Here is a version similar to Rui's, but using ave() and logical
indexing to simplify a bit:
> DF4[with(DF4, ave(as.numeric(var), city, FUN = function(x)length(unique(x)))
> ==1), ]
city wk var
4 city2 1 x
5 city2 2 x
6 city2 3 x
7 city2 4 x
8 city3 1 x
9 city3 2 x
10
Thank you Rudi and Ulrik.
Rudi, your option worked for the small data set but when I applied to
the big data set it taking long and never finished and have to kill
it. I dont know why.
Ulrik's option worked fine for the big data set (> 1.5M records)
and took less than 2 minutes.
These two ar
Using dplyr:
library(dplyr)
# Counting unique
DF4 %>%
group_by(city) %>%
filter(length(unique(var)) == 1)
# Counting not duplicated
DF4 %>%
group_by(city) %>%
filter(sum(!duplicated(var)) == 1)
HTH
Ulrik
On Sat, 18 Mar 2017 at 15:17 Rui Barradas wrote:
> Hello,
>
> I believe this do
Hello,
I believe this does it.
sp <- split(DF4, DF4$city)
want <- do.call(rbind, lapply(sp, function(x)
if(length(unique(x$var)) == 1) x else NULL))
rownames(want) <- NULL
want
Hope this helps,
Rui Barradas
Em 18-03-2017 13:51, Ashta escreveu:
Hi all,
I am trying to find a
Hi all,
I am trying to find a city that do not have the same "var" value.
Within city the var should be the same otherwise exclude the city from
the final data set.
Here is my sample data and my attempt. City1 and city4 should be excluded.
DF4 <- read.table(header=TRUE, text=' city wk var
city1
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
Th
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 si
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 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.
gsub("","
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 wrote:
> Hi all,
>
> I realize that the backslash is an escape character in R, therefore, I
> am trying to replace it with a forwar
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
Dear Pushpa Methekar,
You can use which.max(), as in
> library(car)
> mod <- lm(prestige~ income + education, data=Duncan)
> which.max(abs(rstudent(mod)))
minister
6
In fact, this is exactly what outlierTest() does, as you can see by looking at
car:::outlierTest.lm
Beyond that, I'd en
Hi, this is my function to find rstudent of model which will give me outlier
But I wonder if I could find out the exact no. of outlier in data set.
Like outlierTest() does from car package .
rm.outliers = function(dataset,model){
dataset$rstudent = rstudent(model)
for(i in 1:length(dataset$rstu
Hi,
No problem.
Try:
data.frame(col1=data,col2=as.character(factor(gsub("\\d+","",data),labels=c("positive","negative"))),stringsAsFactors=FALSE)
A.K.
Hello Arun,
Thank you so much, it works great.
But I some sets, the data contains some additional characters
also, like a1, a2, a3..
Hi,
You may try:
dat2 <-
data.frame(col1=data,col2=as.character(factor(data,labels=c("positive","negative"))),stringsAsFactors=FALSE)
A.K.
Hello all,
I have a data something like this;
data<- c("a", "b","b","b","a","a","b","a","b")
and I need to represent all "a"'s as "positive" and "b"
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
Wednesday, March 27, 2013 5:17 PM
> To: Shane Carey
> Cc: r-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,
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:
txt<- "LOI ."
gsub("[.]","%",txt)
#[1] "LOI %"
A.K.
From: Shane Carey
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 o
Hi,
I have a string of text as follows "LOI ."
How do I replace the dot with "(%)"
gsub(".","(%)",LOI .)
gives
"(%)(%)(%)(%)(%)"
Thanks
--
Shane
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__
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https://stat.ethz.c
v_Place)
Students1
# ID Name Fav_Place
#1 101 Andrew Tucson AZ
#2 201 John San Francisco
#3 303 Julie California/New York City
#4 304 Monica New York City
A.K.
- Original Message -
From: Sapana Lohani
To: R help
Cc:
Sent: Mond
ot;ID Name Fav_Place" "101 Andrew Tucson,AZ"
#[3] "201 John SanFrancisco" "303 Julie California/New York City"
#[5] "304 Monica New York City"
A.K.
- Original Message -
From: Sapana Lohani
To: R help
essage -
From: Sapana Lohani
To: R help
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"
Take a look at gsub()
Michael
On Aug 27, 2012, at 6:47 PM, Sapana Lohani 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 with Tucson & N
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]]
__
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'
, yourd
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
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https://stat
On 17.08.2012 02:42, David Winsemius wrote:
On Aug 16, 2012, at 4:33 PM, Kevin Goulding wrote:
Hi, I'm having trouble installing and using the 'foreign' package in R
on a
UNIX machine.
sessionInfo()
R version 2.11.1 (2010-05-31)
sparc-sun-solaris2.10
locale:
[1] C
at
On Aug 16, 2012, at 4:33 PM, Kevin Goulding wrote:
Hi, I'm having trouble installing and using the 'foreign' package in
R on a
UNIX machine.
sessionInfo()
R version 2.11.1 (2010-05-31)
sparc-sun-solaris2.10
locale:
[1] C
attached base packages:
[1] stats graph
Hi, I'm having trouble installing and using the 'foreign' package in R on a
UNIX machine.
> sessionInfo()
R version 2.11.1 (2010-05-31)
sparc-sun-solaris2.10
locale:
[1] C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
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("\\+\\
You are too good :)
Thanks a lot have a nice weekend
B.R
Alex
From: jim holtman
Cc: "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+0_fi
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 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 like
>
>
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 piec
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 strin
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{
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
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 wrote:
> Hi all,
>
> I'm having a problem once again, trying to do something very simple.
> Consider
> the following data frame:
>
> x <- read.table(textConnection("
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=TRU
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", "
Thanks a lot
worked fine
Regards
--- On Thu, 1/6/11, John Kane wrote:
From: John Kane
Subject: Re: [R] Find and remove elemnts of a data frame
To: r-help@r-project.org, "Alaios"
Date: Thursday, January 6, 2011, 3:27 PM
With dataframe xx (naming a data.frame as data.frame is a
With dataframe xx (naming a data.frame as data.frame is a bit dicey
subset(xx, xx[,4]> -45)
--- On Thu, 1/6/11, Alaios wrote:
> From: Alaios
> Subject: [R] Find and remove elemnts of a data frame
> To: r-help@r-project.org
> Received: Thursday, January 6, 2011, 10:07 AM
> De
Dear all,
I have a data frame that is created like that
data.frame(x=CRX[-1],y=CRY[-1],z=CRagent[[1]]$sr)
the output looks like
45 116 162 -30.89105988567164
46 128 79 -42.66296679571184
47 180 195 -30.45626175641315
48 114 83 -45.26843476475688
49 118 73 -46.85389245327003
How can I select
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