x <- rep("000000",2)
y <- c("23/45","67/8")
substr(x,1+nchar(x)-nchar(y), nchar(x)) <- y
x

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Sent from my phone. Please excuse my brevity.

David Winsemius <dwinsemius@comcast.net> wrote:


Copying list one what was sent in reply. Anybody have a better solution?

> On Aug 19, 2011, at 11:57 AM, Vasco Cadavez wrote:
>
>> Thanks,
>>
>> A solution can be by substring to remove the /
>> then numeric will be ok! What you think?
>>
>> How can I remove the /
>
with sub or gsub:

> sprintf("%010.0f", as.integer(gsub("/","", 
c("4/3003","55/333","66/22")) ))
[1] "0000043003" "0000055333" "0000006622"

-- 
David.

>>
>> Thanks
>>
>> Vasco Cadavez	
>>
>> ----- Menssagem Original -----
>> De:
>> "David Winsemius" <dwinsemius@comcast.net>
>>
>> Para:
>> "David Winsemius" <dwinsemius@comcast.net>
>> Cópia:
>> "Vasco Cadavez" <vcadavez@ipbpt>, <r-help@r-project.org>
>> Enviado:
>> Fri, 19 Aug 2011 11:51:08 -0400
>> Assunto:
>> Re: [R] Leading zeros
>>
>>
>>
>> On Aug 19, 2011, at 11:17 AM, David Winsemius wrote:
>>
>> >
>> > On Aug 19, 2011, at 11:12 AM, Vasco Cadavez wrote:
>> >
>> >> Hello,
>> >> I have a dataset with an Id columns like:
>> >> 4/3003
>> >> 55/333
>> >> 66/22
>> >> I want to put leading zeros to get:
>> >> 00000004/3003
>> >> 000000055/333
>> >> 0000000066/22
>> >>
>> >> How can I solve this?
>> >
>> > ?sprintf
>> > ?formatC
>> >
>> I may have been too quick. Padding with leading zeros using sprintf 
>> is
>> described for numeric but not for character types. There are severa

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