similar to: Change line colors based on data values in Lattice

How does one remove a column from a data frame when the name of the column to remove is stored in a variable? For Example: colname <- "LOT" newdf <- subset(olddf,select = - colname) The above statement will give an error, but thats what I'm trying to accomplish. If I had used: newdf <- subset(olddf,select = - LOT) then it would have worked, but as I said the column

As approximately 130 million other US citizens, I am dutifully preparing for the April deadline for filing my income tax returns. Reviewing the WINE AppDB for installable tax software to help out in the process, it appeared that 2nd Story Software's TAX ACT was most likely to work with the WINE version (1.1.17) installed on my Mandriva-equipped HP laptop. As it turned out, TAX ACT was able

Folks, I have a strange situation, which I may have isolated as a bug report. Or, it could just be that there's something about R that I don't know. :-) I have attached the data file and the program file but don't know whether these attachments will make it into the list. Here is my bugreport.R program -- ---------------------------------------------------------------------------

Hi all, I still have problems with the predict function by setting up the values on which I want to predict ie: original df: p1 (193 obs) variates y x1 x2 rm(list=ls()) x1<-rnorm(193) x2<-runif(193,-5,5) y<-rnorm(193)+x1+x2 p1<-as.data.frame(cbind(y,x1,x2)) p1 y x1 x2 1 -0.6056448 -0.1113607 -0.5859728 2 -4.2841793 -1.0432688 -3.3116807 ...... 192

Suppose plotx <- "someName" modx <- "otherName" plotxRange <- c(10,20) modxVals <- c(1,2,3) It often happens I want to create a dataframe or object with plotx or modx as the variable names. But can't understand syntax to do that. I can get this done in 2 steps, creating the data frame and then assigning names, as in newdf <- data.frame( c(1, 2, 3, 4),

Dear R-helpers, I am stuck on an error in R: When I run my code (below), I get this error back: Error in names(x) <- value : 'names' attribute must be the same length as the vector Then when I use traceback(), R gives me back this in return: `colnames<-`(`*tmp*`, value = c(""Item", "Color" ,"Number", "Size")) I'm not exactly

I have a data frame with 155,000 rows. One of the columns represents the user id (of which about 10,000 are unique). I am able to isolate 1000 of these user ids (stored in a list) that I want to eliminate from the data set, but I don't know of an efficient way to do this. Certainly this would be slow: newdf<-df for(i in listofbadusers) { newdf<-subset(tmp,uid!=i) } is there a better

Hi all, Very surprising (to me!) and mystifying result from predict.glm(): the predictions vary depending on whether or not I use ns() or splines::ns(). Reprex follows: library(splines) set.seed(12345) dat <- data.frame(claim = rbinom(1000, 1, 0.5)) mns <- c(3.4, 3.6) sds <- c(0.24, 0.35) dat$wind <- exp(rnorm(nrow(dat), mean = mns[dat$claim + 1], sd = sds[dat$claim + 1])) dat <-

Hello, Everybody: This may not be a "bug", but for me it is an unexpected outcome. A factor variable's levels do not retain their ordering after the levels function is used. I supply an example in which a factor with values "BC" "AD" (in that order) is unintentionally re-alphabetized by the levels function. To me, this is very bad behavior. Would you agree? #

Greetings all: OK, this is bugging the @#@%* out of me. I know the answer is simple and straightforward but for the life of me I cannot find it in the documentation, in the archives, or in my notes (because I know I've encountered this in the past). My problem is: I have a data frame with columns A, B, C, D, and E. A, B, and E are factors and C and D are numeric. I need a new data frame with

Dear All, I need some help arranging data that was imported. The imported data frame looks something like this (the actual file is huge, so this is example data) DF: IDKey X1 Y1 X2 Y2 X3 Y3 X4 Y4 Name1 21 15 25 10 Name2 15 18 35 24 27 45 Name3 17 21 30 22 15 40 32 55 I would like to create a new data frame with the following NewDF: IDKey X Y Name1 21 15 Name1

I am trying to plot a image where the x axis has the units of time. When I issue the image(x,y,z) command with x as a POSIXct object, it fails to put a time stamp on the x axis. Instead I get a warning "Incompatible methods" warning and no dates on my x axis. This example shows my problem: Rmat=t(matrix(data=rnorm(1:500),ncol=10,nrow=50)) tax=seq(ISOdate(2010,4,14,12,0,0),

OK, I need help!! I've been searching, but I don't understand the logic of some this dataframe addressing syntax. What is this type of code called? test [["v3"]] [is.na(test[["v2"]])] <-10 #choose column v3 where column v2 is == 4 and replace with 10 and where is it documented? The code below works for what I want to do (find the non-missing value in a row),

Problems with predict and lines in plotting binomial glm Dear R-helpers I have found quite a lot of tips on how to work with glm through this mailing list, but still have a problem that I can't solve. I have got a data set of which the x-variable is count data and the y-variable is proportional data, and I want to know what the relationship between the variables are. The data was

I'm pasting below a working R file featuring a function I'd like to polish up. I'm teaching regression this semester and every time I come to something that is very difficult to explain in class, I try to simplify it by writing an R function (eventually into my package "rockchalk"). Students have a difficult time with predict and newdata objects, so right now I'm

hi, if I have 20 x 3 data.frame, how to split it into 10 x 6 (moving the lower part of 10x3 to column) or 5 x 12 thanks -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. "Did you always know?" "No, I did not. But I believed..." ---Matrix III

I purchased H&R Block Tax cut Deluxe to do my 2004 taxes. I decided to give The Feb 2005 version of Wine under Fedora 1 a try before rebooting my machine to Windows and actually starting to work on my taxes. Wine seemed to install Tax Cut with out any problems. However, when I tried to start Tax Cut 2004 , it complained that Mozilla did not have an active X Browser and offered to go on

Hi, I was recently misfortunate enough to have to use regular expressions to sort out some data in R. I'm working on a data file which contains taxonomical data of bacteria in hierarchical order. A sample of this file can be generated using: tax.data <- read.table(header=F, con <- textConnection(' G9SS7BA01D15EC Bacteria(100) Cyanobacteria(84) unclassified G9SS7BA01C9UIR

Hi, I have two lists (c and h - see below) containing matrices with similar cases but different values. I want to split these matrices into multiple matrices based on the values in h. So, I did the following: years<-c(1997:1999) for (t in 1:length(years)) { year=as.character(years[t]) h[[year]]<-sapply(colnames(h[[year]]), function(var)

There must be a better way to select the rows after 22-Apr-2004 and before 01-Sep-2004 with a temperature below 65 than this: > before2sw1 <- subset(energy.data, as.Date(start, format="%d-%b-%y") < as.Date("01-Sep-04", format = "%d-%b-%y")) > before2sw2 <- subset(before2sw1, as.Date(start, format="%d-%b-%y") >=