Eri*_*ans 130 python performance for-loop list-comprehension map-function
在Python的性能方面,是一个列表理解,还是比for循环更快的map(),filter()和reduce()等函数？从技术上讲,为什么它们"以C速度运行",而"for循环以python虚拟机速度运行"？
假设在我正在开发的游戏中,我需要使用for循环绘制复杂且巨大的地图.这个问题肯定是相关的,因为如果列表理解确实更快,那么为了避免滞后(尽管代码的视觉复杂性),这将是一个更好的选择.
小智 126
以下是基于经验的粗略指导和有根据的猜测.您应该timeit或者描述您的具体用例以获取更难的数字,这些数字可能偶尔会不同意以下内容.
列表推导通常比精确等效的for循环(实际上构建列表)快一点,很可能是因为它不必append在每次迭代时查找列表及其方法.但是,列表推导仍然会执行字节码级循环:
>>> dis.dis(<the code object for `[x for x in range(10)]`>)
1 0 BUILD_LIST 0
3 LOAD_FAST 0 (.0)
>> 6 FOR_ITER 12 (to 21)
9 STORE_FAST 1 (x)
12 LOAD_FAST 1 (x)
15 LIST_APPEND 2
18 JUMP_ABSOLUTE 6
>> 21 RETURN_VALUE
使用列表推导代替不构建列表的循环,无意义地累积无意义值列表然后抛弃列表,由于创建和扩展列表的开销,因此通常较慢.列表推导不是神奇的,本质上比旧的循环更快.
至于功能列表处理功能:虽然这些都是用C语言编写,并可能超越Python编写的相同的功能,它们是不是一定是最快的选择.如果函数也是用C语言写的,那么预计会有一些加速.但是大多数情况下使用lambda(或其他Python函数),重复设置Python堆栈帧等的开销会减少任何节省.简单地在线执行相同的工作,没有函数调用(例如,列表理解而不是map或filter)通常会稍快一些.
假设在我正在开发的游戏中,我需要使用for循环绘制复杂且巨大的地图.这个问题肯定是相关的,因为如果列表理解确实更快,那么为了避免滞后(尽管代码的视觉复杂性),这将是一个更好的选择.
如果像这样的代码在用非"优化"的Python编写时还不够快,那么很可能没有多少Python级别的微优化能够让它足够快,你应该开始考虑下降到C.微优化通常可以大大加速Python代码,对此有一个较低的(绝对值)限制.而且,甚至在你达到这个上限之前,它就变得更具成本效益(15%的加速比300%加速同样的努力)咬住子弹并写下一些C.
Ant*_*ong 19
如果你查看python.org上的信息,你可以看到这个摘要:
Version Time (seconds)
Basic loop 3.47
Eliminate dots 2.45
Local variable & no dots 1.79
Using map function 0.54
但是你真的应该详细阅读上面的文章,以了解性能差异的原因.
我还强烈建议您使用timeit来计算代码.在一天结束时,可能存在这样的情况,例如,for当满足条件时,您可能需要摆脱循环.它可能比通过调用找出结果更快map.
