On 26/05/2015 9:56 PM, Ista Zahn wrote:
Escape the backslash with another backslash, i.e.,
gsub(\\,/,X:\\Classes\\TT\\Automation, fixed = TRUE)
... and note that if you want to use a regular expression (i.e. fixed =
FALSE), you would need another level of escaping, i.e.
Since the character looks like a Windows file path, you could use
normalizePath() instead of gsub().
normalizePath(X:\\Classes\\TT\\Automation, winslash = /, mustWork =
FALSE)
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie
On 27/05/2015 8:55 AM, Dan Abner wrote:
Hi Ista,
Is there no way to not escape the backslash in the pathway?
You don't need to escape it if you read it from a file, get it from
list.files(), etc. You only need to escape it if you are writing a
literal string in R code.
Duncan Murdoch
The
Hi Ista,
Is there no way to not escape the backslash in the pathway? The
pathway is going to change and will become very long and I need to do
this programmatically. Beside, escaping the backslash defeats the
purpose of using gsub. If I could do this manually each and every
time, I would change
Hi all,
I realize that the backslash is an escape character in R, therefore, I
am trying to replace it with a forward slash. Can someone please
suggest how to get this code to work?
lib-gsub(\,/,X:\Classes\TT\Automation)
Error: unexpected symbol in lib-gsub(\,/,X
Thanks,
Dan
Escape the backslash with another backslash, i.e.,
gsub(\\,/,X:\\Classes\\TT\\Automation, fixed = TRUE)
best,
Ista
On Tue, May 26, 2015 at 9:30 PM, Dan Abner dan.abne...@gmail.com wrote:
Hi all,
I realize that the backslash is an escape character in R, therefore, I
am trying to replace it
Hello,
I have a question and need your help urgently. I am new to R but want to learn
it.
I have several files in a folder which I have imported to R using :
temp = list.files(pattern=*.txt)
myfiles = lapply(temp, read.delim)
The resulting files are on the workspace stored as List[110]. So
Hi,
I have a string of text as follows LOI .
How do I replace the dot with (%)
gsub(.,(%),LOI .)
gives
(%)(%)(%)(%)(%)
Thanks
--
Shane
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
txt- LOI .
gsub([.],%,txt)
#[1] LOI %
A.K.
From: Shane Carey careys...@gmail.com
To: r-help@r-project.org
Sent: Wednesday, March 27, 2013 12:09 PM
Subject: [R] find and replace characters in a string
Hi,
I have a string of text as follows LOI .
How do I
Hello,
The period is a metacharacter so you have to escape it.
The period is escaped with a '\'. In it's turn, '\' is a metacharacter
so it needs to be escaped. Hence the double'\\'.
x - LOI .
gsub(\\., (%), x)
Hope this helps,
Rui Barradas
Em 27-03-2013 16:09, Shane Carey escreveu:
Hi,
-help@r-project.org
Subject: Re: [R] find and replace characters in a string
Hello,
The period is a metacharacter so you have to escape it.
The period is escaped with a '\'. In it's turn, '\' is a metacharacter
so it needs to be escaped. Hence the double'\\'.
x - LOI .
gsub(\\., (%), x
-help@r-project.org
Cc:
Sent: Monday, August 27, 2012 4:19 PM
Subject: [R] find and replace
I have 5 (A,B,C,D,E) columns in my dataframe. I want to replace all x with
y and all a with b within these 5 columns. Can I do it in one step?
Thanks
[[alternative HTML version deleted
#[3] 201 John SanFrancisco 303 Julie California/New York City
#[5] 304 Monica New York City
A.K.
- Original Message -
From: Sapana Lohani lohani.sap...@ymail.com
To: R help r-help@r-project.org
Cc:
Sent: Monday, August 27, 2012 7:47 PM
Subject: [R] find and replace
Hi
City
#4 304 Monica New York City
A.K.
- Original Message -
From: Sapana Lohani lohani.sap...@ymail.com
To: R help r-help@r-project.org
Cc:
Sent: Monday, August 27, 2012 7:47 PM
Subject: [R] find and replace
Hi,
My data frame (Students) is
ID Name Fav_Place
101 Andrew
I have 5 (A,B,C,D,E) columns in my dataframe. I want to replace all x with
y and all a with b within these 5 columns. Can I do it in one step?
