我試着重複IPython%time的功能,但出於某種奇怪的原因,某些函數的測試結果是可怕的。timeit的奇怪結果
IPython的:
In [11]: from random import shuffle
....: import numpy as np
....: def numpy_seq_el_rank(seq, el):
....: return sum(seq < el)
....:
....: seq = np.array(xrange(10000))
....: shuffle(seq)
....:
In [12]: %timeit numpy_seq_el_rank(seq, 10000//2)
10000 loops, best of 3: 46.1 µs per loop
的Python:
from timeit import timeit, repeat
def my_timeit(code, setup, rep, loops):
result = repeat(code, setup=setup, repeat=rep, number=loops)
return '%d loops, best of %d: %0.9f sec per loop'%(loops, rep, min(result))
np_setup = '''
from random import shuffle
import numpy as np
def numpy_seq_el_rank(seq, el):
return sum(seq < el)
seq = np.array(xrange(10000))
shuffle(seq)
'''
np_code = 'numpy_seq_el_rank(seq, 10000//2)'
print 'Numpy seq_el_rank:\n\t%s'%my_timeit(code=np_code, setup=np_setup, rep=3, loops=100)
,其輸出:
Numpy seq_el_rank:
100 loops, best of 3: 1.655324947 sec per loop
正如你所看到的,在Python我做了100圈,而不是10000(並獲得比ipython慢35000倍,結果),因爲它需要很長時間。任何人都可以解釋爲什麼python的結果如此之慢?
UPD: 這裏是cProfile.run('my_timeit(code=np_code, setup=np_setup, rep=3, loops=10000)')
輸出:
30650 function calls in 4.987 seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
1 0.000 0.000 4.987 4.987 <string>:1(<module>)
1 0.000 0.000 0.000 0.000 <timeit-src>:2(<module>)
3 0.001 0.000 4.985 1.662 <timeit-src>:2(inner)
300 0.006 0.000 4.961 0.017 <timeit-src>:7(numpy_seq_el_rank)
1 0.000 0.000 4.987 4.987 Lab10.py:47(my_timeit)
3 0.019 0.006 0.021 0.007 random.py:277(shuffle)
1 0.000 0.000 0.002 0.002 timeit.py:121(__init__)
3 0.000 0.000 4.985 1.662 timeit.py:185(timeit)
1 0.000 0.000 4.985 4.985 timeit.py:208(repeat)
1 0.000 0.000 4.987 4.987 timeit.py:239(repeat)
2 0.000 0.000 0.000 0.000 timeit.py:90(reindent)
3 0.002 0.001 0.002 0.001 {compile}
3 0.000 0.000 0.000 0.000 {gc.disable}
3 0.000 0.000 0.000 0.000 {gc.enable}
3 0.000 0.000 0.000 0.000 {gc.isenabled}
1 0.000 0.000 0.000 0.000 {globals}
3 0.000 0.000 0.000 0.000 {isinstance}
3 0.000 0.000 0.000 0.000 {len}
3 0.000 0.000 0.000 0.000 {method 'append' of 'list' objects}
1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler' objects}
29997 0.001 0.000 0.001 0.000 {method 'random' of '_random.Random' objects}
2 0.000 0.000 0.000 0.000 {method 'replace' of 'str' objects}
1 0.000 0.000 0.000 0.000 {min}
3 0.003 0.001 0.003 0.001 {numpy.core.multiarray.array}
1 0.000 0.000 0.000 0.000 {range}
300 4.955 0.017 4.955 0.017 {sum}
6 0.000 0.000 0.000 0.000 {time.clock}
您是否嘗試過分析代碼以查看緩慢發生的位置? – Soviut
@Soviut,不,我不知道如何配置timeit.repeat函數,因爲它對我來說太複雜了 – KgOfHedgehogs
是否有某些原因讓你傳遞字符串代碼塊而不是簡單地運行IPython中的相同測試? – Soviut