CODEWITHSUNDEEP
pythonglobal-variablesnameerrortimeit
Bentley4
Bentley4

Reputation: 11008

Timeit, NameError: global name is not defined. But I didn't use a global variable

I'd like to measure the execution speed of the following code:

def pe1():
    l = []
    for i in range(1000):
        if i%3 == 0 or i%5 == 0:
            l.append(i)
    print sum(l)

I stored this code under pe1m.py . Now I'd like to test the speed of file with the python interpreter. I did:

import timeit
import pe1m

t = timeit.Timer(stmt = 'pe1m.pe1()')
t.timeit()

but I get:

File "<stdin>", line 1, in <module>
  File "/usr/lib/python2.7/timeit.py", line 195, in timeit
    timing = self.inner(it, self.timer)
  File "<timeit-src>", line 6, in inner
NameError: global name 'pe1m' is not defined

But I don't have any global variables.

Upvotes: 11

Views: 9476

Answers (2)

Adebiyi Abdurrahman
Adebiyi Abdurrahman

Reputation: 191

You can also try this

>>>def pe1():
>>>    l = []
>>>    for i in range(1000):
>>>        if i%3 == 0 or i%5 == 0:
>>>            l.append(i)
>>>    print(sum(l))
>>>
>>>from  timeit import time it
>>>timeit('pe1()',setup="from __main__ import pe1",number=100) # Run 100 times
>>>timeit('pe1()',setup="from __main__ import pe1")  # run default times of 1000000

Upvotes: 7

yak
yak

Reputation: 9041

Try this:

t = timeit.Timer(stmt='pe1()', setup='from pe1m import pe1')

timeit.Timer object doesn't know about the namespace you're calling it in so it can't access the pe1m module that you imported.

The setup argument is a statement executed in the context of the timed statement, they share the same namespace so whatever you define there, will be accessible in stmt.

Upvotes: 17

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