小智 12
你特别询问map(),filter()和reduce(),但我想你一般都想了解函数式编程.我自己测试了计算一组点内所有点之间距离的问题,函数式编程(使用内置itertools模块中的starmap函数)结果比for循环略慢(长度为1.25倍) , 事实上).这是我使用的示例代码:
import itertools, time, math, random
class Point:
def __init__(self,x,y):
self.x, self.y = x, y
point_set = (Point(0, 0), Point(0, 1), Point(0, 2), Point(0, 3))
n_points = 100
pick_val = lambda : 10 * random.random() - 5
large_set = [Point(pick_val(), pick_val()) for _ in range(n_points)]
# the distance function
f_dist = lambda x0, x1, y0, y1: math.sqrt((x0 - x1) ** 2 + (y0 - y1) ** 2)
# go through each point, get its distance from all remaining points
f_pos = lambda p1, p2: (p1.x, p2.x, p1.y, p2.y)
extract_dists = lambda x: itertools.starmap(f_dist,
itertools.starmap(f_pos,
itertools.combinations(x, 2)))
print('Distances:', list(extract_dists(point_set)))
t0_f = time.time()
list(extract_dists(large_set))
dt_f = time.time() - t0_f
功能版本比程序版本更快吗？
def extract_dists_procedural(pts):
n_pts = len(pts)
l = []
for k_p1 in range(n_pts - 1):
for k_p2 in range(k_p1, n_pts):
l.append((pts[k_p1].x - pts[k_p2].x) ** 2 +
(pts[k_p1].y - pts[k_p2].y) ** 2)
return l
t0_p = time.time()
list(extract_dists_procedural(large_set))
# using list() on the assumption that
# it eats up as much time as in the functional version
dt_p = time.time() - t0_p
f_vs_p = dt_p / dt_f
if f_vs_p >= 1.0:
print('Time benefit of functional progamming:', f_vs_p,
'times as fast for', n_points, 'points')
else:
print('Time penalty of functional programming:', 1 / f_vs_p,
'times as slow for', n_points, 'points')
我修改了@Alisa 的代码并用来cProfile说明为什么列表理解更快：
from functools import reduce
import datetime
def reduce_(numbers):
return reduce(lambda sum, next: sum + next * next, numbers, 0)
def for_loop(numbers):
a = []
for i in numbers:
a.append(i*2)
a = sum(a)
return a
def map_(numbers):
sqrt = lambda x: x*x
return sum(map(sqrt, numbers))
def list_comp(numbers):
return(sum([i*i for i in numbers]))
funcs = [
reduce_,
for_loop,
map_,
list_comp
]
if __name__ == "__main__":
# [1, 2, 5, 3, 1, 2, 5, 3]
import cProfile
for f in funcs:
print('=' * 25)
print("Profiling:", f.__name__)
print('=' * 25)
pr = cProfile.Profile()
for i in range(10**6):
pr.runcall(f, [1, 2, 5, 3, 1, 2, 5, 3])
pr.create_stats()
pr.print_stats()
结果如下：
=========================
Profiling: reduce_
=========================
11000000 function calls in 1.501 seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
1000000 0.162 0.000 1.473 0.000 profiling.py:4(reduce_)
8000000 0.461 0.000 0.461 0.000 profiling.py:5(<lambda>)
1000000 0.850 0.000 1.311 0.000 {built-in method _functools.reduce}
1000000 0.028 0.000 0.028 0.000 {method 'disable' of '_lsprof.Profiler' objects}
=========================
Profiling: for_loop
=========================
11000000 function calls in 1.372 seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
1000000 0.879 0.000 1.344 0.000 profiling.py:7(for_loop)
1000000 0.145 0.000 0.145 0.000 {built-in method builtins.sum}
8000000 0.320 0.000 0.320 0.000 {method 'append' of 'list' objects}
1000000 0.027 0.000 0.027 0.000 {method 'disable' of '_lsprof.Profiler' objects}
=========================
Profiling: map_
=========================
11000000 function calls in 1.470 seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
1000000 0.264 0.000 1.442 0.000 profiling.py:14(map_)
8000000 0.387 0.000 0.387 0.000 profiling.py:15(<lambda>)
1000000 0.791 0.000 1.178 0.000 {built-in method builtins.sum}
1000000 0.028 0.000 0.028 0.000 {method 'disable' of '_lsprof.Profiler' objects}
=========================
Profiling: list_comp
=========================
4000000 function calls in 0.737 seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
1000000 0.318 0.000 0.709 0.000 profiling.py:18(list_comp)
1000000 0.261 0.000 0.261 0.000 profiling.py:19(<listcomp>)
1000000 0.131 0.000 0.131 0.000 {built-in method builtins.sum}
1000000 0.027 0.000 0.027 0.000 {method 'disable' of '_lsprof.Profiler' objects}
恕我直言：
reduce并且map通常很慢。不仅如此，与使用列表相比sum，在map返回的迭代器上使用速度很慢sumfor_loop 使用 append，这当然在某种程度上很慢sum速度更快map我写了一个测试速度的简单脚本,这就是我发现的.实际上for循环在我的情况下是最快的.这真让我感到惊讶,请查看贝娄(正在计算平方和).
