On Mon, Oct 19, 2009 at 1:58 PM, Emile van Sebille <em...@fenx.com> wrote:
> On 10/19/2009 12:20 PM Alan Gauld said...
>
>
>> "Sander Sweers" <sander.swe...@gmail.com> wrote
>>
>> mylist = ['John', 'Canada', 25, 32, 'right']
>>>> a = [item.upper() for item in mylist if type(item) == type('good')]
>>>>
>>>
>>> Usually it is recommended to use hasattr() instead of type()
>>> hasattr(s, 'upper')
>>>
>>
>> Nope, they do completely different things
>> I think you might be thinking of isinstance() which can be used instead of
>> type(). I see you use hasattr as a means of testing for a method but that is
>> still different from testing type - the method names might be the same but
>> the functions be completely different in effect!
>>
>> returned this: ['JOHN', 'CANADA', 'RIGHT']
>>>> I was expecting this: ['JOHN', 'CANADA', 25, 32, 'RIGHT']
>>>> So, actually the "if" acted like a filter.
>>>>
>>>
>> It is intended to be used as a filter.
>>
>> In order to use a list comprehension I created this function instead.
>>>> def upperfy(item)
>>>> try:
>>>> item = item.upper()
>>>> except AttributeError:
>>>> pass
>>>> return item
>>>>
>>>
>> I would move return item under the except and remove the pass, other
>>> might disagree on this.
>>>
>>
>> I would :-)
>> Doing that would result in None being returned for each successful
>> conversion. The OPs code is correct (even if unnecessary)
>>
>> a = [upperfy(item) for item in mylist]
>>>>
>>>
>> a = [item.upper() if type(item) == str else item for item in mylist]
>>
>> should do it I think.
>>
>
> or even
>
> a = [ str(item).upper() for item in mylist ]
>
> Emile
>
>
>
>> HTH,
>>
>>
>>
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Lambda can save the day to keep everything on one line, and leave variable
type the same:
mylist = ['John', 'Canada', 25, 32, 'right']
new_list = [(lambda y: y.upper() if hasattr(y, 'upper') else y)(a) for a in
x]
>> ['JACK', 'CANADA', 25, 32, 'RIGHT']
Vince
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