[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
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