blocks|key|595967|text|您可以通过反汇编程序运行它来找出：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|595968|import+dis

def+test():
++++name+=+'test'
++++tdict+=+{}
++++tdict[name]+=+0
++++tdict[name]+%2B=+1
++++tdict[name]+%2B=+1

dis.dis(test)|code-block|syntax|javascript|595969|运行下面的代码，我们会得到：|595970|+13+++++++++++0+LOAD_CONST+++++++++++++++1+('test')
++++++++++++++3+STORE_FAST+++++++++++++++0+(name)

+14+++++++++++6+BUILD_MAP++++++++++++++++0
++++++++++++++9+STORE_FAST+++++++++++++++1+(tdict)

+15++++++++++12+LOAD_CONST+++++++++++++++2+(0)
+++++++++++++15+LOAD_FAST++++++++++++++++1+(tdict)
+++++++++++++18+LOAD_FAST++++++++++++++++0+(name)
+++++++++++++21+STORE_SUBSCR++++++++

+16++++++++++22+LOAD_FAST++++++++++++++++1+(tdict)
+++++++++++++25+LOAD_FAST++++++++++++++++0+(name)
+++++++++++++28+DUP_TOPX+++++++++++++++++2
+++++++++++++31+BINARY_SUBSCR+++++++
+++++++++++++32+LOAD_CONST+++++++++++++++3+(1)
+++++++++++++35+INPLACE_ADD+++++++++
+++++++++++++36+ROT_THREE+++++++++++
+++++++++++++37+STORE_SUBSCR++++++++

+17++++++++++38+LOAD_FAST++++++++++++++++1+(tdict)
+++++++++++++41+LOAD_FAST++++++++++++++++0+(name)
+++++++++++++44+DUP_TOPX+++++++++++++++++2
+++++++++++++47+BINARY_SUBSCR+++++++
+++++++++++++48+LOAD_CONST+++++++++++++++3+(1)
+++++++++++++51+INPLACE_ADD+++++++++
+++++++++++++52+ROT_THREE+++++++++++
+++++++++++++53+STORE_SUBSCR++++++++
+++++++++++++54+LOAD_CONST+++++++++++++++0+(None)
+++++++++++++57+RETURN_VALUE++++++++|595971|看起来，在这种情况下，每次我们尝试访问LOAD_FAST来执行增量操作时，它都会加载tdict和name的值，因此答案似乎是否定的。|offset|length|style|CODE|595972|entityMap^0|0|0|0|0|J|9|16|5|1C|4|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]|$M|Z|N|10|O|P]|$M|11|N|12|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|R|$]]

You can run it through the disassembler to find out:

<pre><code>import dis

def test():
 name = 'test'
 tdict = {}
 tdict[name] = 0
 tdict[name] += 1
 tdict[name] += 1

dis.dis(test)
</code></pre>

Running this we get:

<pre><code> 13 0 LOAD_CONST 1 ('test')
 3 STORE_FAST 0 (name)

 14 6 BUILD_MAP 0
 9 STORE_FAST 1 (tdict)

 15 12 LOAD_CONST 2 (0)
 15 LOAD_FAST 1 (tdict)
 18 LOAD_FAST 0 (name)
 21 STORE_SUBSCR 

 16 22 LOAD_FAST 1 (tdict)
 25 LOAD_FAST 0 (name)
 28 DUP_TOPX 2
 31 BINARY_SUBSCR 
 32 LOAD_CONST 3 (1)
 35 INPLACE_ADD 
 36 ROT_THREE 
 37 STORE_SUBSCR 

 17 38 LOAD_FAST 1 (tdict)
 41 LOAD_FAST 0 (name)
 44 DUP_TOPX 2
 47 BINARY_SUBSCR 
 48 LOAD_CONST 3 (1)
 51 INPLACE_ADD 
 52 ROT_THREE 
 53 STORE_SUBSCR 
 54 LOAD_CONST 0 (None)
 57 RETURN_VALUE 
</code></pre>

It looks like, in this case, that <code>LOAD_FAST</code> is loading up the values of <code>tdict</code> and <code>name</code> each time we try to access it to perform the increment, so the answer would appear to be no.

blocks|key|811460|text|这种类型的优化不可能仅仅通过检查代码来实现。您的名字dict可以引用实现__setitem__的用户定义对象，而不是本机字典，并且该方法必须被调用三次。在运行时，复杂的实现可以注意到名称的实际值，并进行优化，但在运行之前无法在不破坏某些Python语义的情况下完成此操作。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|811461|entityMap^0|Q|4|10|B|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

That type of optimization isn't possible simply by inspecting the code. Your name <code>dict</code> could refer not to a native dictionary, but a user-defined object that implements <code>__setitem__</code>, and that method has to be called three times. At runtime, a sophisticated implementation could note the actual value of the name, and make an optimization, but it can't be done before runtime without breaking some Python semantics.

