similar to: tapply on a matrix?

Dear R-helpers, I have found a slightly annoying problem when trying to plot lines on graphs. I first created my data using tapply, thus:- y1=as.vector(fit1$coef$random$id) x1=tapply(o1,id,median,na.rm=T) x2=tapply(o2,id,median,na.rm=T) #then I plot the data, thus:- plot(x1[x2==0],y[x2==0]) #if I now try to fit the linear regression, R 'hangs up'

I am trying to analyse some data and was given R code to do this with but there seem to be errors in the code. My level of knowledge is improving but still limited. The details are; Data on clover lines; Lines.txt attached. Comma seperations Code: options(digits=3) clover <- read.table("Lines.txt",header=T,sep=",") vnames <- names(clover);nv <- length(vnames)

Hi all, SETUP: I have pairwise data on 22 chromosomes. Data matrix X for a given chromosome looks like this: 1 13 58 1.12 6 142 56 1.11 18 307 64 3.13 22 320 58 0.72 Where column 1 is person ID 1, column 2 is person ID 2, column 3 can be ignored, and column 4 is how much chromosomal sharing those two individuals have in some small portion of the chromosome. There are 9000 individual people, and

Hola! > rm(list=ls(all=TRUE)) > x <- 1:20 > y <- 1+x+rnorm(x) > xy.coords(y ~ x,NULL) ... expected output, correct, but when called from inside lowess: > lowess(y ~ x) Error in xy.coords(x, y) : x and y lengths differ > debug(xy.coords) > lowess(y ~ x) debugging in: xy.coords(x, y) ... long listing deleted if (is.language(x)) { if (inherits(x,

Dear all, I searched the mail archives and the R site and found no guidance (tried "merge", "cbind" and terms like "coalesce" with no success). There surely is a way to coalesce (like in SQL) columns in a dataframe, right? For example, I would like to go from a dataframe with two columns to one with only one as follows: From Name.x Name.y nx1 ny1 nx2 NA

On 'aggregate data.frame', the URL should be https://stat.ethz.ch/pipermail/r-help/2016-May/438631.html . vector(typeof(ans)) (or vector(storage.mode(ans))) has length zero and can be used to initialize array. Instead of if(missing(default)) , if(identical(default, NA)) could be used. The documentation could then say, for example: "If default = NA (the default), NA of appropriate

I'm studying alone the R language for data preparation. I found a course at MIT for data preparation that uses python but I'm using R to learning. The first exercise is the preparation of data from a database that shows the contributions made to candidates for U.S. president. The database is described in FORMART

Here's something related to last week's apply() problem: R> x <- matrix(1:20, nc = 4) R> x [,1] [,2] [,3] [,4] [1,] 1 6 11 16 [2,] 2 7 12 17 [3,] 3 8 13 18 [4,] 4 9 14 19 [5,] 5 10 15 20 R> tapply(x, row(x), table) [1] Numeric,4 Numeric,4 Numeric,4 Numeric,4 Numeric,4 ??? In S, > tapply(x, row(x), table)

Dear all, I'm wanting to do a series of comparisons among 4 categorical variables: a <- aggregate(y, list(var1, var2, var3, var4), sum) This gets me a very nice 2-dimensional data frame with one column per variable, BUT, as help for aggregate says, <<empty subsets are removed>>. I don't see in help(aggregate) how I can change this. In contrast, a <- tapply(y,

Does anyone have something like tapply that is extremely fast for matrices when there is a very large number of levels of the grouping variable? I'm referring to, for example, tapply(x, grouping.variable, function.operating.on.submatrix) where x is a matrix and the submatrix is a subset of the rows of x. The grouping variable's length equals the number of rows of x. -- Frank E

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"). .....................................................

Dear R-developers, when tapply() is invoked on factors that have empty levels, it returns NA. This behaviour is in accord with the tapply documentation, and is reasonable in many cases. However, when FUN is sum, it would also seem reasonable to return 0 instead of NA, because "the sum of an empty set is zero, by definition." I'd like to raise a discussion of the possibility of an

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:

Hi, I'm writing for a little help. I have a dataframe with same NA value and I'd like to obtain the means of the value of a coloumn grouped by the levels of a factor coloumn of the datframe. I'm using the function "tapply" but I see that if only a NA value is present the result is NA. There is an option to have the correct result or I must use an other function? Thanks of

Hei, i am trying to plot the means of two variables (d13C and d15N), by 2 grouping factors (Species and Year) that i obtained by the function tapply. I would like to plot with different colours according to the Year and show the "Species" as data labels. My data looks like this: Species d13C d13N Year "Species1" 14,4 11.5 2009 "Species2"

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 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)

Hi all, the following problem is still beyond my R-knowledge: I have one data vector containing the signal from 4 channels that are measured subsequently and in repeating cycles (with one factor vector for cycle and one for channel identification). To extract the mean of each channel during each cycle tapply is the method of choice. However, I cannot use the whole measuring period for each

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){

The "no factor combination" case is distinguishable by 'tapply' with simplify=FALSE. > D2 <- data.frame(n = gl(3,4), L = gl(6,2, labels=LETTERS[1:6]), N=3) > D2 <- D2[-c(1,5), ] > DN <- D2; DN[1,"N"] <- NA > with(DN, tapply(N, list(n,L), FUN=sum, simplify=FALSE)) A B C D E F 1 NA 6 NULL NULL NULL NULL 2 NULL NULL 3 6