search for: ifelse

...t;clone" (/plant genotype) on the survival of eggs (due to predation). Each plant was provided with three different sizes of clutches (45,15,5) and had pseudo-replications of size 15 and 5 on it. _Question 1: right code for nested Anova_ Here part of the code I used: clone1<-as.factor(ifelse(clone=="Gudrun",1,ifelse(clone=="Loden",2,ifelse(clone=="021",3,4)))) plant1<-as.factor(ifelse(plant==1,1,ifelse(plant==2,2,ifelse(plant==3,3,ifelse(plant==4,4,ifelse(plant==5,5,ifelse(plant==6,6, ifelse(plant==7,7,ifelse(plant==8,8,ifelse(plant==9,9,ifelse(plant=...

Hi R users, Is there a way to simplify this instruction: ifelse(B=="0","A", ifelse(B=="1","A", ifelse(B=="2","A", ifelse(B=="3","A", ifelse(B=="4","A", ifelse(B=="5","A", ifelse(B=="6","A", ifelse(B=="7","A&q...

I'm trying to create a new column in my data.frame where subjects are categorized depending on values on four other columns. In any other case I would just nest a few ifelse statements, however, in this case i have 4*6*2*3=144 combinations and i get weird 'context overflow' errors. So I wonder if there is a more efficient way of doing this. For illustrational purposes, let's say i have: x<-c(1,0,0,1,0,0,1,0,0,1) y<-c(1,3,2,3,2,1,2,3,2,3) z<-c(1,...

...hat are 1 million or longer and need to repeat that process several hundred times. The for loop works but is slow. If there is a faster way, please let me know. probs <- c(.1, .3, .2, .4, .7, .9, .3, .4, .5, .6) p10 <- 0.6 p00 <- 0.4 vec1 <- rep(0, 10) for (i in 2:10) { vec1[i] <- ifelse(vec1[i-1] == 0, ifelse(probs[i]<p10, 0, 1), ifelse(probs[i]<p00, 0, 1)) } Thanks GG [[alternative HTML version deleted]]

I have a problem with ifelse(), I do not understand how it works. > X<-c(2,2,1,1,0,0) > str(X) num [1:6] 2 2 1 1 0 0 > Y<-ifelse(X>0,1,0) > Y [1] 1 1 1 1 0 0 > Can some one explain what is going on, I do not understand what ifelse is doing in this case. Can someone explain the output Y. Thanks --...

# This is more about trying to find a more effecient way to code some simple vectorized computations using ifelse(). # Say you have some vector representing a factor with a number of levels (6 in this case), representing the location that samples were collected. Population <- gl( n=6, k=5,length=120, labels =c("CO", "CN","Ga","KO", "Mw", "Ng"))...

...en 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 things ineffectively. Below I've used a "ifelse" function to make an assignment to the variable FUT.direction, but what I'd really like to do is make several assignments based on the same "ifelse" logic. Since the "ifelse" wants to create a vector the same size as the logical condition it doesn't seem obvious ho...

dear list: I need help to achieve a simpler code to complete a task I'm performing. here is an example: dat<-expand.grid(a=seq(1,5),b=seq(1000,1005)) I want to add a new column dat$c in that: t1<-ifelse(dat$a==1&dat$b==1001,1001,0) t2<-ifelse(dat$a==2&dat$b==1002,1001,0) t3<-ifelse(dat$a==3&dat$b==1003,1001,0) t4<-ifelse(dat$a==1&dat$b==1002,1002,0) t5<-ifelse(dat$a==2&dat$b==1003,1002,0) t6<-ifelse(dat$a==1&dat$b==1003,1003,0) dat$c<-t1+t2+t3+t4+t5+t6 M...

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$statu...

I have found a strange "ifelse" behaviour (I think) This works: > ifelse(T,1+1,1+2) [1] 2 > ifelse(F,1+1,1+2) [1] 3 Maybe I missed something about R internals, but why > ifelse(T,print("hello"),print("goodbye")) [1] "hello" [1] "hello" > ifelse(F,print("hello&quot...

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 and it did work as expected. > ddd <...

...#39; 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 otherwise. I use the 'ifelse' function to do this but when I do, I find > ifelse(i==0.33,1,0) [1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 even though the third and eighth elements of 'i' clearly equal 0.33. Unless I have missed something, 'ifelse' should return a vector that looks like, [1] 0 0 1 0 0 0 0 1 0 0...

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: http://n4.nabble.com...

...) if(x<=1/10){y[i]=1} else if(x<=2/10){y[i]=2} else if(x<=4/10){y[i]=3} else if(x<=7/10){y[i]=4} else if(x<=9/10){y[i]=5} else{y[i]=6} } bar<-barplot(table(y)) table<-table(y) return(list(bar,table)) } mydie2() but I also want to try to create the same dice stimulation using ifelse() statement, but all hell break loose when I attempt to use the command below (the barplot only shows the results to be 6), can anyone tell me what went wrong please: mydie2<-function(n=1000,y=NULL,...){ for(i in 1:n){ x<-sample(1:6,n,replace=TRUE,prob=c(1,1,2,3,2,1)/10) x=runif(n) ifelse((x&...

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 propose...

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, n...

...ng 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 case). In the past I have used ifelse() but this tends to get fairly redundant and messy with a large amount of sub-groupings (z). I've created a sample dataset below. Can anyone recommend a better way of achieving what I am currently achieving with ifelse()? A long series of ifelse statements makes me think that there is something...

...ore cogent phrasing of a previous post. I'm trying to compute a z-score to rows in a large dataframe based on values in another dataframe. Here's the script (that does not work). 2 questons, 1) Anyone know of a more elegant way to calculate the "rounded" age value than the nested ifelse's I've used? 2) how to reference the lookup table based on computed indices? Thanks Jon # Define tables DSTzlook <- read.table("/Users/jween/Documents/ResearchProjects/ABC/data/DSTz.txt", header=TRUE, sep="\t", na.strings="NA", dec=".", strip.w...

...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. > >>> > >>> > >>> > >>>>test <- c(TRUE, TRUE, TRUE, FALSE, FALSE, FALSE, FALSE) ifelse(test, > >>>>0, 1:4) > >>> > >>>[1] 0 0 0 4 1 2 3 > >>> > > > > &...

...gamos que tenemos tres variables: V1 <- c (47, 71, 41, 23, 83, 152, 82, 8, 160, 18) V2a <- c (NA, 36, 15, 5, 56, 18, NA, 5, NA, 5) V2b <- c (37, NA, 15, NA, NA, NA, 90, NA, 161, NA) Supongamos que tengo la expresión (que no puedo asignarlo a ninguna variable): (ifelse (is.na (V2a) & ! is.na (V2b), V2b, V2a)) Supongamos que tengo que utilizar esta expresión dos o más veces y no puedo utilizar ni un xapply () ni un bucle. Por ejemplo: V3 <- (((ifelse (is.na (V2a) & ! is.na (V2b), V2b, V2a)) - V1)/V1)*100 V4 <- ifelse (! is.na ((((ifelse (is.na (V...