Speed-up for loops

BartC bartc at freeuk.com
Sun Sep 5 13:28:47 CEST 2010

"David Cournapeau" <cournape at gmail.com> wrote in message
news:mailman.455.1283665528.29448.python-list at python.org...
> On Thu, Sep 2, 2010 at 7:02 PM, Michael Kreim <michael at perfect-kreim.de>
> wrote:

>> imax = 1000000000
>> a = 0
>> for i in xrange(imax):
>>    a = a + 10
>> print a

>> Unfortunately my Python Code was much slower [than Matlab] and I do not
>> understand why.
> Getting the above kind of code fast requires the interpreter to be
> clever enough so that it will use native machine operations on a int
> type instead of converting back and forth between internal
> representations.

Writing for i in xrange(1000000000) you'd think would give it a clue, but it
makes no difference.

> Matlab since version 6 I believe, has a JIT to do
> just that. There is no mature JIT-like implementation of python which
> will give you the same speed up for this exact case today.

>> Or do I have to live with the fact that Matlab beats Python in this
>> example?
> Yes. Without a JIT, python cannot hope to get the same kind of speeds
> for this kind of examples.
> That being said, neither matlab nor matlab are especially good at
> doing what you do in your example - for this exact operation, doing it
> in C or other compiled languages will be at least one order of
> magnitude faster.

One order of magnitude (say 10-20x slower) wouldn't be so bad. That's what
you might expect for a dynamically typed, interpreted language.

But on my machine this code was more like 50-200x slower than C, for
unaccelerated Python.

> Generally, you use matlab's vectorized operations,
> and in that case, numpy gives you similar performances (sometimes
> faster, sometimes slower, but in the same ballpark in general).

That would simply be delegating Python to a scripting language. It would be
nice if you could directly code low-level algorithms in it without relying
on accelerators, and not have to wait two and a half minutes (or whatever) 
for a simple test to complete.


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