This is very interesting. Thanks for sharing.
On Fri, Oct 13, 2023 at 10:25 AM Richard O'Keefe wrote:
>
> ?ifelse
> 'ifelse' returns a value with the same shape as 'test' which is
> filled with elements selected from either 'yes' or 'no' depending
> on whether the element of 'test' is '
?ifelse
'ifelse' returns a value with the same shape as 'test' which is
filled with elements selected from either 'yes' or 'no' depending
on whether the element of 'test' is 'TRUE' or 'FALSE'.
This is actually rather startling, because elsewhere in the
S (R) language, operands are normal
Às 21:22 de 12/10/2023, Christofer Bogaso escreveu:
Hi,
Following expression returns only the first element
ifelse(T, c(1,2,3), c(5,6))
However I am looking for some one-liner expression like above which
will return the entire vector.
Is there any way to achieve this?
___
What a strange question... ifelse returns a vector (all data in R is vectors...
some have length 1, but length zero is also possible, as are longer vectors)
that is exactly as long as the logical vector that you give it, filled with
elements from the respective positions in the vectors supplied
how about
if(T) c(1,2,3) else c(5,6)
?
On 2023-10-12 4:22 p.m., Christofer Bogaso wrote:
Hi,
Following expression returns only the first element
ifelse(T, c(1,2,3), c(5,6))
However I am looking for some one-liner expression like above which
will return the entire vector.
Is there any way
HI Jim
thank you so much! This is amazing answer!!!
Ana
On Sat, Jun 13, 2020 at 4:09 AM Jim Lemon wrote:
>
> Right, back from shopping. Since you have fourteen rows containing NAs
> and you only want seven, we can infer that half of them must go. As
> they are neatly divided into seven rows in
On 2020-06-13 19:09 +1000, Jim Lemon wrote:
> Right, back from shopping. Since you have fourteen rows containing NAs
> and you only want seven, we can infer that half of them must go. As
> they are neatly divided into seven rows in which only one NA appears
> and seven in which two stare meaningles
Dear Ana,
pmax could also fit here.
pmax(b$FLASER, b$PLASER, na.rm = TRUE)
Bests,
Mark
> --
>
> Message: 21
> Date: Sat, 13 Jun 2020 19:09:11 +1000
> From: Jim Lemon
> To: sokovic.anamar...@gmail.com
> Cc: Rasmus Liland , r-help
>
Right, back from shopping. Since you have fourteen rows containing NAs
and you only want seven, we can infer that half of them must go. As
they are neatly divided into seven rows in which only one NA appears
and seven in which two stare meaninglessly out at us. I will assume
that the latter are the
Great idea!
Here it is:
> b[is.na(b$FLASER) | is.na(b$PLASER),]
FID IID FLASER PLASER pheno
1: fam1837 G1837 1 NA 2
2: fam2410 G2410 NA NA 2
3: fam2838 G2838 NA 2 2
4: fam3367 G3367 1 NA 2
5: fam3410 G3410 1 NA 2
6: fam
Since you have only a few troublesome NA values, if you look at them,
or even better, post them:
b[is.na(b$FLASER) | is.na(b$PLASER),]
perhaps we can work out the appropriate logic to get rid of only the
ones you don't want.
