Eric Texier <[EMAIL PROTECTED]> wrote:
> 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
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