performance question

[Eric Texier]

I need speed here. What will be the fastest method or does it matter?

(for the example 'a' is only 3 values for the clarity of the example)
a = [1,3,4.] ##


method1:

f.write("vec %f %f %f \n" % (a[0],a[1],a[2]))

method2:

f.write("vec " + str(a[0]) + " " + str(a[1]) + " " + str(a[2]) + "\n")

also it there a relevant speed difference between making few small write
instead of 1 bigger one.

Thanks for any feed back,
Eric

[Alex Martelli]

Re: performance question

Eric Texier <texiereric@yah oo.comwrote:

I need speed here. What will be the fastest method or does it matter?
>
(for the example 'a' is only 3 values for the clarity of the example)
a = [1,3,4.] ##
>
>
method1:
>
f.write("vec %f %f %f \n" % (a[0],a[1],a[2]))
>
method2:
>
f.write("vec " + str(a[0]) + " " + str(a[1]) + " " + str(a[2]) + "\n")
>
also it there a relevant speed difference between making few small write
instead of 1 bigger one.

Learn to use the timeit module from the standard library, particularly
via the handy -mtimeit commandline switch, and you can measure
performance issues for yourself. E.g., on my laptop:

brain:~ alex$ python -mtimeit -s"a=[1,3,4];f=open('/dev/null','w')"
'f.write("vec %f %f %f \n" % (a[0],a[1],a[2]))'
100000 loops, best of 3: 5.64 usec per loop

brain:~ alex$ python -mtimeit -s"a=[1,3,4];f=open('/dev/null','w')"
'f.write("vec " + str(a[0]) + " " + str(a[1]) + " " + str(a[2]) + "\n")'
100000 loops, best of 3: 3.36 usec per loop

So, the ugly "method 2" is about 2.3 microseconds faster than the nicer
"method 1" -- when the items of a are ints which method 1 widens to
floats while method 2 doesn't (the results are different between the
methods). When they're floats to start with...:

brain:~ alex$ python -mtimeit -s"a=[1.,3.,4.];f=open('/dev/null','w')"
'f.write("vec %f %f %f \n" % (a[0],a[1],a[2]))'
100000 loops, best of 3: 5.45 usec per loop

brain:~ alex$ python -mtimeit -s"a=[1.,3.,4.];f=open('/dev/null','w')"
'f.write("vec " + str(a[0]) + " " + str(a[1]) + " " + str(a[2]) + "\n")'
100000 loops, best of 3: 6.26 usec per loop

then method 1 accelerates a little bit and method 2 slows down a lot, so
method 1 is actually about 0.8 microseconds faster.

Make sure you do your measurements with data that well represents your
actual application needs, and -mtimeit will serve you well (if you don't
care about that microsecond or two either way, which is often the case,
then choose the nicer metod 1, of course:-).


Alex

[Raymond Hettinger]

Re: performance question

[Eric Texier]

I need speed here. What will be the fastest method or does it matter?

Follow Alex's advice and use the timeit module, but do not generalize
from too small examples; otherwise, the relative timings will be
thrown-off by issues like the time to lookup "write" and "a" and "str"
(all of which will be faster if made a local). Likewise, do the
timings with the actual expected vector length.

(for the example 'a' is only 3 values for the clarity of the example)
a = [1,3,4.] ##
f.write("vec %f %f %f \n" % (a[0],a[1],a[2]))

It's a waste of time to lookup and repack with (a[0],a[1],a[2]).
Instead, try:
f.write('vec %f %f %f' % tuple(a))

f.write("vec " + str(a[0]) + " " + str(a[1]) + " " + str(a[2]) + "\n")

Often, it is better to join than to make successive concatenations:

f.write('vec' + ' '.join(map(str, a)))

also it there a relevant speed difference between making few small write
instead of 1 bigger one.

Yes. One big one is faster than many small.


Raymond