[Chicago] constant length string manipulation

[Cosmin Stejerean]

In case everyone forgot, this is Python not perl: there is only one  
way to do it (it just so happens to be my way)

This is turning into an exercise of writing obfuscated code.

Cosmin Stejerean (m)


On Dec 2, 2007, at 7:48 AM, Massimo Di Pierro  
<mdipierro at cs.depaul.edu> wrote:

> 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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