On May 1, 8:01 pm, Yves Dorfsman <[EMAIL PROTECTED]> 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).
>
> 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 ?
>
> Thanks.
>
> #!/usr/bin/python
>
> l = [ { 'colour': 'black', 'num': 0},
> { 'colour': 'brown', 'num': 1},
> { 'colour': 'red', 'num': 2},
> { 'colour': 'orange', 'num': 3},
> { 'colour': 'yellow', 'num': 4},
> { 'colour': 'green', 'num': 5},
> { 'colour': 'blue', 'num': 6},
> { 'colour': 'violet', 'num': 7},
> { 'colour': 'grey', 'num': 8},
> { 'colour': 'white', 'num': 9}
> ]
>
> 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 ]
>
Looks like m1 is the cleanest; if you really want to run list-
comprehension once, one possible way:
>>> p = [ (e['colour'], e['num']) for e in l ]
>>> import operator
>>> map(operator.itemgetter(0), p)
['black', 'brown', 'red', 'orange', 'yellow', 'green', 'blue',
'violet', 'grey', 'white']
>>> map(operator.itemgetter(1), p)
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>>
Karthik
> --
> Yves.http://www.SollerS.ca
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