在生成器表达式(test1)周围使用[]似乎比将其放在list()(test2)中要好得多。当我只是将列表传递给list()以进行浅拷贝(test3)时,速度就不存在了。这是为什么?
证据:
from timeit import Timer
t1 = Timer("test1()", "from __main__ import test1")
t2 = Timer("test2()", "from __main__ import test2")
t3 = Timer("test3()", "from __main__ import test3")
x = [34534534, 23423523, 77645645, 345346]
def test1():
[e for e in x]
print t1.timeit()
#0.552290201187
def test2():
list(e for e in x)
print t2.timeit()
#2.38739395142
def test3():
list(x)
print t3.timeit()
#0.515818119049
机器:64位AMD,Ubuntu 8.04,Python 2.7(r27:82500)
答案 0 :(得分:34)
嗯,我的第一步是独立设置两个测试,以确保这不是由例如定义函数的顺序。
>python -mtimeit "x=[34534534, 23423523, 77645645, 345346]" "[e for e in x]"
1000000 loops, best of 3: 0.638 usec per loop
>python -mtimeit "x=[34534534, 23423523, 77645645, 345346]" "list(e for e in x)"
1000000 loops, best of 3: 1.72 usec per loop
果然,我可以复制这个。好的,下一步是查看字节码,看看实际发生了什么:
>>> import dis
>>> x=[34534534, 23423523, 77645645, 345346]
>>> dis.dis(lambda: [e for e in x])
1 0 LOAD_CONST 0 (<code object <listcomp> at 0x0000000001F8B330, file "<stdin>", line 1>)
3 MAKE_FUNCTION 0
6 LOAD_GLOBAL 0 (x)
9 GET_ITER
10 CALL_FUNCTION 1
13 RETURN_VALUE
>>> dis.dis(lambda: list(e for e in x))
1 0 LOAD_GLOBAL 0 (list)
3 LOAD_CONST 0 (<code object <genexpr> at 0x0000000001F8B9B0, file "<stdin>", line 1>)
6 MAKE_FUNCTION 0
9 LOAD_GLOBAL 1 (x)
12 GET_ITER
13 CALL_FUNCTION 1
16 CALL_FUNCTION 1
19 RETURN_VALUE
请注意,第一个方法直接创建列表,而第二个方法创建一个genexpr
对象并将其传递给全局list
。这可能是开销所在。
另请注意,差异大约是一微秒,即极其微不足道。
这仍然适用于非平凡的列表
>python -mtimeit "x=range(100000)" "[e for e in x]"
100 loops, best of 3: 8.51 msec per loop
>python -mtimeit "x=range(100000)" "list(e for e in x)"
100 loops, best of 3: 11.8 msec per loop
以及不那么简单的地图功能:
>python -mtimeit "x=range(100000)" "[2*e for e in x]"
100 loops, best of 3: 12.8 msec per loop
>python -mtimeit "x=range(100000)" "list(2*e for e in x)"
100 loops, best of 3: 16.8 msec per loop
和(虽然不太强烈)如果我们过滤列表:
>python -mtimeit "x=range(100000)" "[e for e in x if e%2]"
100 loops, best of 3: 14 msec per loop
>python -mtimeit "x=range(100000)" "list(e for e in x if e%2)"
100 loops, best of 3: 16.5 msec per loop
答案 1 :(得分:9)
list(e for e in x)
不是列表解析,而是创建genexpr
对象(e for e in x)
并将其传递给list
工厂函数。据推测,对象创建和方法调用会产生开销。
答案 2 :(得分:2)
在python list
中,必须在模块中查找名称,然后在内置中查找。虽然你无法改变列表理解意味着列表调用必须只是标准的查找+函数调用,因为它可以被重新定义为其他东西。
查看为理解而生成的vm代码,可以看到它在内联调用时是内联的。
>>> import dis
>>> def foo():
... [x for x in xrange(4)]
...
>>> dis.dis(foo)
2 0 BUILD_LIST 0
3 DUP_TOP
4 STORE_FAST 0 (_[1])
7 LOAD_GLOBAL 0 (xrange)
10 LOAD_CONST 1 (4)
13 CALL_FUNCTION 1
16 GET_ITER
>> 17 FOR_ITER 13 (to 33)
20 STORE_FAST 1 (x)
23 LOAD_FAST 0 (_[1])
26 LOAD_FAST 1 (x)
29 LIST_APPEND
30 JUMP_ABSOLUTE 17
>> 33 DELETE_FAST 0 (_[1])
36 POP_TOP
37 LOAD_CONST 0 (None)
40 RETURN_VALUE
>>> def bar():
... list(x for x in xrange(4))
...
>>> dis.dis(bar)
2 0 LOAD_GLOBAL 0 (list)
3 LOAD_CONST 1 (<code object <genexpr> at 0x7fd1230cf468, file "<stdin>", line 2>)
6 MAKE_FUNCTION 0
9 LOAD_GLOBAL 1 (xrange)
12 LOAD_CONST 2 (4)
15 CALL_FUNCTION 1
18 GET_ITER
19 CALL_FUNCTION 1
22 CALL_FUNCTION 1
25 POP_TOP
26 LOAD_CONST 0 (None)
29 RETURN_VALUE
答案 3 :(得分:1)
您的test2大致相当于:
def test2():
def local():
for i in x:
yield i
return list(local())
呼叫开销解释了处理时间的增加。