我正在尝试使用python的timeit模块,看起来好像在timeit源代码中有错误(尽管这看起来并不正确)。
以下是正在运行的代码的片段:
def recordCuckoo(amtElements, loadFactor):
'''
Determines the average lookup speed in seconds of a cuckoo hash table
with @amtElements elements and a load factor of @loadFactor
'''
mySetup = '''
import Statistics
import random
import hashingLibrary
from CuckooHashing import *
'''
controlStatement = "Statistics.timeCuckooControl(" + str(amtElements) + "," + str(loadFactor) + ")"
testStatement = "Statistics.timeCuckoo(" + str(amtElements) + "," + str(loadFactor) + ")"
controlTime = timeit.timeit(controlStatement, setup=mySetup, number=1)
testTime = timeit.timeit(testStatement, setup=mySetup, number=1)
lookupTime = (testTime - controlTime)/1000000
print ("The average lookup time for a cuckoo table with {0} elements and a load factor of {1} was:".format(amtElements, loadFactor))
print (lookupTime)
return lookupTime
if __name__ == "__main__":
recordCuckoo(100, 0.5)
运行时收到以下错误:
Traceback (most recent call last):
File "C:\Python34\CuckooHashing\Statistics.py", line 308, in <module>
recordCuckoo(100, 0.5)
File "C:\Python34\CuckooHashing\Statistics.py", line 267, in recordCuckoo
controlTime = timeit.timeit(controlStatement, setup=mySetup, number=1)
File "C:\Python34\lib\timeit.py", line 213, in timeit
return Timer(stmt, setup, timer).timeit(number)
File "C:\Python34\lib\timeit.py", line 122, in __init__
code = compile(src, dummy_src_name, "exec")
File "<timeit-src>", line 9
_t0 = _timer()
^
IndentationError: unindent does not match any outer indentation level
我知道错误很可能发生在键盘和主席之间,但我收到的错误似乎表明timeit模块中存在不正确的空格/标签。发生了什么???
答案 0 :(得分:1)
您可以像这样定义mySetup
变量:
mySetup = '''
import Statistics
import random
import hashingLibrary
from CuckooHashing import *
'''
如果您只是考虑到这一点,那根本不是问题。但是,这些行实际上出现在函数声明中:
def recordCuckoo(amtElements, loadFactor):
mySetup = '''
import Statistics
import random
import hashingLibrary
from CuckooHashing import *
'''
实际上,mySetup
的内容如下:
'''
import Statistics
import random
import hashingLibrary
from CuckooHashing import *
'''
正如您所看到的,import
行前面有一个缩进,这使得它们无效(因为它们在没有缩进的情况下执行)。所以你应该以不同的方式设置setup变量:
def recordCuckoo(amtElements, loadFactor):
mySetup = '''
import Statistics
import random
import hashingLibrary
from CuckooHashing import *
'''
或者像他这样的东西:
def recordCuckoo(amtElements, loadFactor):
mySetup = '\n'.join((
'import Statistics',
'import random',
'import hashingLibrary',
'from CuckooHashing import *'
))
答案 1 :(得分:0)
以下测试有效:
timeit.timeit('a+3', setup='a=1', number=10000)
timeit.timeit('a+3 -float(2\n)', setup='a=1', number=10000)
timeit.timeit(' a+3 -float(2\n)', setup='a=1', number=10000)
但这会因你的错误而失败:
timeit.timeit('a+3', setup=' a=1', number=10000)
请注意设置中的空格。您可以通过传递mySetup.strip()
来消除它。
在运行有问题的函数之前打印变量总是一个好主意,但是你很困惑,错误绝对是神秘的。