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
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