Question

I was playing around with Python’s collection.deque and wrote the following benchmark:

#!/usr/bin/python

import timeit

if __name__=='__main__':
    number = 1000000
    for r in (1,10,100,1000,5000,10000,100000):
        print r
        print timeit.timeit("while x: x.pop(0);",
                            "x = list(range("+str(r)+"))",
                            number=number)
        print timeit.timeit("while x: x.popleft();",
                            "from collections import deque; x = deque(range("+str(r)+"))",
                             number=number)

This will pop(0)/popleft() from a list/deque with various sizes. The results are:

1
0.0801048278809
0.0822219848633
10
0.081041097641
0.080836057663
100
0.0806250572205
0.0807700157166
1000
0.081248998642
0.082062959671
5000
0.0976719856262
0.0825741291046
10000
0.157499074936
0.0825819969177
100000
16.0247170925
0.097559928894

My question is: Why is the performance for small deques and lists (~1000 elements) almost the same?

Answer 1

The timeit module runs the setup code once, and then the timed code number times (in this case, number==1000000). In your case this looks like (for the list case):

x = list(range(r))
#timer is started here
for iteration in xrange(1000000):
    while x: x.pop(0)
#timer is stopped here

As you can see, only the first iteration does anything, and the other 999999 iterations will just check the size of x once since it will be empty. This check will take approximately the same amount of time for lists and deques.

For small lists/deques, the first iteration is short relative to the other 999999 iterations combined, so you are not really measuring anything relevant, and get similar times.

If you use number==1 you won’t have this problem. Another option is to have the timed code push and pop an item so that the list/deque is left at the same size.