scope which is the first one looked in.
def fast(iterators, sum=sum):
for i in iterator:
yield sum(i)
now sum is right in the local scope.
If you are worried about the performance implication of copying an object reference, you are probably using the wrong language :)
Do what's more readable to you. In this case, it depends on how long is the init method.
I'm not sure I fully understand your question, but keep in mind that assigning self.myObj = myObj
will simply assign a reference so it isn't likely to slow things down much. My guess is that idiom is being used to save the programmer from typing the word self
a bunch of times.
There is really no significant difference either way. One reason to create a local variable is if you are going to use the variable in __init__ a lot, you don't have to repeat the self.
so much. e.g.
def __init__(self, someargs):
self.myObj = OtherClass()
self.myDict = {}
self.myDict[1] = self.myObj
self.myDict[2] = self.myObj
self.myDict[3] = self.myObj
self.myDict[4] = self.myObj
self.myObj = myObj
self.myDict = myDict
v.s.
def __init__(self, someargs):
obj = OtherClass()
d = {}
d[1] = obj
d[2] = obj
d[3] = obj
d[4] = obj
self.myObj = obj
self.myDict = d
I won't worry performance so much unless you have a good reason to.
Assigning a reference doesn't take too much time, but typing 'self' each time does take you time.
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