On May 1, 10:50 pm, George Sakkis <[EMAIL PROTECTED]> wrote:
> On May 1, 11:46 pm, Carsten Haese <[EMAIL PROTECTED]> wrote:
>
>
>
> > Yves Dorfsman wrote:
>
> > > In the following script, m1() and m2() work fine. I am assuming m2() is
> > > faster although I haven't checked that (loops through the list twice
> > > instead of once).
>
> > Well, let's check it:
>
> > $ python -m timeit -s "import x" "x.m1()"
> > 100000 loops, best of 3: 6.43 usec per loop
>
> > $ python -m timeit -s "import x" "x.m2()"
> > 100000 loops, best of 3: 8.34 usec per loop
>
> > As it turns out, m1 is faster than m2. The reason is that list
> > comprehensions do the loop in C, whereas the for-append pattern does the
> > loop on the Python level.
>
> > > Now what I am trying to do is something like m3(). As currently written
> > > it does not work, and I have tried different ways, but I haven't managed
> > > to make it work.
>
> > > Is there a possibility ? Or is m2() the optimum ?
>
> > > [...]
> > > def m1():
> > > colours = [ e['colour'] for e in l ]
> > > nums = [ e['num'] for e in l ]
>
> > > def m2():
> > > colours = []
> > > nums = []
> > > for e in l:
> > > colours.append(e['colour'])
> > > nums.append(e['num'])
>
> > > #def m3():
> > > # colours, nums = [ e['colour'], e['num'] for e in l ]
>
> > m3 doesn't work because you're building a list of 10 color/number pairs
> > that you're trying to unpack that into just two names. The working
> > "derivative" of m3 is m1, which is the most natural, fastest and
> > clearest solution to your problem.
>
> Another alternative is:
>
> from operator import itemgetter
>
> def m3():
> colours, nums = zip(*map(itemgetter('colour','num'), l))
>
> It's slower than m1() but faster than m2(); it's also the most
> concise, especially if you extract more than two keys.
>
> George
Why deal with zip and unpacking? This seems more obvious to me:
colours, nums = map(itemgetter('colour'), l), map(itemgetter('num'),
l)
Matt
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