Jim
On Sat, Jun 13, 2020 at 12:50 PM Ana Marija wrote:
>
> Hi Rasmus,
Hi Rasmus,
thank you for getting back to be, the command your provided seems to
add all 11 NAs to 2s
> b$pheno <-
+ ifelse(b$PLASER==2 |
+ b$FLASER==2 |
+ is.na(b$PLASER) |
+ is.na(b$PLASER) & b$FLASER %in% 1:2 |
+ is.na
On 2020-06-13 11:30 +1000, Jim Lemon wrote:
> On Fri, Jun 12, 2020 at 8:06 PM Jim Lemon wrote:
> > On Sat, Jun 13, 2020 at 10:46 AM Ana Marija wrote:
> > >
> > > I am trying to make a new column
> > > "pheno" so that I reduce the number
> > > of NAs
> >
> > it looks like those two NA values in
>
Obviously my guess was wrong. I thought you wanted to impute the value
of "pheno" from FLASER if PLASER was missing. From just your summary
table, it's hard to guess the distribution of NA values. My guess that
the two undesirable NAs were cases where PLASER was missing and FLASER
was 2. My tactic
Hi Jim,
I tried it:
> b$pheno<-ifelse(b$PLASER==2 | b$FLASER==2 |is.na(b$PLASER) & b$FLASER ==
> 2,2,1)
> table(b$pheno,exclude = NULL)
12
859 828 11
> b$pheno<-ifelse(b$PLASER==2 | b$FLASER==2 |is.na(b$FLASER) & b$PLASER ==
> 2,2,1)
> table(b$pheno,exclude = NULL)
12
859
Hi Ana,
>From your desired result, it looks like those two NA values in PLASER
are the ones you want to drop.
If so, try this:
b$pheno<-ifelse(b$PLASER==2 | b$FLASER==2 |
is.na(b$PLASER) & b$FLASER == 2,2,1)
and if I have it the wrong way round, swap FLASER and PLASER in the
bit I have added.
J
Hi
another possible version
b$pheno <- ((b$FLASER==2) | (b$PLASER==2))+1
Cheers
Petr
> -Original Message-
> From: R-help On Behalf Of Rui Barradas
> Sent: Monday, May 4, 2020 8:32 PM
> To: sokovic.anamar...@gmail.com; r-help
> Subject: Re: [R] if else statement
>
Your ifelse expression looks fine. What goes wrong with it?
On Tue, 5 May 2020 at 05:16, Ana Marija wrote:
>
> Hello,
>
> I have a data frame like this:
>
> > head(b)
>FID IID FLASER PLASER
> 1: fam1000 G1000 1 1
> 2: fam1001 G1001 1 1
> 3: fam1003 G1003 1
Hello,
Here is a way, using logical indices.
b$pheno <- NA
b$pheno[b$FLASER == 1 & b$PLASER == 1] <- 1
b$pheno[b$FLASER == 2 | b$PLASER == 2] <- 2
Hope this helps,
Rui Barradas
Às 18:15 de 04/05/20, Ana Marija escreveu:
Hello,
I have a data frame like this:
head(b)
FID IID FLA
Thank you for the tip about table function, it seems correct:
> table(b$FLASER, b$PLASER, exclude = NULL)
1 2
1836 6916
2 14 708
0 45 28
> table(b$pheno,exclude = NULL)
12
836 828 34
On Mon, May 4, 2020 at 12:45 PM Jeff Newmiller
wrote:
> T
To expand on Patrick's response...
You can use the expand.grid function to generate a test table containing all
combinations. However, we would not be in a position to verify that the results
you get when you apply your logic to the test table are what you want... you
know the requirements much
"I tried this but I am not sure if this is correct:"
Does it provide the expected result for all possible combinations of 1/2/NA
for both variables?
On Mon, May 4, 2020 at 1:16 PM Ana Marija
wrote:
> Hello,
>
> I have a data frame like this:
>
> > head(b)
>FID IID FLASER PLASER
> 1: f
Hello,
It's just a sequence of ifelse instructions.
dat <- read.table(text = "
A B
1 1
1 0
0 1
0 0
", header = TRUE)
dat$A1 <- ifelse(dat$A == 1 & dat$B == 1, 1, 0)
dat$A2 <- ifelse(dat$A == 1 & dat$B == 0, 1,
Thanks very much for your detailed reply to my post. Very helpful/useful
tool(s) you’ve provide me. Best wishes, B.
