我更新了代码和时间。
我正在尝试改进我的代码中的一个函数性能。我必须生成一个带有随机元素的列表。然而,列表的不同部分必须被来自不同集合的元素填充。以下是一段示例代码。我必须逐个生成数百万个这样的列表。
函数foo1是最快的,但它不能满足我的需求。它只是用于性能参考。函数foo2和foo3能够满足我的需求,但处理时间几乎是foo1的三倍。
Python 2.7.9 (默认,Feb 10 2015, 03:29:19). [GCC 4.2.1 兼容苹果 LLVM 6.0 (clang-600.0.56)] 在Darwin上。numpy.version '1.8.1'
import numpy
import timeit
_ops_1 = ["-123.456", "3.1416", "1", "2"]
_ops_2 = ["ABC", "XYZ", 'A', 'B', 'C']
size = 10
def foo1 ():
return numpy.random.choice(_ops_1 + _ops_2, 5*size)
def foo2 ():
return list(numpy.concatenate((numpy.random.choice(_ops_1, 2*size),
numpy.random.choice(_ops_1 + _ops_2, size),
numpy.random.choice(_ops_2, 2*size)), 0))
def foo3 ():
return numpy.random.choice(_ops_1, 2*size).tolist() + \
numpy.random.choice(_ops_1 + _ops_2, size).tolist() + \
numpy.random.choice(_ops_2, 2*size).tolist()
### Suggested by Divakar
def random_choice_replace_True(arr,size):
return numpy.take(arr,numpy.random.randint(0,len(arr),size))
def foo4 ():
return random_choice_replace_True(_ops_1, 2*size).tolist() + \
random_choice_replace_True(_ops_1 + _ops_2, size).tolist() + \
random_choice_replace_True(_ops_2, 2*size).tolist()
### 2nd suggestion by Divakar
def random_choice_replace_True_idx(arr,size):
return numpy.array(arr)[numpy.random.randint(0,len(arr),size)]
def foo5 ():
return random_choice_replace_True_idx(_ops_1, 2*size).tolist() + \
random_choice_replace_True_idx(_ops_1 + _ops_2, size).tolist() + \
random_choice_replace_True_idx(_ops_2, 2*size).tolist()
###########
setup = '''import numpy
_ops_1 = ["-123.456", "3.1416", "1", "2"]
_ops_2 = ["ABC", "XYZ", 'A', 'B', 'C']
size = 10'''
# As required, Number was increased to 10 million to get closer to actual timings
timeit.timeit(foo1, setup=setup, number=10000000)
timeit.timeit(foo2, setup=setup, number=10000000)
timeit.timeit(foo3, setup=setup, number=10000000)
timeit.timeit(foo4, setup=setup, number=10000000)
timeit.timeit(foo5, setup=setup, number=10000000)
我机器上的运行时间为:
目前,我会采纳Divakar提出的第二个建议,这个建议很不错。但其他建议也欢迎!timeit.timeit(foo1, setup=setup, number=10000000) 235.22050380706787
timeit.timeit(foo2, setup=setup, number=10000000) 760.1884841918945
timeit.timeit(foo3, setup=setup, number=10000000) 560.77258586883545
timeit.timeit(foo4, setup=setup, number=10000000) 388.69550228118896
timeit.timeit(foo5, setup=setup, number=10000000) 252.32089233398438