Ordered dictionary?

[Josiah Carlson]

Carmine Moleti wrote:

> Hi Dragos,
> 
> 
>>def filterList(p_list, p_value):
>>    l_len = len(p_list)
>>    l_temp = map(None, (p_value,)*l_len, p_list)
>>    l_temp = filter(lambda x: x[1].find(x[0])==0, l_temp)
>>    return map(operator.getitem, l_temp, (-1,)*len(l_temp))
> 
> 
>>phones_dict = {'jansen' : '0000', 'xxx' : '1111', 'jan2' : '2222'}
>>names_list = filterList(phones_dict.keys(), 'jan')
>>phones_list = map(phones_dict.get, names_list)
> 
> 
>>This extracts from the dictionary the telephone(values) numbers for
>>names(keys) starting with 'jan'...
> 
> 
> Why you didn't used the string.startswith(...) method?
> 
> I wrote this:
> 
> d={'carmine':'123456','carmela':'4948399','pippo':'39938303'}
> for name,number in d.items():
>     if name.startswith('car'):
>         print name,number
> 
> This also extract from the dictionay all the (name,number) pairs whose
> name starts with a given substring ('car' in the example).
> 
> Thanks for your answer

Something strange is that for short 'other' strings
string[:len(other)] == other
#is faster than
string.startswith(other)

Maybe find is faster than startswith.

  - Josiah