from functools import reduce
import datetime
def time_it(func, numbers, *args):
start_t = datetime.datetime.now()
for i in range(numbers):
func(args[0])
print (datetime.datetime.now()-start_t)
def square_sum1(numbers):
return reduce(lambda sum, next: sum+next**2, numbers, 0)
def square_sum2(numbers):
a = 0
for i in numbers:
i = i**2
a += i
return a
def square_sum3(numbers):
sqrt = lambda x: x**2
return sum(map(sqrt, numbers))
def square_sum4(numbers):
return(sum([int(i)**2 for i in numbers]))
time_it(square_sum1, 100000, [1, 2, 5, 3, 1, 2, 5, 3])
time_it(square_sum2, 100000, [1, 2, 5, 3, 1, 2, 5, 3])
time_it(square_sum3, 100000, [1, 2, 5, 3, 1, 2, 5, 3])
time_it(square_sum4, 100000, [1, 2, 5, 3, 1, 2, 5, 3])
0:00:00.302000 #Reduce
0:00:00.144000 #For loop
0:00:00.318000 #Map
0:00:00.390000 #List comprehension
为Alphii 答案添加一个转折，实际上 for 循环将是第二好的并且比它慢 6 倍map
from functools import reduce
import datetime
def time_it(func, numbers, *args):
start_t = datetime.datetime.now()
for i in range(numbers):
func(args[0])
print (datetime.datetime.now()-start_t)
def square_sum1(numbers):
return reduce(lambda sum, next: sum+next**2, numbers, 0)
def square_sum2(numbers):
a = 0
for i in numbers:
a += i**2
return a
def square_sum3(numbers):
a = 0
map(lambda x: a+x**2, numbers)
return a
def square_sum4(numbers):
a = 0
return [a+i**2 for i in numbers]
time_it(square_sum1, 100000, [1, 2, 5, 3, 1, 2, 5, 3])
time_it(square_sum2, 100000, [1, 2, 5, 3, 1, 2, 5, 3])
time_it(square_sum3, 100000, [1, 2, 5, 3, 1, 2, 5, 3])
time_it(square_sum4, 100000, [1, 2, 5, 3, 1, 2, 5, 3])
主要的变化是消除了缓慢的sum调用，以及int()在最后一种情况下可能不必要的调用。实际上，将 for 循环和 map 放在相同的术语中使其成为事实。请记住，lambda 是函数式概念，理论上不应该有副作用，但是，它们可能会产生副作用，例如添加到a. 在这种情况下，Python 3.6.1、Ubuntu 14.04、Intel(R) Core(TM) i7-4770 CPU @ 3.40GHz 的结果
0:00:00.257703 #Reduce
0:00:00.184898 #For loop
0:00:00.031718 #Map
0:00:00.212699 #List comprehension
小智 5
我设法修改了@alpiii 的一些代码，发现列表理解比 for 循环快一点。这可能是由于int()，列表理解和 for 循环之间不公平。
from functools import reduce
import datetime
def time_it(func, numbers, *args):
start_t = datetime.datetime.now()
for i in range(numbers):
func(args[0])
print (datetime.datetime.now()-start_t)
def square_sum1(numbers):
return reduce(lambda sum, next: sum+next*next, numbers, 0)
def square_sum2(numbers):
a = []
for i in numbers:
a.append(i*2)
a = sum(a)
return a
def square_sum3(numbers):
sqrt = lambda x: x*x
return sum(map(sqrt, numbers))
def square_sum4(numbers):
return(sum([i*i for i in numbers]))
time_it(square_sum1, 100000, [1, 2, 5, 3, 1, 2, 5, 3])
time_it(square_sum2, 100000, [1, 2, 5, 3, 1, 2, 5, 3])
time_it(square_sum3, 100000, [1, 2, 5, 3, 1, 2, 5, 3])
time_it(square_sum4, 100000, [1, 2, 5, 3, 1, 2, 5, 3])
0:00:00.101122 #Reduce
0:00:00.089216 #For loop
0:00:00.101532 #Map
0:00:00.068916 #List comprehension
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