From: William Dunlap [mailto:wdun...@tibco.com]
Sent: Wednesday, September 21, 2016 10:48 AM
To: Crombie, Burnette N
Cc: r-help@r-project.org
Subject: Re: [R] if/else help
If
ishes - B
-Original Message-
From: MacQueen, Don [mailto:macque...@llnl.gov]
Sent: Wednesday, September 21, 2016 11:26 AM
To: Crombie, Burnette N ; r-help@r-project.org
Subject: Re: [R] if/else help
Hopefully this is not a homework question.
The other responses are fine, but I would sugges
> On Sep 21, 2016, at 8:26 AM, MacQueen, Don wrote:
>
> Hopefully this is not a homework question.
>
> The other responses are fine, but I would suggest the simplest way to do
> exactly what you ask is
>
>
> if (!exists('r4')) r4 <- data.frame(a=0, b=0, c=0, d='x')
>
>
> The exists() functi
Hopefully this is not a homework question.
The other responses are fine, but I would suggest the simplest way to do
exactly what you ask is
if (!exists('r4')) r4 <- data.frame(a=0, b=0, c=0, d='x')
The exists() function requires a character string for its first argument,
i.e., the name of the
If you write your code as functions you can avoid the nasty
'if(exists("x"))x<-...' business this by writing default values for
arguments to your function. They will be computed only when
they are used. E.g.,
analyzeData <- function(a=0, b=0, c=0, d="x", r4 = data.frame(a, b, c, d)) {
summa
> On Sep 20, 2016, at 12:31 PM, Crombie, Burnette N wrote:
>
> If a data.frame (r4) does not exist in my R environment, I would like to
> create it before I move on to the next step in my script. How do I make that
> happen? Here is what I want to do from a code perspective:
>
> if (exists(r
Get rid of the commas? Get rid of the get() function call? Get rid of the
cbind() function call? Post using plain text format so the HTML doesn't screw
up code? Read the Posting Guide? All of these ideas have merit IMHO...
--
Sent from my phone. Please excuse my brevity.
On September 20, 2016
> jim holtman
> on Sun, 22 May 2016 16:47:06 -0400 writes:
> if you want to use 'ifelse', here is a way:
hmm, why should he want that ?
The OP did mention that it's about somewhat large objects, so
efficiency is one of the considerations :
ifelse() is often convenient and nicely
if you want to use 'ifelse', here is a way:
> k =
+ structure(c(0.0990217544905328, 1.60623837694539, -0.104331330281166,
+ -2.91485614212114, -0.108388742328104, -1.41670341534772,
-1.70609114096417,
+ 2.92018951284015, 0.201868946570178, 0.907637296638577,
-0.403004972105994,
+ -2.4771801580322
Try this:
> sign(ifelse(abs(k)<=1.5, 0, k))
C1 C2 C3 C4
A 0 0 0 0
B 1 0 0 0
C 0 -1 0 0
D -1 1 -1 -1
On Sun, May 22, 2016 at 2:00 PM Adrian Johnson
wrote:
> Hi group:
> I am having difficulty with if else condition. I kindly request some help.
>
> I have a matrix k
>
> > k
>
> On May 22, 2016, at 11:23 AM, Adrian Johnson
> wrote:
>
> Thank you both Dylan and Wray.
>
> since my matrix is quite large and for simplicity in downstream
> operation, i will use sign function. thanks a lot.
>
> On Sun, May 22, 2016 at 2:12 PM, Dylan Keenan wrote:
>> Try this:
>>
>>> si
Thank you both Dylan and Wray.
since my matrix is quite large and for simplicity in downstream
operation, i will use sign function. thanks a lot.
