similar to: ?mean

I found myself wanting to average a vector [vec] within each level of a factor [Fac], returning a vector of the same length as vec. After a while I realised that lm1 <- lm(vec ~ Fac) fitted(lm1) did what I want. But there must be another way to do this, and it would be good to be able to apply other functions than mean() in this way. Cheers, Murray -- Dr Murray Jorgensen

This inconsistency recently came to my attention: > df <- data.frame(A = 1:10, B = rnorm(10)) > min(df) [1] -1.768958 > max(df) [1] 10 > mean(df) [1] NA Warning message: In mean.default(df) : argument is not numeric or logical: returning NA I recall the times where `mean(df)` would give `colMeans(df)` and this behaviour was deemed inconsistent. It seems though that the change has

Dear R users, The Sweave code below runs fine, as it is. However, an error occurs when the line 'library(xtable)' is uncommented: Error: chunk 1 Error in "label<-"(`*tmp*`, value = "month") : no applicable method for "label<-" Is anybody aware of this and knows a workaround? Thanks, Sander. *******************

R version 2.10.1 (2009-12-14) Copyright (C) 2009 The R Foundation for Statistical Computing ISBN 3-900051-07-0 R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type 'license()' or 'licence()' for distribution details. Natural language support but running in an English locale R is a collaborative project with

Hello list. I feel like an idiot. There exists a method called expand.grid which, from the documentation, appears to do just what I want, but then it doesn''t, and I can''t get it to behave. Given a dataframe dfr<-data.frame(c1=c("a", "b", NA, "a", "a"), c2=c("d", NA, "d", "e", "e"),

Hi R-users, I have a simple question for R heavy users. If I have a data frame like this dfr <- data.frame(id=1:16, categ=rep(LETTERS[1:4], 4), var3=c(8,7,6,6,5,4,5,4,3,4,3,2,3,2,1,1)) dfr <- dfr[order(dfr$categ),] and I want to score values or points in variable named "var3" following this kind of logic: 1. the highest value of var3 within category (variable named

I am using xYplot to plot data and CIs. How do I add several lines to the figure? _____________________________ Professor Michael Kubovy University of Virginia Department of Psychology USPS: P.O.Box 400400 Charlottesville, VA 22904-4400 Parcels: Room 102 Gilmer Hall McCormick Road Charlottesville, VA 22903 Office: B011 +1-434-982-4729 Lab: B019

Hi, I am facing the problem that I want to plot a histogram chart set freq to true and overlay with normal or weibull or exponential distribution curve. The sample code is shown as below: >samp<-c(-8.2262,-8.2262,-8.2262,-8.20209,-8.09294,-8.07321,-8.07321, -8.07321,-8.07175,-8.04948,-8.04948,-8.04948,-8.03848,-8.03848,

Dear R users, I have a dataframe with two columns: first column is date data (e.g. 1/1/2000 with character format: daily data from 1/1/1970 till 31/12/2003) and second column is temperature value. Now I'd like to calculate mean for each month in a year (i.e. May 2001, June 1997) and mean for each month in all of years. As the number of days in some months is different from others I could

I would like to sum/mean/min a list of lists and numbers to return the related lists. -1+2*c(1,1,0)+2+c(-1,10,-1) returns c(2,13,0) but sum(1,2*c(1,1,0),2,c(-1,10,-1)) returns 15 not a list. Using the suggestions of Gabor Grothendieck, Reduce('+',list(-1,2*c(1,1,0),2,c(-1,10,-1))) returns what we want, c(2,13,0). However, it seems that this way does not work to mean/min. So, how to

Given a long zoo matrix, the goal is to "sweep" out a statistic from the entire length of the sequences. longzoomatrix<-zoo(matrix(rnorm(720),ncol=6),as.yearmon(outer(1900,seq(0,length=120)/12,"+"))) cnames<-c(12345,23456,34567,45678,56789,67890) colnames(longzoomatrix)<-cnames longzoomatrix[1:24,] 12345 23456 34567 45678

Hi Gabor, In strictly reading the help files for both nrow() and row(), the 'x' argument in the former case is "a vector, array, data frame, or NULL.", whereas in the latter case it is "a matrix-like object, that is one with a two-dimensional dim.". Thus, I would expect row() to fail on a >= 3-dimensional array, as your example shows. In reading the help file for

Dear all, Sorry to bother you with one more newbie question. I have a dataobject with several hundreds of columns. I want to remove columns with a mean of the column values below a certain value: > a<- c(1,2,3,4,5,6) > b<-c(2,4,6,8,10,12) > c<- c(3,6,9,12,15,18) > test<- as.matrix(cbind(a, b, c)) > mean(a) [1] 3.5 > mean(b) [1] 7 > mean(c) [1] 10.5 Say the

This is my 2007 New Year wishlist for R features: 1. Matrix Multiplication Enhance matrix multiplication to work with multidimensional arrays such that the last dimension of the first multiplicand must equal the first dimension of the second. See: https://www.stat.math.ethz.ch/pipermail/r-devel/2006-July/038497.html 2. Grid - logical-valued function as first arg of grid.edit -

I would be incredibly grateful to anyone who'll help me translate some SAS code into R code. Say for example that I have a dataset named "dat1" that includes five variables: wshed, site, species, bda, and sla. I can calculate with the following SAS code the mean, CV, se, and number of observations of "bda" and "sla" for each combination of

The last two lines of example(delayedAssign) give this: > e <- (function(x, y = 1, z) environment())(1+2, "y", {cat(" HO! "); pi+2}) > (le <- as.list(e)) # evaluates the promises $x <promise: 0x032b31f8> $y <promise: 0x032b3230> $z <promise: 0x032b3268> which contrary to the comment appears unevaluated. Is the comment wrong or is it supposed to

The first component name has backticks around it and the second does not. Though not wrong, it seems inconsistent. list(a = 1, b = 2) ## $`a` ## [1] 1 ## ## $b ## [1] 2 R.version.string ## [1] "R version 3.5.1 Patched (2018-07-02 r74950)" -- Statistics & Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com

The fact that it is consistent with the documentation is not the point. The point is that the design itself is inconsistent. On Sun, Sep 8, 2024 at 8:27?AM Marc Schwartz <marc_schwartz at me.com> wrote: > > Hi Gabor, > > In strictly reading the help files for both nrow() and row(), the 'x' argument in the former case is "a vector, array, data frame, or

In ?exists it says: inherits: should the enclosing frames of the environment be searched? I believe what it is saying is that if inherits is TRUE and it fails to find the variable it will look in the parent environment and the parent of the parent, etc. (as opposed to looking in the calling frame next and the caller of the caller, etc.) Now I thought that standard terminology in R was: 1.

Consider this: > y <- matrix(1:8, ncol=2) > is.matrix(y[-c(1,2),]) [1] TRUE > is.matrix(y[-c(1,2,3),]) [1] FALSE > is.matrix(y[-c(1,2,3,4),]) [1] TRUE It seems like an inconsistent behaviour: - with 2 or more rows we have a matrix - with 1 row we do not have a matrix and - with 0 rows we have a matrix again I just stumbled on this behaviour, because I had a problem with my