replace last 3 characters of string

I would like to replace the last tree characters of the values of a certain
column in a dataframe.

This replacement should only take place if the last three characters
correspond to the value "/:/" and they should be replaced with ""(blank)

I cannot perform a simple gsub because the characters /:/ might also be
present somewhere else in the string values and then they should not be
replaced.

Keep reading up on regular expressions, this tends to pay off. Here we
use the fact that you can achor a regexp to the end of a string:

R> aString <- "abc:def:::"
R> gsub(":::$", "", aString)
[1] "abc:def"
R> 

Hth, Dirk

Three out of two people have difficulties with fractions.

Thx
I could imagine it was so simple.:)
Bert

 
-----Original Message-----
From: Dirk Eddelbuettel [mailto:edd at master.debian.org] On Behalf Of Dirk
Eddelbuettel
Sent: 04 April 2011 01:39
To: Bert Jacobs
Cc: r-help at r-project.org
Subject: Re: [R] replace last 3 characters of string

I would like to replace the last tree characters of the values of a

certain

column in a dataframe.

This replacement should only take place if the last three characters
correspond to the value "/:/" and they should be replaced with ""(blank)

I cannot perform a simple gsub because the characters /:/ might also be
present somewhere else in the string values and then they should not be
replaced.

Keep reading up on regular expressions, this tends to pay off. Here we
use the fact that you can achor a regexp to the end of a string:

R> aString <- "abc:def:::"
R> gsub(":::$", "", aString)
[1] "abc:def"
R> 

Hth, Dirk

Three out of two people have difficulties with fractions.

I would like to replace the last tree characters of the values of a
certain
column in a dataframe.

Besides the mentioned standard method: I found the subset of string
operations in Hadley Wickhams stringr package helpful. They have a much more
consistent interface compared to the standard interface.

str_sub: accepts negative positions, which are calculated from the left of
the last character.

Dieter



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