On Sun, May 22, 2016 at 2:12 PM, Dylan Keenan wrote:
> Try this:
>
>> sign(ifelse(abs(k)<=1.5, 0, k))
>
>
> C1 C2 C3 C4
> A 0 0 0 0
> B 1 0 0
Hi Adrian I'm not sure that you need to use the ifelse here. You can simply
assign values ina vector or matrix using a simple condition -- here is a simple
example:
v<-c(4,5,6,7)
v1<-v
v1[]<-0
v1[v<5]<--1
v1[v>6]<-1
v1
Nick
>
> On 22 May 2016 at 18:58 Adrian Johnson wrote:
>
>
> Hi
Ted: You are either being deliberately obtuse or playing Devil's
advocate or just stirring. It is clear from his/her posts that the OP
has limited understanding of both R and statistics. Your sophisticated
philosophising about the possibility of "three sexes" is very unlikely
to have anyth
8:23:05 AM PDT, Val wrote:
> >>>>> Hi All,
> >>>>>
> >>>>>
> >>>>> Yes I need to change to numeric because I am preparing a data set
> >>>>> for
> >>>>> further analysis. The variable to be changed
>>>>> numeric
>>>>> (in this case, sex) will be a response variable. Some records have
>>>>> missing
>>>>> observation on sex and it is blank.
>>>>> id sex
>>>>> 1
>>>>> 2
>>>&g
on sex and it is blank.
> >>> > id sex
> >>> > 1
> >>> > 2
> >>> > 3 M
> >>> > 4 F
> >>> > 5 M
> >>> > 6 F
> >>> > 7 F
> >>&g
F
>>> >
>>> >I am reading the data like this
>>> >
>>> >mydata <- read.csv(header=TRUE, text=', sep=", ")
>>> > id sex
>>> > 1 NA
>>> > 2 NA
>>> > 3 M
>>
>> > 7 F
>> >
>> >The data set is huge (>250,000)
>> >
>> >
>> >I want the output like this
>> >
>> > id sexsex1
>> > 1 NA0
>> > 2 NA 0
>> > 3 M 1
>> &
F 2
> > 7 F 2
> >
> >Thank you in advance
> >
> >
> >On Sat, Oct 31, 2015 at 5:59 AM, John Kane wrote:
> >
> >> In line.
> >>
> >> John Kane
> >> Kingston ON Canada
> >>
> >>
> >> > -
2
>
>Thank you in advance
>
>
>On Sat, Oct 31, 2015 at 5:59 AM, John Kane wrote:
>
>> In line.
>>
>> John Kane
>> Kingston ON Canada
>>
>>
>> > -Original Message-
>> > From: valkr...@gmail.com
>> > Sent:
ada
>
>
> > -Original Message-
> > From: valkr...@gmail.com
> > Sent: Fri, 30 Oct 2015 20:40:03 -0500
> > To: istaz...@gmail.com
> > Subject: Re: [R] If else
> >
> > I am trying to change the mydata$sex from character to numeric
>
> Wh
In line.
John Kane
Kingston ON Canada
> -Original Message-
> From: valkr...@gmail.com
> Sent: Fri, 30 Oct 2015 20:40:03 -0500
> To: istaz...@gmail.com
> Subject: Re: [R] If else
>
> I am trying to change the mydata$sex from character to numeric
W
I am trying to change the mydata$sex from character to numeric
I want teh out put like
id sex
1 NA 0
2 NA 0
3 M 1
4 F 2
5 M1
6 F 2
7 F2
mydata$sex1 <- 0
if(mydata$sex =="M " ){
mydata$sex1<-1
} else {
mydata$sex1<-2
Using numeric for missing sounds like asking for trouble. But if you
must, something like
mydata$confusingWillCauseProblemsLater <-
ifelse(
is.na(mydata$sex),
0,
as.numeric(factor(mydata$sex,
levels = c("M", "F"
should do it.
Best,
Ista
On Fri, Oct 30, 20
Dear all,
All works well. Thank you so much for your help.
D## Function 1
wet_dry1 <- function(x,thresh=0.1)
{ for(column in 1:dim(x)[2]) x[,column] <- ifelse(x[,column]>=thresh,1,0)
return(x)
}
wet_dry1(dt)
## Function 2
wet_dry2 <- ( dt >= 0.1)*1
wet_dry2
wet_total <- colSums(wet_dry2)
p
Your f1() has an unneeded for loop in it.
f1a <- function(mat) mat > 0.1, 1, 0)
would do the same thing in a bit less time.