blocks|key|811498|text|不，因为这是行不通的，你甚至用你自己的代码演示了这一点--这实际上并不等同：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|811499|>>>+a+=+{}
>>>+name+=+'x'
>>>+a[name]+=+0
>>>+a[name]+%2B=+1
>>>+a[name]+%2B=+1
>>>+a[name]+#+ok+no+suprises+so+far
2
>>>+a+=+{}
>>>+a[name]+=+0
>>>+x+=+a[name]+#+x+is+now+literally+`0`,+not+some+sort+of+reference+to+a[name]
>>>+x
0
>>>+x+%2B=+1
>>>+x+%2B=+1
>>>+a[name]+#+so+this+never+changed
0
>>>|code-block|syntax|javascript|811500|Python没有C语言的“引用”。您所想的只适用于可变类型，如list。这是Python的一个非常基本的属性，在编写Python时，您可能会忘记C语言教给您的有关变量的所有内容。|offset|length|style|CODE|811501|entityMap^0|0|0|V|4|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

No, because that would not work, and you even demonstated that with your own code - it's actually not equivalent:

<pre><code>&gt;&gt;&gt; a = {}
&gt;&gt;&gt; name = 'x'
&gt;&gt;&gt; a[name] = 0
&gt;&gt;&gt; a[name] += 1
&gt;&gt;&gt; a[name] += 1
&gt;&gt;&gt; a[name] # ok no suprises so far
2
&gt;&gt;&gt; a = {}
&gt;&gt;&gt; a[name] = 0
&gt;&gt;&gt; x = a[name] # x is now literally `0`, not some sort of reference to a[name]
&gt;&gt;&gt; x
0
&gt;&gt;&gt; x += 1
&gt;&gt;&gt; x += 1
&gt;&gt;&gt; a[name] # so this never changed
0
&gt;&gt;&gt;
</code></pre>

Python does not have C-ish "references". What you had in mind would work only for mutable types such as <code>list</code>. This is a very fundamental property of Python and you should probably forget everything C taught you about variables when programming Python.

blocks|key|811519|text|将您的两个示例更改为如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|811520|#v1.py
di+=+{}
name+=+"hallo"
di[name]+=+0
for+i+in+range(2000000):
++++di[name]+%2B=+1|code-block|syntax|javascript|811521|和|811522|#v2.py
di+=+{}
name+=+"hallo"
di[name]+=+0
value+=+di[name]
for+i+in+range(2000000):
++++value+%2B=+1|811523|你可以在下面的测试中看到，v2更快，但是pypy更快:-)|811524|$+time+python2.7+v1.py
real++++0m0.788s
user++++0m0.700s
sys+++++0m0.080s

$+time+python2.7+v2.py
real++++0m0.586s
user++++0m0.490s
sys+++++0m0.090s

$+time+pypy+v1.py
real++++0m0.203s
user++++0m0.210s
sys+++++0m0.000s

$+time+pypy+v2.py
real++++0m0.117s
user++++0m0.080s
sys+++++0m0.030s|811525|所以:优化单个解释器的代码并不好(例如，我还没有测试过Jython...)，但是当someone优化解释器时，这是非常好的……|offset|length|811526|entityMap|0|LINK|mutability|MUTABLE|url|http://pypy.org/people.html^0|0|0|0|0|0|0|15|7|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|14|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|16|8|@]|9|@$Q|17|R|18|1|19]]|A|$]]|$1|S|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]]|T|$U|$5|V|W|X|A|$Y|Z]]]]

Changing your two examples into something like this:

<pre><code>#v1.py
di = {}
name = "hallo"
di[name] = 0
for i in range(2000000):
 di[name] += 1
</code></pre>

and 

<pre><code>#v2.py
di = {}
name = "hallo"
di[name] = 0
value = di[name]
for i in range(2000000):
 value += 1
</code></pre>

You can see in the following tests, that v2 is faster, but pypy is much faster :-)

<pre><code>$ time python2.7 v1.py
real 0m0.788s
user 0m0.700s
sys 0m0.080s

$ time python2.7 v2.py
real 0m0.586s
user 0m0.490s
sys 0m0.090s

$ time pypy v1.py
real 0m0.203s
user 0m0.210s
sys 0m0.000s

$ time pypy v2.py
real 0m0.117s
user 0m0.080s
sys 0m0.030s
</code></pre>

SO: it's not good to optimize code for a single interpreter (I have not tested Jython for example...), but it's great when <a href="http://pypy.org/people.html" rel="nofollow">someone</a> optimizes the interpreter...

Consider the following code snippet:

<pre><code>dict [name] = 0
dict [name] += 1
dict [name] += 1
</code></pre>

Does the python interpreter automatically recognise the repeated references to the dictionary value and use a cached local reference instead?, somewhat akin to the aliasing optimisations of C/C++, becoming something like so:

<pre><code>value = dict [name]
value = 0
value += 1
value += 1
</code></pre>

Obviously, it's not a big deal to do this manually but I'm curious if it's actually necessary. any insight, feedback, etc is appreciated.

Python interpreted code optimisation

解释器

Python

考虑下面的代码片段：dict [name] = 0dict [name] += 1dict [name] += 1python解释器是否会自动识别对字典值的重复引用，并使用缓存的本地引用，这有点类似于C/C++的别名优化，如下所示：value = dict [name]value = 0value += 1value ...

问Python解释型代码优化
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