I'm a newbie to python and I have a question that I can't answer it. Please help me!
For example, I have an array in python:
index: 1 | 2 | 3 | 4 | 5 | 6 |
value: a | b | c | d | e | f |
I can read value from the 5th slot of this array by using 2 ways:
print array[4]
print array[-2]
So what is the fastest way to access into the 5th position on this array?
Thank you so much.
I don't actually know. It would seem to me that they're equivalently fast, but that's what profiling is for, so let's do it!
import timeit
setup = '''the_list = ['a','b','c','d','e','f']'''
timeit.timeit('the_list[4]', setup)
timeit.timeit('the_list[-2]', setup)
for me, those values were:
0.12980895690317062
and
0.12949802723105344
Which shows positive indexing ~3/10000 of a second faster over 10,000 iterations.
They're both very close to constant time.
You may find this helpful:
https://wiki.python.org/moin/TimeComplexity
When I run this many times and average the results, a[-2] actually runs a bit faster...but the difference is so small it probably won't matter even on huge data sets.
from datetime import datetime
a = ['a', 'b', 'c', 'd', 'e', 'f']
startTime = datetime.now()
for i in xrange(0,10000000):
a[4]
print(datetime.now()-startTime)
startTime = datetime.now()
for i in xrange(0,10000000):
a[-2]
print(datetime.now()-startTime)
Just timeit:
c:>py -m timeit -s "x=[1,2,3,4,5,6]" "y=x[4]" 10000000 loops, best of 3: 0.0458 usec per loop
c:>py -m timeit -s "x=[1,2,3,4,5,6]" "y=x[4]" 10000000 loops, best of 3: 0.0459 usec per loop
c:>py -m timeit -s "x=[1,2,3,4,5,6]" "y=x[4]" 10000000 loops, best of 3: 0.0489 usec per loop
c:>py -m timeit -s "x=[1,2,3,4,5,6]" "y=x[-2]" 10000000 loops, best of 3: 0.0448 usec per loop
c:>py -m timeit -s "x=[1,2,3,4,5,6]" "y=x[-2]" 10000000 loops, best of 3: 0.045 usec per loop
c:>py -m timeit -s "x=[1,2,3,4,5,6]" "y=x[-2]" 10000000 loops, best of 3: 0.045 usec per loop
Negligible difference. In these timings it looks like the negative value is a few nanoseconds faster (on my system anyway).