However, I think that a simple
mat > 0.1
would be preferable. The resulting TRUEs and FALSEs
are easier to interpret than the 1s and 0s that f1a()
produces and arithme
I'm sorry, but I have to take issue with this particular use case of
ifelse(). When the goal is to generate a logical vector, ifelse() is
very inefficient. It's better to apply a logical condition directly to
the object in question and multiply the result by 1 to make it
numeric/integer rather than
Thank you jim.
On Saturday, June 6, 2015, Jim Lemon wrote:
> Hi rosalinazairimah,
> I think the problem is that you are using "if" instead of "ifelse". Try
> this:
>
> wet_dry<-function(x,thresh=0.1) {
> for(column in 1:dim(x)[2]) x[,column]<-ifelse(x[,column]>=thresh,1,0)
> return(x)
> }
> we
Hi rosalinazairimah,
I think the problem is that you are using "if" instead of "ifelse". Try this:
wet_dry<-function(x,thresh=0.1) {
for(column in 1:dim(x)[2]) x[,column]<-ifelse(x[,column]>=thresh,1,0)
return(x)
}
wet_dry(dt)
and see what you get.
Also, why can I read your message perfectly w
Please do not post in HTML. It made your posting unreadable. R-help is a plain
text list and when it removes all the HTML tags often the result is gibberish
Have a look at
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
and http://adv-r.had.co.nz/Reproduci
Hi
I will not inspect your function as it is corrupted by HTML posting.
If your data frame is named rain
newrain <- (rain>.1)*1
gives you new data frame with reqired coding.
However I am not sure, what do you want to do next. Do you want to merge those
2 data frames so as coded column is besi
ogy
> Texas A&M University
> College Station, TX 77840-4352
>
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of David
> Winsemius
> Sent: Tuesday, December 2, 2014 2:50 PM
> To: Jefferson Ferreira-Ferreira
> Cc: r-help
ege Station, TX 77840-4352
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of David Winsemius
Sent: Tuesday, December 2, 2014 2:50 PM
To: Jefferson Ferreira-Ferreira
Cc: r-help@r-project.org
Subject: Re: [R] if else for cumulative sum error
On Dec 2, 2014, at 12:26 PM
On Dec 2, 2014, at 12:26 PM, Jefferson Ferreira-Ferreira wrote:
> Thank you for replies.
>
> David,
>
> I tried your modified form
>
> for (i in 1:seq_along(rownames(dadosmax))){
No. it is either 1: or seq_along(...). in this case perhaps
1:(nrow(dadosmax)-44 would be safer
You do not
Thank you for replies.
David,
I tried your modified form
for (i in 1:seq_along(rownames(dadosmax))){
dadosmax$enchday[i] <- if ( (sum(dadosmax$above[i:(i+44)])) >= 45) 1 else
0
}
However, I'm receiving this warning:
Warning message:
In 1:seq_along(rownames(dadosmax)) :
numerical expression
On Tue, Dec 2, 2014 at 12:08 PM, Jefferson Ferreira-Ferreira <
jeco...@gmail.com> wrote:
> Hello everybody;
>
> I'm writing a code where part of it is as follows:
>
> for (i in nrow(dadosmax)){
> dadosmax$enchday[i] <- if (sum(dadosmax$above[i:(i+44)]) >= 45) 1 else 0
> }
>
Without some test d
On Dec 2, 2014, at 10:08 AM, Jefferson Ferreira-Ferreira wrote:
> Hello everybody;
>
> I'm writing a code where part of it is as follows:
>
> for (i in nrow(dadosmax)){
> dadosmax$enchday[i] <- if (sum(dadosmax$above[i:(i+44)]) >= 45) 1 else 0
> }
>
> That is for each row of my data frame, su
e(1:3, .Label = c("samas4", "samas5", "samas6"
), class = "factor")), .Names = c("FID", "IID"), class = "data.frame",
row.names = c(NA,
-3L))
> -Original Message-
> From: Kate Ignatius [mailto:kate.ignat...@g
Ooops,
I edited the code wrong to make it more easier for interpretation and
got X and Y's mixed up. Try this:
for(i in length(1:(nrow(X{
Y$IID1new <- ifelse((as.character(Y[,2]) == as.character(X[,i]) &
Y$IID1new != ''), as.character(as.matrix(X[,(nrow(X)+i)])),'')
}
The second should
Hi
Please, be more clear in what do you want. I get many errors trying your code
and your explanation does not help much.