Thanks
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
I am making the assumption that all the columns are character and not factors:
for (i in c(A, B, C, D, E)){
yourdf[[i]] - ifelse(yourdf[[i]] == 'x'
, 'y'
, ifelse(yourdf[[i]] == 'a'
, 'b'
, yourdf[[i]]
Hi,
My data frame (Students) is
ID Name Fav_Place
101 Andrew Phoenix AZ
201 John San Francisco
303 JulieCalifornia / New York
304 Monica New York
How can I replace Phoenix with Tucson New York with New York City in the df?
Thanks
[[alternative HTML version deleted]]
Take a look at gsub()
Michael
On Aug 27, 2012, at 6:47 PM, Sapana Lohani lohani.sap...@ymail.com wrote:
Hi,
My data frame (Students) is
ID Name Fav_Place
101 Andrew� Phoenix AZ
201 John San Francisco
303 JulieCalifornia / New York
304 Monica� New York
How can I replace Phoenix
Dear all,
I would like to search in a string for the second occurrence of a symbol and
replace the symbol after it
For example my strings look like
sta_+1+0_field2ndtry_$01.cfg
I want to find the digit that comes after the second +, in that case is zero
and then over a loop create the
try this:
x - c('sta_+1+0_field2ndtry_$01.cfg'
+ , 'sta_+1+0_field2ndtry_$01.cfg'
+ , 'sta_+1-0_field2ndtry_$01.cfg'
+ , 'sta_+1+0_field2ndtry_$01.cfg'
+ )
# find matching fields
values - grep([^+]*\\+[^+]*\\+0, x, value = TRUE)
# split into two pieces
If the length of the fists part is constant (the sta_+1+ part) the
you can use substr()
On 2 December 2011 13:30, Alaios ala...@yahoo.com wrote:
Dear all,
I would like to search in a string for the second occurrence of a symbol and
replace the symbol after it
For example my strings look
You are too good :)
Thanks a lot have a nice weekend
B.R
Alex
From: jim holtman jholt...@gmail.com
Cc: R-help@r-project.org R-help@r-project.org
Sent: Friday, December 2, 2011 1:51 PM
Subject: Re: [R] find and replace string
try this:
x - c('sta_+1
You've been given a workable solution already, but here's a one-liner:
x - c('sta_+1+0_field2ndtry_$01.cfg' ,
'sta_+B+0_field2ndtry_$01.cfg' , 'sta_+1+0_field2ndtry_$01.cfg' ,
'sta_+9+0_field2ndtry_$01.cfg')
sapply(1:length(x), function(i)gsub(\\+(.*)\\+., paste(\\+\\1\\+,
Hi all,
I'm having a problem once again, trying to do something very simple. Consider
the following data frame:
x - read.table(textConnection(locus1 locus2 locus3
A T C
A T NA
T C C
A T G), header = TRUE)
closeAllConnections()
I am trying to make a new data frame, replacing A with A/A, T with
Josh, you've made it far too complicated. Here's one simpler way (note
that I changed your read.table statement to make the values NOT factors,
since I wouldn't think you want that).
x - read.table(textConnection(locus1 locus2 locus3
+ A T C
+ A T NA
+ T C C
+ A T G), header = TRUE, as.is=TRUE)
Try this:
xNew - as.data.frame(mapply(paste, x, x, sep = /))
xNew[is.na(x)] - NA
xNew
On Thu, Feb 17, 2011 at 2:54 PM, Josh B josh...@yahoo.com wrote:
Hi all,
I'm having a problem once again, trying to do something very simple.
Consider
the following data frame:
x -
Hi,
You could use car::recode to change the levels of the factors,
library(car)
transform(x, locus1 = recode(locus1, 'A' = 'A/A' ; else = 'T/T'),
locus2 = recode(locus2, 'T'='T/T' ; 'C' = 'C/C'),
locus3 = recode(locus3, 'C'='C/C' ; 'G' = 'G/G'))
HTH,
You may write as this:
for (i in 1:nrow(x)){
for (j in 1:ncol(x)){
if (!is.na(x[i, j])) {
if(x[i, j] == 'A') {x2[i, j] - 'A/A'} else{
if(x[i, j] == 'T') {x2[i, j] - 'T/T'} else{
if(x[i, j] == 'G') {x2[i, j] - 'G/G'} else{
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