similar to: rep within an ifelse statment

This is a "long" way; i.e., not necessarily efficient: > qs2 [1] 2 1 1 4 4 4 1 1 1 4 2 4 3 1 4 3 3 2 4 3 > qs9 [1] 4 4 1 3 4 3 1 3 1 4 1 2 3 3 4 4 1 4 2 3 > decision <- function(a, b) { + if (a == 1 || b == 1) return(1) + if (a == 2 || b == 2) return(2) + if (a == 3 || b == 3) return(3) + if (a == 4 || b == 4) return(4) + NA + } > mapply(decision,

To anyone who can help: I am trying to do an analysis of variance with a very unbalanced design and a few empty cells. How do I get R to use type III and more importantly type IV sums of squares? Thanks Deven Hamilton Department of Sociology University of Washington

Hi, Given > sec98 <- factor(rep(1:2,3), labels=c("A", "B")) > sec99 <- factor(rep(2:1,3), labels=c("A", "B")) > sec99[c(2,5)] <- NA > sec00 <- factor( c( rep(1,3), rep(2,3) ), labels=c("A", "B")) > sec00[c(2,4)] <- NA > sec1 <- ifelse(!is.na(sec99), sec99, ifelse(!is.na(sec00), sec00, NA

Hi all, I have time series data in a matrix format 20 rows x 205 columns and have been trying to replace outliers with NA. My first column contains the outliers threshold (3 Standard deviations) for each row - here's a bit of the first row sd3 V1 V2 V3 V4 V5 V6 V7 V8 V9 1 13.03267 1797157 75 84 58 -1.958649 0.048775 2.056198 8.063622

Dear R-helper, I am working on a very large data frame and I am trying to add a new column and write in it with certain conditions. I have try to use this code with the data frame p : ID = 0 p[,"newColumn"]<- ifelse (p$flagFoehn3_durr == 1, ifelse(p$Guetsch == 0, ID <<- ID ++ , ID ) , 0 ) What I am trying to do

dear R experts--- t <- 1:30 f <- function(t) { cat("f for", t, "\n"); return(2*t) } g <- function(t) { cat("g for", t, "\n"); return(3*t) } s <- ifelse( t%%2==0, g(t), f(t)) shows that the ifelse function actually evaluates both f() and g() for all values first, and presumably then just picks left or right results based on t%%2.

I'm a fairly new R user and while I have a solution to my problem I'm wondering if there is a better one. In SAS it's common to use if/then logic along with a "do" statement to make several things happen. While I could do the same thing in R using a "for" loop, and "if" and {}, I've read that loops are less common in R and I wonder if I'm doing

Related to the length of 'ifelse' result, I want to say that "example of different return modes" in ?ifelse led me to perceive a wrong thing in the past. ## example of different return modes: yes <- 1:3 no <- pi^(0:3) typeof(ifelse(NA, yes, no)) # logical typeof(ifelse(TRUE, yes, no)) # integer typeof(ifelse(FALSE, yes, no)) # double As

Full_Name: Woolton Lee Version: 2.1 OS: windows Submission from: (NULL) (128.118.224.46) I did the following ('g' and 'h' are both numeric vectors) > i <- abs(g-h) creating a vector 'i' with values, > i [1] 0.08 0.00 0.33 0.00 0.00 0.00 0.00 0.33 0.00 0.00 0.08 0.08 0.20 0.00 0.13 Now, I want to create a new vector =1 whenever 'i' = 0.33 and =0

> Thomas Lumley wrote: > > On Tue, 31 May 2005, Duncan Murdoch wrote: > > > > > >>M??kinen Jussi wrote: > >> > >>>Dear All, > >>> > >>>I luckily found the following feature (or problem) when tried to > >>>apply > >>>ifelse-function to an ordered data. > >>> > >>> >

When I run this code from an R-script: ddd = 360 + round ( atan2(-u,-v) / d2r ) print(class(ddd)) print(ddd) ifelse ( ddd>360, ddd-360, ddd ) print(ddd) I get this output: [1] "numeric" [1] 461 213 238 249 251 [1] 461 213 238 249 251 Why does ifelse not change the 461 to 101? I recreated the vector ddd and ran the same ifelse code

Hello I want to make some variables with the ifelse-function, but i don't know how to do it. I want to make these five variables; b2$PRRSvac <- ifelse(b2$status=='A' | b2$status=='Aa',1,0) b2$PRRSdk <- ifelse(b2$status=='B' | b2$status=='Bb',1,0) b2$sanVac <- ifelse(b2$status=='C' | b2$status=='sanAa',1,0) b2$sanDk <-

Hi R-helpers, Please see the below R output. The problem is that after running the ifelse statement, data$SOCIAL_STATUS is converted from a factor to a character. Is there some way I can avoid this conversion? Thanks in advance, Mark Na > str(data) 'data.frame': 2100 obs. of 11 variables: $ DATE : Factor w/ 5 levels "4-Jun-09","7-May-09",..: 1 1 1 1 1

I propose a patch to ifelse that leverages anyNA(test) to achieve an improvement in performance. For a test vector of length 10, the change nearly halves the time taken and for a test of length 1 million, there is a tenfold increase in speed. Even for small vectors, the distributions of timings between the old and the proposed ifelse do not intersect. The patch does not intend to change the

Could someone help me with the following code snippet. The results are not what I expect: > Sheet1$Claims[1:10] [1] NA 1 2 NA NA NA NA NA NA NA > Sheet1[1:10,"SubmissionStatus"] [1] Declined Bound Bound Bound Bound Bound Declined Dead Declined [10] Not Taken Levels: Bound Dead Declined Not Taken > Sheet1$Claimsnum <- NA >

Hi all, Using R v2.7.1, platform i386-pc-mingw32 Can someone please shed some light on the behaviour of ifelse() for me? My intent is to calc relative proportions of z$b, at the same time subsetting z$b based on z$a. I could attack the problem other ways (suggestions welcome) but I am also intrigued by the _order_ in which ifelse seems to assign values, and how recycling works. For instance,

I need use different parameters of distribution for different case to generate random value, but I use ifelse, the generated value is fixed without change. Here is example data1 y x 1 1 2 2 2 1 3 3 2 4 4 3 5 5 3 6 6 1 7 7 2 8 8 1 9 9 1 10 10 3 11 11 3 12 12 2 ifelse(data1$x==1,rnorm(1,2,1),ifelse(data1$x==2,rnorm(1,-2,1),rnorm(1,110,1))) [1] -1.8042172 0.8478681

Dear R users, I have a question how to use 2 "ifelse" to sort my data. Such as from 11 to 20 assign to A; 6 to 10 assign to B, and the rest of them assign to C a<-1:20 tt<-ifelse(a>10, "A",no=ifelse( 5< a <=10, "B", "C")) Many Thanks Chunhao -- View this message in context:

Hello all, This is one of those "Is there a better way to do this questions". Say I have a dataframe (df) with a grouping variable (z). This is my base data. Now I know that there is a higher order level of grouping that exist for my group variable. So what I want to do is create a new column that express that higher order level of grouping based on values in the sub-group (z in this

Hello, All, Mac OS 10.6.5 R64 2.11.1 This works as expected: f1 = c(0.084, 0.099, 0) data= data.frame(f1) data$f1=with(data,ifelse(f1==0, 0.0001, f1)) data f1 1 0.0840 2 0.0990 3 0.0001 Substituting ''f1==0'' with ''T'' produces the expected result: f1 = c(0.084, 0.099, 0) data= data.frame(f1) data$f1=with(data,ifelse(T, 0.0001, f1)) data f1 1 1e-04