> for(i in length(1:(2*nrow(X{
+ Y$IID1new <- ifelse((as.character(Y[,2]) == as.characterXl[,i]) &
X$IID1new != '') , as.character(as.matrix(X[,(2*nrow(X)+i)])),'')
On Nov 19, 2013, at 5:30 PM, Gary Dong wrote:
> Dear R users,
>
> I am a R beginner and I am having trouble in using "If Else" in R. Here is
> an example:
>
> ## create a data frame
>
> a<-c(1,2,3,4)
> b<-c(2,0,5,0)
> ab<-data.frame(cbind(a,b))
>
> ##calculate c, which is the ratio between a
Which means you should use the ifelse function...
ab$c <- ifelse( ab$b>0, ab$a/ab$b, 0 )
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Hi,
Try:
within(ab, {c <- ifelse(b>0, a/b,0)})
A.K.
On Tuesday, November 19, 2013 8:51 PM, Gary Dong wrote:
Dear R users,
I am a R beginner and I am having trouble in using "If Else" in R. Here is
an example:
## create a data frame
a<-c(1,2,3,4)
b<-c(2,0,5,0)
ab<-data.frame(cbind(a,b))
##c
Hi,
On Nov 19, 2013, at 9:41 PM, Nick Matzke wrote:
> Hi,
>
> This would be an issue with if() as well as if/else. ab$b has 4
> numbers in it, so ab$b > 0 evaluates to "TRUE TRUE FALSE TRUE" or
> whatever. if() can only take a single true or false. Cheers! Nick
>
As a follow up, you could ma
Hi,
This would be an issue with if() as well as if/else. ab$b has 4
numbers in it, so ab$b > 0 evaluates to "TRUE TRUE FALSE TRUE" or
whatever. if() can only take a single true or false. Cheers! Nick
On Tue, Nov 19, 2013 at 8:30 PM, Gary Dong wrote:
> Dear R users,
>
> I am a R beginner and I a
y by:
dat1$t.tr<- as.character(dat1$t.tr)
A.K.
- Original Message -
From: "Fethe, Michael"
To: Dennis Murphy ; "R-help@r-project.org"
Cc:
Sent: Monday, August 26, 2013 8:45 AM
Subject: Re: [R] If else loop problem: the condition has length > 1 and only
689 -43.47 645.53
> #4 Pto 72 624 -68.17 555.83
> #5 Pto 72 666 -68.17 597.83
> #6 Pto 24 620 -29.39 590.61
>
>
> #or ?merge()
>
> A.K.
>
>
>
>
> - Original Message -
> From: Bert Gunter
> To: arun
> Cc: &q
a-j)? This code
produces :
In Ops.factor(signal2, 10.465507) : - not meaningful for factors
Thanks for the input.
____________
From: Dennis Murphy
Sent: Monday, August 26, 2013 4:59 AM
To: Fethe, Michael
Subject: Re: [R] If else loop problem: the condition has length > 1 and
nces@r-
> project.org] On Behalf Of arun
> Sent: Monday, August 26, 2013 8:32 AM
> To: Fethe, Michael
> Cc: R help
> Subject: Re: [R] If else loop problem: the condition has length > 1 and
> only the first element will be used
>
> HI,
>
> It may be better to provide
Cc: "Fethe, Michael" ; R help
Sent: Monday, August 26, 2013 10:09 AM
Subject: Re: [R] If else loop problem: the condition has length > 1 and only
the first element will be used
Suggestion:
Don't do the ifelse stuff below.
