检查密钥是否存在的更有效/更快/更好的方法是哪种?
if 'subject' in request.POST:
subject = request.POST['subject']
else:
// handle error
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要么
try:
subject = request.POST['subject']
except KeyError:
// handle error
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Mik*_*ham 38
后者(try/except
)形式通常是更好的形式.
try
块非常便宜但是捕获异常可能更昂贵.对dict的收容检查往往便宜,但并不比没有便宜.我怀疑效率会有平衡,具体取决于'subject'
实际存在的频率.然而,这并不重要,因为过早的优化是无用的,分散注意力,浪费和无效.你会选择更好的解决方案.
如果代码实际上是形式的
if 'subject' in request.POST:
subject = request.POST['subject']
else:
subject = some_default
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然后你真正想要的是request.POST.get('subject', some_default)
.
小智 5
我使用 .get() 方法——这是更好的方法。
Python 2.5.2 (r252:60911, Jul 22 2009, 15:33:10)
[GCC 4.2.4 (Ubuntu 4.2.4-1ubuntu3)] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import dis
>>> def f1(key, d):
... if key in d:
... return d[key]
... else:
... return "default"
...
>>> dis.dis(f1)
2 0 LOAD_FAST 0 (key)
3 LOAD_FAST 1 (d)
6 COMPARE_OP 6 (in)
9 JUMP_IF_FALSE 12 (to 24)
12 POP_TOP
3 13 LOAD_FAST 1 (d)
16 LOAD_FAST 0 (key)
19 BINARY_SUBSCR
20 RETURN_VALUE
21 JUMP_FORWARD 5 (to 29)
>> 24 POP_TOP
5 25 LOAD_CONST 1 ('default')
28 RETURN_VALUE
>> 29 LOAD_CONST 0 (None)
32 RETURN_VALUE
>>> def f2(key, d):
... return d.get(key, "default")
...
>>> dis.dis(f2)
2 0 LOAD_FAST 1 (d)
3 LOAD_ATTR 0 (get)
6 LOAD_FAST 0 (key)
9 LOAD_CONST 1 ('default')
12 CALL_FUNCTION 2
15 RETURN_VALUE
>>> def f3(key, d):
... try:
... return d[key]
... except KeyError:
... return "default"
...
>>> dis.dis(f3)
2 0 SETUP_EXCEPT 12 (to 15)
3 3 LOAD_FAST 1 (d)
6 LOAD_FAST 0 (key)
9 BINARY_SUBSCR
10 RETURN_VALUE
11 POP_BLOCK
12 JUMP_FORWARD 23 (to 38)
4 >> 15 DUP_TOP
16 LOAD_GLOBAL 0 (KeyError)
19 COMPARE_OP 10 (exception match)
22 JUMP_IF_FALSE 11 (to 36)
25 POP_TOP
26 POP_TOP
27 POP_TOP
28 POP_TOP
5 29 LOAD_CONST 1 ('default')
32 RETURN_VALUE
33 JUMP_FORWARD 2 (to 38)
>> 36 POP_TOP
37 END_FINALLY
>> 38 LOAD_CONST 0 (None)
41 RETURN_VALUE
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