similar to: lines.formula() problem when data argument is missing (PR#13296)

Dear all, I noticed the following bug in points.formula > library(DAAG) > data(roller) > fm <- lm(depression ~ weight, data=roller) > plot( depression ~ weight, data=roller, type="n") > abline(fm) > attach(roller) > points( depression~weight, subset=1:7) > points( depression~weight, subset=8:10, col="blue") Error in if (length(x) == l) x[s] else x :

Hi all, Can anyone explain why the following use of the subset() function produces a different outcome than the use of the "[" extractor? The subset() function as used in density(subset(mydf, ht >= 150.0 & wt <= 150.0, select = c(age))) appears to me from documentation to be equivalent to density(mydf[mydf$ht >= 150.0 & mydf$wt <= 150.0, "age"])

Here's a simple example: > x <- rnorm(20) > y <- 1 + 10*x + rnorm(20) > plot(y ~ x, subset=2*(1:10),xlab="In",ylab="Out") Error in length(x) == l : comparison (1) is possible only for vector types The problem seems to be in this section of the code for plot.formula: if (!missing(subset)) { s <- eval(m$subset, data, parent.frame()) l

Dear list, When I try to use lines.formula with subset and another argument I get an error. e.g. x<-1:5 y<-c(1,3,NA,2,5) plot(y~x, type="n") # set up frame lines(y~x, subset=!is.na(y)) # works OK lines(y~x, type="o", col="blue") # works OK # but lines(y~x, subset=!is.na(y), col="red") # gives an error: Error in if (length(x) == l) x[s] else x :

I could use some help understanding how nls parses the formula argument to a model.frame and estimates the model. I am trying to utilize the functionality of the nls formula argument to modify garchFit() to handle other variables in the mean equation besides just an arma(u,v) specification. My nonlinear model is y<-nls(t~a*sin(w*2*pi/365*id+p)+b*id+int,data=t1,

I could use some help understanding how nls parses the formula argument to a model.frame and estimates the model. I am trying to utilize the functionality of the nls formula argument to modify garchFit() to handle other variables in the mean equation besides just an arma(u,v) specification. My nonlinear model is y<-nls(t~a*sin(w*2*pi/365*id+p)+b*id+int,data=t1,

I''ve having issues with some part of Mongrel start up process stalling with versions of Mongrel greater than 1.1 Using Mongrel Gem version 1.1 both ./script/server and ./merb boot fine But from gem version 1.1.1 both processes stall immediately I can get it to continue by attempting to force quit with CTRL -C system info Mac OS X 10.5.5 Ruby 1.8.7 (2008-05-31 patchlevel 0)

I could use some help understanding how nls parses the formula argument to a model.frame and estimates the model. I am trying to utilize the functionality of the nls formula argument to modify garchFit() to handle other variables in the mean equation besides just an arma(u,v) specification. My nonlinear model is y<-nls(t~a*sin(w*2*pi/365*id+p)+b*id+int,data=t1,

Hi All, I'm fiddling with an program to read a text file containing periods that SAS uses for missing values. I know that if I had the original SAS data set instead of a text file, R would handle this conversion for me. Data frames do not allow missing values in their indices but vectors do. Why is that? A search of the error message points out the problem and solution but not why they

Hi folks, I''m new to Rails. I picked up Simply Rails 2 by Patrick Lenz a few weeks ago and I finally sat down to get stuck into it. I have Rails installed on my iMac (OS X Leopard) and I can load up the Rails Welcome screen at localhost:3000 without any issues. My problem is with running IRB within Terminal. If I try to run: irb> 1 I receive the error: dyld: NSLinkModule() error

Sorry I had typo in previous email, this typo corrected version: Dear R experts I have simple question, please execuse me: #example data, the real data consists of 20000 pairs of variables K1 <- c(1,2,1, 1, 1,1); K2 <- c(1, 1,2,2, 1,2); K3 <- c(3, 1, 3, 3, 1, 3) M1a <- rep( K1, 100); M1b <- rep(K2, 100) M2a <- rep(K1, 100); M2b <- rep(K1, 100) M3a <- rep(K1, 100); M3b

On converting character variables to ordered factors, regression result has strange names. Is it possible to obtain same variable names with and without intercept? Thanks, Naresh mydf <- data.frame(date = seq.Date(as.Date("2024-01-01"), as.Date("2024-03-31"), by = 1)) mydf[, "wday"] <- weekdays(mydf$date, abbreviate = TRUE) mydf.work <- subset(mydf, !(wday

Hello! I have a data frame with dates. I need to create a new "month" that starts on the 20th of each month - because I'll need to aggregate my data later by that "shifted" month. I wrote the code below and it works. However, I was wondering if there is some ready-made function in some package - that makes it easier/more elegant? Thanks a lot! # Example data:

I'm writing a function and keep getting the following error message. myfunc <- function(lst) { lst <- list(roots = c("car insurance", "auto insurance"), roots2 = c("insurance"), prefix = c("cheap", "budget"), prefix2 = c("low cost"), suffix = c("quote", "quotes"), suffix2 = c("rate",

Hi All, I've been successfully using the with function for analyses and the transform function for multiple transformations. Then I thought, why not use "with" for both? I ran into problems & couldn't figure them out from help files or books. So I created a simplified version of what I'm doing: rm( list=ls() ) x1<-c(1,3,3) x2<-c(3,2,1) x3<-c(2,5,2)

Dear R experts, I am trying to arrange multiple plots, creating one graph for each size1 factor variable in my data frame, and each plot has the median price on the y-axis and the size2 on the x-axis grouped by clarity: library(ggplot2) df <- data.frame(price=matrix(sample(1:1000, 100, replace = TRUE), ncol = 1)) df$size1 = 1:nrow(df) df$size1 = cut(df$size1, breaks=11)

Hello, Try the following. icol <- which(grepl("flag", names(mydf))) mydf[icol] <- lapply(mydf[icol], function(x){ is.na(x) <- x == 0 x }) mydf # A A_flag B B_flag #1 8 10 5 12 #2 7 NA 6 9 #3 10 1 2 NA #4 1 NA 1 5 #5 5 2 0 NA Hope this helps, Rui Barradas On 11/22/2017 10:34 AM, Massimo Bressan

##data y1 <- matrix(c(3,1,0,1,0,1,1,0,0,0,1,0,0,0,1,1,0,1,1,1), nrow = 5, byrow = TRUE) y2 <- matrix(c (3,0,10,3,3,0,0,1,1,0,0,0,0,0,1,0,1,0,0,2,1,0,1,1,0,2,1,1,4,1), nrow = 5, byrow = TRUE) y1 <- as.data.frame(y1) y2 <- as.data.frame(y2) rownames(y1) <- rownames(y2) <- paste("site", 1:5, sep = "") colnames(y1) <-

...well, I don't think this is exactly the expected result (see my post) to be noted that the columns affected should be "A" and "B" thanks for the help max ----- Messaggio originale ----- Da: "Rui Barradas" <ruipbarradas at sapo.pt> A: "Massimo Bressan" <massimo.bressan at arpa.veneto.it>, "r-help" <r-help at

I'm working with some data, and am trying to generate it in the following format. state city zipcode I like pizza 0 0 0 I live in Denver 0 1 0 All the fun stuff is in Alaska 1 0 0 he lives in 66062