See ?switch instead.
-- Bert
On Sun, Aug 25, 2013 at 1
Suggestion:
Don't do the ifelse stuff below.
See ?switch instead.
-- Bert
On Sun, Aug 25, 2013 at 11:32 PM, arun wrote:
> HI,
>
> It may be better to provide an example dataset using ?dput().
> dput(head(dataset),20)
>
> Try:
> signal3<- ifelse(t.tr=="Pst 24", signal2-17.29, ifelse(t.tr=="Pst
HI,
It may be better to provide an example dataset using ?dput().
dput(head(dataset),20)
Try:
signal3<- ifelse(t.tr=="Pst 24", signal2-17.29, ifelse(t.tr=="Pst 48", signal2
- 43.93256, etc.))
A.K.
- Original Message -
From: "Fethe, Michael"
To: "r-help@r-project.org"
Cc:
Sent:
Hi
What about creating a new column with same factor and just chase its levels.
dat2$col2<-dat2$col1
> levels(dat2$col2)
[1] "high""low" "Neutral"
> levels(dat2$col2)<-c("H", "L","N")
> dat2
col1 col2
1 highH
2 NeutralN
3 NeutralN
4 lowL
5 highH
6
It seems to me that the "recode()" function from the "car" package
is what you need.
cheers,
Rolf Turner
On 13/08/12 13:07, Sachinthaka Abeywardana wrote:
The thing is I have about 10 cases. I saw the ifelse statement but was
wondering if there was a cleaner method of doing it. Th
ot;low"="L"')
dat2
# col1 col2
#1 high H
#2 Neutral N
#3 Neutral N
#4 low L
#5 high H
#6 low L
#7 low L
#8 Neutral N
#9 Neutral N
#10 high H
A.K.
____________
From: Sachinthaka Abeywardana
To:
_
From: Sachinthaka Abeywardana
To: arun
Cc: R help
Sent: Sunday, August 12, 2012 9:07 PM
Subject: Re: [R] if else elseif for data frames
The thing is I have about 10 cases. I saw the ifelse statement but was
wondering if there was a cleaner method of doing it. The coding will get re
On Sun, Aug 12, 2012 at 8:07 PM, Sachinthaka Abeywardana
wrote:
> The thing is I have about 10 cases. I saw the ifelse statement
Note that there is no "ifelse" statement: there is only nested if/else of forms
if
else if
else if
else
> but was
> wondering if
On Aug 12, 2012, at 5:43 PM, Sachinthaka Abeywardana wrote:
Hi all,
It seems like I cannot use normal 'if' for data frames. What would
be the
best way to do the following.
if data$col1='high'
data$col2='H'
else if data$col1='Neutral'
data$col2='N'
else if data$col='low'
data$col2='
Maybe the cut function?
?cut
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Pl
You might look at the 'recode' that is part of the 'car' package. You
can also setup a dataframe with two columns; one with the current
value and one with the new value you want, then you can use 'merge' or
'match' to create your new column. If you have provided some sample
data, I could have pro
Hi,
Try this:
dat1<-data.frame(col1=c(rep("high",3),rep("Neutral",3),rep("low",4)))
dat1$col2<-ifelse(dat1$col1=="high",dat1$col2<-"H",ifelse(dat1$col1=="Neutral",dat1$col2<-"N","L"))
dat1
col1 col2
1 high H
2 high H
3 high H
4 Neutral N
5 Neutral N
6 Neutral
The thing is I have about 10 cases. I saw the ifelse statement but was
wondering if there was a cleaner method of doing it. The coding will get
really messy when I write all 10 cases.
Cheers,
Sachin
On Mon, Aug 13, 2012 at 11:04 AM, arun wrote:
> Hi,
> Try this:
> dat1<-data.frame(col1=c(rep("h
It is well explained here
http://www.burns-stat.com/pages/Tutor/R_inferno.pdf page 67.
