[Tutor] why is list.pop(0) slow?

[Wari Wahab]

I've been playing around with very large lists and found out by chance 
that pop(0) is 8 times slower than pop(). Is it really that bad? 
Performance improved when the list i have is reversed and used pop(). 
See this little snippet of code for example:

---
"""
This shows that poping from the back of the list is faster then the 
beginning
of the list
"""
import profile

def generate(num = 2000):
     return [[] for p in xrange(num)]

def popper(l):
     while l:
         l.pop()

def pooper(l):
     while l:
         l.pop(0)

if __name__ == '__main__':
     print "###### Time taken to generate list"
     profile.run('generate(50000)')
     l = generate(50000)
     print "###### Time taken to pop at the end of the list"
     profile.run('popper(l)')
     l = generate(50000)
     print "###### Time taken to pop at the beginning of the list"
     profile.run('pooper(l)')

"""
This prints:

###### Time taken to generate list
          3 function calls in 0.260 CPU seconds

    Ordered by: standard name

    ncalls  tottime  percall  cumtime  percall filename:lineno(function)
         1    0.040    0.040    0.230    0.230 <string>:1(?)
         1    0.190    0.190    0.190    0.190 pooperpoper.py:7(generate)
         1    0.030    0.030    0.260    0.260 profile:0(generate(50000))
         0    0.000             0.000          profile:0(profiler)


###### Time taken to pop at the end of the list
          3 function calls in 0.260 CPU seconds

    Ordered by: standard name

    ncalls  tottime  percall  cumtime  percall filename:lineno(function)
         1    0.000    0.000    0.260    0.260 <string>:1(?)
         1    0.260    0.260    0.260    0.260 pooperpoper.py:10(popper)
         1    0.000    0.000    0.260    0.260 profile:0(popper(l))
         0    0.000             0.000          profile:0(profiler)


###### Time taken to pop at the beginning of the list
          3 function calls in 8.870 CPU seconds

    Ordered by: standard name

    ncalls  tottime  percall  cumtime  percall filename:lineno(function)
         1    0.000    0.000    8.870    8.870 <string>:1(?)
         1    8.870    8.870    8.870    8.870 pooperpoper.py:14(pooper)
         1    0.000    0.000    8.870    8.870 profile:0(pooper(l))
         0    0.000             0.000          profile:0(profiler)
"""



-- 
Regards: Wari Wahab
RoughingIT - http://roughingit.subtlehints.net
PyBlosxom  - http://roughingit.subtlehints.net/pyblosxom