similar to: More on tapply and factors

Dear List, I'd appreciate you guidance for obtaining the desired result shown below, by combining tapply(x, g, mean) and g in the example. Basically, I'm trying to create a vector whose values are based on the result from tapply(x, g, mean) but that follow the pattern and length given by the factor g. Of course I'm looking for a generic solution (i.e, not something that just work

For block effects with small variance, lmer will sometimes estimate the variance as being very close to zero and issue a warning. I don't have a problem with this -- I've explored things a bit with some simulations (see below) and conclude that this is probably inevitable when trying to incorporate random effects with not very much data (the means and medians of estimates are plausibly

Hello, I use the function 'aggregate' a lot. One small annoyance is that it is necessary to name the factors in the 'by' list to get the names in the resulting data.frame (else, they appear as Group.1, Group.2...etc). For example, I am forced to write: aggregate(y,list(f1=f1,f2=f2),mean) instead of aggregate(y,list(f1,f2),mean) (for two factors with short names, it is not such

Dear List, Shouldn't result1 and result2 be equal in the following case? Note that log$RequestID is a factor. That is, is.factor(log$RequestID) yields TRUE. result1 <- tapply(log$Flag,factor(log$RequestID),sum) result2 <- tapply(log$Flag,log$RequestID,sum) Yet, when I summarize the output, I get the following: summary(result1) Min. 1st Qu. Median Mean 3rd Qu.

Greetings, I am attempting to run a function, which produces a vector and requires two input variables, across two nested factor levels. I can do this using by(X, list(factor1, factor2), function), however I haven't found a simple way to extract the list output into an organized vector form. I can do this using nested loops but it isn't exactly an optimal approach. Thank you

I have two nested data frames: a<-rnorm(6) b<-rnorm(9) f1<-c("x1","x2","x3")) f2<-c("y1","y2") id<-c(1:6) a_df<-data.frame(cbind(id,f1,"y1",a)) id<-c(1:9) b_df<-data.frame(cbind(id,f1,"y2",b)) I want to preserve id and f1, but want to collapse f2 and take the corresponding mean values of a and b.

Hello, I'd like to produce a ggplot where the order of factors within facets is based on the average of another variable. Here's a reproducible example. My problem is that the factors are ordered similarly in both facets. I would like to have, within each facet of `f1', boxplots for 'x' within each factor `f2', where the boxplots are ordered based on the average of x

Hi, (Please use ?dput() to share the example dataset. Avoid using images to show dataset. Also, please read the posting guide esp. regarding home work, assignments etc.) res <- sapply(Gene[,-1],function(x) tapply(x,list(Gene$Genotype),mean)) #or res2 <-? aggregate(.~Genotype, data=Gene,mean) #or library(plyr) ?res3 <- ddply(Gene,.(Genotype),numcolwise(mean)) identical(res2,res3)

Is there a conventional way to test for nested factors? I.e., if 'a' and 'b' are lists of same-length factors, does each level specified by 'a' correspond to exactly one level specified by 'b'? The function below seems to suffice, but I'd be happy to know of a more succinct solution, if it already exists. Thanks, Tim. --- "%nested.in%" <-

Dear R-helpers, Please have a look at the following. f1 is the same as f2 except that it has some values replaced by NA. But it's corresponding file is slightly bigger than the file containing f2. Could someone please tell me if this is an anomaly ? > load("file1") > ls() [1] "f1" > load("file2") > ls() [1] "f1" "f2" > >

The issue is not with boxplot, but with split. boxplot.formula() calls boxplot(split(split(mf[[response]], mf[-response]), ...), but look at what split() returns when there are empty levels in the factor: > f <- factor(gl(3, 6), levels=1:5) > y <- rnorm(f) > split(y, f) $"1" [1] 0.4832124 1.1924811 0.3657797 1.7400198 0.5577356 0.9889520 $"2" [1] -1.1296642

Warm greetings to you all. Using the tapply function below: data<-read.table("FD1month",col.names = c("Dates","count")) x=data$count f<-factor(data$Dates) AB<- tapply(x,f,mean) I made a simple calculation. The result, stored in AB, is of the form below. But an effort to write AB to a file as a data frame fails. When I use the write table, it only produces

?s 01:43 de 29/03/2024, Ogbos Okike escreveu: > Dear Rui, > Thanks again for resolving this. I have already started using the version > that works for me. > > But to clarify the second part, please let me paste the what I did and the > error message: > >> set.seed(2024) >> data <- data.frame( > + Date = sample(seq(Sys.Date() - 5, Sys.Date(), by = "1

Dear Rui, Thanks again for resolving this. I have already started using the version that works for me. But to clarify the second part, please let me paste the what I did and the error message: > set.seed(2024) > data <- data.frame( + Date = sample(seq(Sys.Date() - 5, Sys.Date(), by = "1 days"), 100L, + TRUE), + count = sample(10L, 100L, TRUE) + ) > > # coerce

Norm Josephy <NJOSEPHY@bentley.edu> writes: > Sir: > > I hope you do not mind me sending e-mail directly to you > rather than to the developer's list. I wish to avoid > unnecessary traffic on the list when the issue I raise may > have already been fixed. As a matter of principle: I *do* mind. One of the purposes of having a list is that not everyone has the time

Norm Josephy <NJOSEPHY@bentley.edu> writes: > Sir: > > I hope you do not mind me sending e-mail directly to you > rather than to the developer's list. I wish to avoid > unnecessary traffic on the list when the issue I raise may > have already been fixed. As a matter of principle: I *do* mind. One of the purposes of having a list is that not everyone has the time

Hello all, I'm new to R (and the S language in general) so go easy on me if this is really simple. Given a data.frame df which looks like this: f1 f2 f3 f4 c1 c2 1 y y a b 10 20 2 n y b a 20 20 3 n n b b 8 10 4 y n a a 30 5 I'd like to aggregate it by the factors f1 and f2 (or f2 and f3, or any other combination of the three) and compute the sum of c1 and c2 (as separate values). I

Full_Name: George Leigh Version: 1.8.1 OS: Windows 2000 Submission from: (NULL) (203.25.1.208) The following example gives the correct answer when the first argument of tapply is a numeric vector, but an incorrect answer when it is a factor. If the function used by tapply is "length", the type and contents of the first argument should make no difference, provided it has the same

Hi, all. The help file for tapply says that if simplify is true, and the result of the calculation is always a scalar, then tapply will return a vector. Nonetheless: > t1 <- tapply(runif(10), rep(1:5, 2), mean) > is.vector(t1) [1] FALSE > is.array(t1) [1] TRUE > I have found this in version 0.65.1 on an SGI running Irix 6.5, and on a Mac running Linux-PPC. I've also

After some fiddling around with the tapply command, I discovered that the factors (the INDEX argument) given to tapply must be specified in fastest-cycling first order. The following code shows how I discovered my error: (R version 1.2.1) -o-o-o-o-o- x <- as.data.frame(list(data=c(-9,0,3,1,-9,1,0,-9,0,3,1,-9,1,0), subj=c(rep(1,7),rep(2,7)),