And in your solution the lambda is evaluated at each comparaison of the sort algorithm isn't it?
So your code seems less productive than the bengt's code which apply the same code as the lambda only one time by entry in the list.
Cyril
On 8/12/05, George Yoshida <[EMAIL PROTECTED]> wrote:
Bengt Richter wrote:
>>>>[name for dec,name in sorted((int(nm.split('.')[1]),nm) for nm in namelist)]
>
> ['test.1', 'test.2', 'test.3', 'test.4', 'test.10', 'test.15', 'test.20']
Giving a key argument to sorted will make it simpler::
>>> sorted(namelist, key=lambda x:int(x.rsplit('.')[-1]))
-- george
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