[Chicago] constant length string manipulation

[Massimo Di Pierro]

c'mon. No loops... and why limit to ' ' padding?

import re
f=lambda a,n,c=' ': re.compile('.{%i}'%n).findall(a+c*n)[0]
f('hello world',5)

Massimo

On Dec 2, 2007, at 9:42 AM, Carl Karsten wrote:

> def s5(s):
>    if len(s)<>5:
>      s=s5((s+' ')[:5])
>    return s
>
> print s5('Hello Worlds').__repr__()
> print s5('He').__repr__()
>
> The :5 blows the elegance.
>
> I thought I could +1 and -1 at the same time and have it converge  
> on 5, but that
> didn't work.
>
> Carl K
>
>
>
> Jeff Hinrichs - DM&T wrote:
>> Actually, this kind of problem just screams recursion
>>
>> def s5(s):
>>       if len(s) < 5: s=s5(s+' ')
>>       if len(s) > 5: s=s5(s[:-1])
>>       return s
>>
>> print s5('Hello Worlds')
>>
>>
>> On Dec 2, 2007 12:26 AM, Massimo Di Pierro  
>> <mdipierro at cs.depaul.edu> wrote:
>>> f=lambda a,n: (a[:n]+' '*n)[:n] # theta(n) for every a
>>> f('hello world',5)
>>>
>>>
>>> On Dec 2, 2007, at 12:14 AM, Cosmin Stejerean wrote:
>>>
>>>> It's cheating. What if he wanted 10 instead of 5? Hardcoding  
>>>> spaces is
>>>> a bad idea.
>>>>
>>>> Cosmin Stejerean (m)
>>>>
>>>>
>>>> On Dec 1, 2007, at 9:29 PM, Carl Karsten <carl at personnelware.com>
>>>> wrote:
>>>>
>>>>> string concatenation is icky.
>>>>>
>>>>> you have too many 5s.
>>>>>
>>>>> the line is too long.
>>>>>
>>>>> your variables are too short.
>>>>>
>>>>> there are no comments.
>>>>>
>>>>> and it isn't OOP.
>>>>>
>>>>> Carl K
>>>>>
>>>>> Massimo Di Pierro wrote:
>>>>>> what's wrong with
>>>>>>
>>>>>> b=(a[:5]+'     ')[:5]
>>>>>>
>>>>>> Massimo
>>>>>>
>>>>>> On Nov 30, 2007, at 9:25 AM, Lukasz Szybalski wrote:
>>>>>>
>>>>>>> On Nov 29, 2007 3:29 PM, Cosmin Stejerean <cstejerean at gmail.com>
>>>>>>> wrote:
>>>>>>>> b.ljust(5)[:5]
>>>>>>> a='1234567890' or a='123'
>>>>>>> b=a.ljust(5)[:5]
>>>>>>>
>>>>>>> so I guess ljust will do.
>>>>>>> Thanks.
>>>>>>> Lucas
>>>>>>> _______________________________________________
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>>>>>>
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>>
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