On Sat, Jan 21, 2012 at 9:56 PM, Ery Arias-Castro wrote:
> Hello,
>
> This example seems strange to me:
>
> > if (2 > 3) print('Yes'); else print('No')
> Error: unexpected 'else' in " else"
>
> > {if (2 > 3) p
Don't use the ';'
> if (2 > 3) print('Yes'); else print('No')
Error: unexpected 'else' in " else"
No suitable frames for recover()
> if (2 > 3) print('Yes') else print('No')
[1] "No"
>
'else' is part of the 'if'. The ';' implies the start of a new
statement. You do not need the ';' when writing
Hi jost87,
Try using a good text editor with syntax highlighting and
parenthesis/brace matching (personal preference is Emacs + ESS, others
like Rstudio, and there are tons more). I find at least one
mismatched pair of parentheses/braces in those expressions.
Cheers,
Josh
On Fri, Jan 13, 2012
On 11/15/2011 06:46 PM, Kevin Burton wrote:
What is wrong with the following?
x<- 1:2
if(x[1]> 0)
{
if(x[2]> 0)
{
print("1& 2> 0")
}
else
{
Hi,
Check out this R-help thread from 2007.
http://tolstoy.newcastle.edu.au/R/e2/help/07/06/19513.html
Cheers,
Ben
On Nov 15, 2011, at 6:46 PM, Kevin Burton wrote:
> What is wrong with the following?
>
>
>
> x <- 1:2
>
> if(x[1] > 0)
>
> {
>
>if(x[2] > 0)
>
>
On Tue, Nov 15, 2011 at 3:46 PM, Kevin Burton wrote:
> What is wrong with the following?
AFAIK the else needs to follow the end brace of if {} on the same
line, at least at the main level.
Peter
>
>
>
> x <- 1:2
>
> if(x[1] > 0)
>
> {
>
> if(x[2] > 0)
>
> {
>
>
Hi Peter:
Beware that "gnashing of teeth" business. My Dad, a dentist, said that
a couple of his patients who were musicians did this when they played
and ground their teeth down so much he had to make them dentures!
;-)
Cheers,
Bert
On Tue, Jul 5, 2011 at 2:53 PM, peter dalgaard wrote:
>
> On
On Tue, Jul 5, 2011 at 2:53 PM, peter dalgaard wrote:
>
> On Jul 5, 2011, at 22:27 , gary engstrom wrote:
>
>> I am trying to use if...else loop
>
> Argh!
>
> A loop goes _around_ and around. "for", "repeat", "while".
>
> "if ... else" is a _branching_ construct.
Nominated for a fortune.
>
> -
On Jul 5, 2011, at 22:27 , gary engstrom wrote:
> I am trying to use if...else loop
Argh!
A loop goes _around_ and around. "for", "repeat", "while".
"if ... else" is a _branching_ construct.
--
Peter Dalgaard
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederi
Gary,
Was the second half of my message this morning not clear enough?
It wasn't clear from your original message that you were using isin()
from the prob package, rather than using isin() as pseudocode, so I'd
written a function to do that part. But the second half of my message
went through the
Hi Gary,
A solution in two pieces.
First, you need to be able to match the rows of your data frame. There
might be a more elegant way to do it, but I couldn't think of one that
gave the option of ordering or not, so I wrote a function:
isin <- function(dd, tomatch, ordered=TRUE) {
# find toma
Thanks alot B77S. That was a critical post.
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Thanks for the answers. They were very helpful. I did it like below.
data$z <- ifelse(data$x == "A" & data$y == "A", "NA", "B").
That is enough for today. I guess I learned a enough today. Thanks to
everbody who tried to help.
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Hi ozgrerg,
I am not a "good" R user, but according to me you can do this;
data <- data.frame(x=c("A","A","B","A"), y=c("A","B","B","B"))
for (i in 1:nrow(data)){
if(data$x[i]=="B"||data$y[i]=="B") z[i]=c("B")
else
z[i]=c("NA")
}
cbind(data,z)
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