search for: ifelses

Hi, I tried to do a nested Anova with the attached Data. My response variable is "survivors" and I would like to know the effect of (insect-egg clutch) "size", "position" (of clutch on twig) and "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

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

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",

I would like to know whether there is a faster way to do the below operation (updating vec1). My objective is to update the elements of a vector (vec1), where a particular element i is dependent on the previous one. I need to do this on vectors that 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

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 -- View this message in context:

# 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",

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

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)

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 <-

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"),print("goodbye")) [1] "goodbye" [1]

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

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

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 people, I want to make a biased dice using the sample() function and print out the results after n number of runs, I've successfully generated the dice using the following command: 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) if(x<=1/10){y[i]=1} else if(x<=2/10){y[i]=2} else if(x<=4/10){y[i]=3} else

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

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

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

Hi This is (hopefully) a bit more 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

> 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. > >>> > >>> >

Hola a todos: Se me ha planteado un problema que no está ligado a ningún problema concreto. Es más teórico. Supongamos 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