similar to: plot data from tapply

Dear list, I have a data.frame with x, y values and a 3-level factor "group", say. I want to create a new column in this data.frame with the values of y scaled to 1 by group. Perhaps the example below describes it best: > x <- seq(0, 10, len=100) > my.df <- data.frame(x = rep(x, 3), y=c(3*sin(x), 2*cos(x), > cos(2*x)), # note how the y values have a different

Hi folk, I tried this and it works just perfectly tapply(iris[,1],iris[5],mean) but, how to obtain a single table from multiple variables? In tapply x is an atomic object so this code doesn't work tapply(iris[,1:4],iris[5],mean) Thanx and great summer holidays Gianandrea -- View this message in context: http://www.nabble.com/multiple-tapply-tp18868063p18868063.html Sent from the R help

I need to compute the mean and the standard deviation of a data set and would like to have the results in one table/data frame. I call tapply() two times and do then merge the resulting tables to have them all in one table. Is there any way to tell tapply() to use the functions mean and sd within one function call? Something like tapply(data$response, list(data$targets, data$conditions), c(mean,

dear R experts: has someone written a function that returns the results of by() as a data frame? ??of course, this can work only if the output of the function that is an argument to by() is a numerical vector. presumably, what is now names(byobject) would become a column in the data frame, and the by object's list elements would become columns. it's a little bit like flattening the by()

Hi everyone suppose I have a <- array(1:256,rep(4,4)) and want to access a[1,2,3,1] by the vector c(1,2,3,1). As per yesterday, I can use do.call(): a[1,2,3,1] == do.call("[",c(list(a),c(1,2,3,1))) Now how do I apply the above technique (or indeed any other technique!) to get a[1,2,3,] [1] 37 101 165 229 from a vector like c(1,2,3,0) or c(1,2,3,NULL) or c(1,2,3,NA)?

Hi R help, I'm a fairly experienced R user but this manipulation has me stumped, please help: DATA id<-rep(LETTERS[1:5],20) distance<-rnorm(1:100, mean = 100) bearing<-sample(1:360,100,replace=T) month<-sample(1:12,100,replace=T) I have a dataset with records of individuals (id) , each with a distance (distance) & direction (bearing) recorded for each month (month). I want

Hello! I have encountered a really weird problem. Maybe you've encountered it before? I have a large data frame "importances". It has one factor ($A) with 3 levels: 3, 9, and 15. $B is a regular numeric variable. Below I am picking a really small sub-frame (just 3 rows) based on "indices". "indices" were chosen so that all 3 levels of A are present:

I'm sure I can put this together from the various 'apply's and split, but I wonder if anyone has a quick incantation: E.g. I can do tapply( data, groups, mean) but how can I do something like: tapply( list(data,weights), groups, weighted.mean ) ? (or: mapply is to sapply as ? is to tapply ) Thanks for your help. -- View this message in context:

Hello, I'm trying to use tapply to find group means in a function. It works outside of a function, but I get the error message from the following code: "Error in tapply(index, cluster, mean) : arguments must have same length." Any suggestions? Thanks. eric d <- data.frame(cbind(cluster=1:2, value1=1:10, value2=11:20)) d FindClusterTraits <- function(framename, index){

Hello, I am using tapply to pull out data by the day of week and then perform functions (e.g. mean). I would like to have the number of values used for the calcuation for the functions, sorted by each day of week. A number of entries in any given column are NAs. I have tried the following code and simple variants with no luck. for (i in 1:length(a[1,])){ x<-tapply(a[,i],a[,1],mean,

I'm learning how to use tapply. Now I'm having a go at the following code in which dati contains almost 600 lines, Pot - numeric - are the capacities of power plants and SGruppo - text - the corresponding six technologies ("CCC", "CIC","TGC", "CSC","CPC", "TE"). .....................................................

I have the following data.frame: index <- c("a","a","b","b","b") alpha <- c(1,2,3,4,5) beta <- c(2,3,4,5,6) table <-data.frame(index,alpha,beta) I'm now interested in getting means of alpha and beta for each of the index values and do a tapply() for each of the columns, e.g. means.alpha <- tapply(table$alpha, index,mean)

Hello, I want to conduct normality test to a series of data and get the p-value for each subset. I am using the following codes, but it does not work. tapply(re, list(reg, ast), pvalue(shapiro.test)) Could anyone give me some advice? Many thanks. -- View this message in context: http://www.nabble.com/tapply-tf3851631.html#a10910748 Sent from the R help mailing list archive at Nabble.com.

Hi, How do I get labels onto the output from tapply? I need the labels to merge the output with another data frame. Thanks. eric d <- data.frame(cbind(x=1:3, y=1:10)) a <- with(d, tapply(y, x, sum)) [[alternative HTML version deleted]]

Is the following behaviour of tapply not disappointing? Problem with tapply occurs when dealing with na.rm when an argument additional to na.rm is sent to the applied function (here quantile). Any comment? Thank you, Philippe Lambert > x <- c(12,10,12,2,4,11,3,7,2,1,18,7,NA,NA,7,5) > fac <- gl(4,4,16) > # Works fine > tapply(x,fac,quantile,na.rm=T) $"1" 0% 25%

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.

Hi everyone, Is it possible to use tapply(x,y,mean) if not all groups of x by y are of the same length (for example if you have one missing observation)? I tried tapply(x,y,mean,na.omit=T) but it doesn't work! Steffi -- --------------------------------- Stefanie von Felten Doktorandin ETH Z?rich Institut f?r Pflanzenwissenschaften ETH Zentrum, LFW A 2 Telefon: 044 632 85 97 Telefax: 044

I'm interested in a version of tapply() that operates with shingles instead of factors. For instance: x <- c(1,1,2,2,3,3) y <- c(1,1,1,0,0,0) s <- shingle(x,intervals=cbind(c(0.5,1.5),c(2.5,3.5))) # the following function should exist! tapply.shingle(x,s,mean) # returns the vector c(0.75,0.25) I've written such a function as follows: tapply.shingle <-

The following message is provided by Erik Please provide the reproducible code to do this. Generate a sample data set using the random data generating functions and show us what you'd like, we can then more easily help. ctu at bigred.unl.edu wrote: > Hi, > How about using "subset"? > x1<-tapply(subset(years, length(area)>20), function(x) length(unique(x))) >

Dear List, why does this not work? df <- data.frame(var1 = c(3,2,1), var2 = c(6,5,4), var3 = c(9,8,7), fac = c('A', 'A', 'B')) tapply(cbind(df$var1, df$var2, df$var3), df$fac, mean) Thank you, Stefan