search for: 100,0,1

R-list, 1. Given a formula (f) w variables referencing some data set (dat), is there any easier/faster way than this to get the names (in character form) of the variables on the RHS of '~' ? dat <- data.frame(x1 = x1 <- rnorm(100,0,1), x2 = x2 <- rnorm(100,0,1), y = x1 + x2 + rnorm(100,0,1)) f <- y ~ x1 + x2 mf <- model.frame(f, data=dat) mt <- attr(mf, "terms") predvarnames <- attr(mt, "term.labels") > predvarnames [1] "x1" "x2" ----- 2. Also, is there a...

...) and npudensbw(). This confusion will only be generated in >=3 dimensions. I marked the four places with confusion1-4. I think there should be some kind of correspondence in those four places,but cannot figure them out.Thanks very much for chewing on this. #simulated dataset: d x1<-c(runif(100,0,1),runif(50,0.67,1));y1<-c(runif(100,0,1),runif(50,0.67,1));d1<-data.frame(x1,y1);colnames(d1)<-c("x","y") x2<-c(runif(100,0,1),runif(50,0.33,0.67));y2<-c(runif(100,0,1),runif(50,0.33,0.67));d2<-data.frame(x2,y2);colnames(d2)<-c("x","y"...

...I got this error message whenever it chose a model with an interaction term X_p :X_q and if I put X_q before X_p in the upper scope formula, like X_q*X_p***. Shouldn't X_p:X_q be the same as X_q:X_p? So why am I getting this error message? Any help is greatly appreciated. > x1<-rnorm(100,0,1) > x2<-rnorm(100,0,1) > x3<-rnorm(100,0,1) > y<-x1+x2+2*x3+2*x1*x3+rnorm(100,0,1) > datasim<-data.frame(cbind(y,x1,x2,x3)) > steplm<-step(lm(y~1,data=datasim),scope=list(upper=~x1*x2*x3,lower=~1),direction="forward",trace=1) Start: AIC= 274.81 y ~ 1...

Dear R users, Can anyone please tell me how to generate a large number of samples in R, given certain distribution and size. For example, if I want to generate 1000 samples of size n=100, with a N(0,1) distribution, how should I proceed? (Since I dont want to do "rnorm(100,0,1)" in R for 1000 times) Thanks for help Debbie _________________________________________________________________ Looking to change your car this year? Find car news, revi...

Hello, I am a new member, and I need your help. For my work (thesis), I use the package rpart to construct trees. But, to continu my studies, I need to calcule the 'variable importance'. But, I don't find a program to do this in the implementation of R. Is there someone who know if there exist a program which calculate the 'variable importance', notion defined by Breiman and

.... Some points (or parts of the line) are not shown on the graph because they lie beyond the scale of the axis. Is there a way to overcome this so all points (or the entire line) are shown on the graph? Here's an example of my problem: colors = c("red", "blue") plot(x=rnorm(100,0,1), y=runif(100), type="p", pch=21, col = colors[1]) # if I add these points not all of them are shown on the graph lines(x=rnorm(100,3,1), y=runif(100), type="p", pch=24, col = colors[2]) Thanks in advance, Martin

Hello, all; I get an unexpected result when trying to plot a probability histogram with R1.9.1 on windows xp: #with the following code: > x <- runif(100,0,1) > hist(x) > hist(x, freq=F) > h <- hist(x, freq=F) > summary(h) # Length Class Mode #breaks 11 -none- numeric #counts 10 -none- numeric #intensities 10 -none- numeric #density 10 -none- numeric #mids 10 -none- numeric #xname...

Hi R-helpers, I have a list containing 10 elements, each of which is a dataframe. I wish to add a new column to each list element (dataframe) containing the product of the last two columns of each dataframe. I'd appreciate any pointers, thanks! Mark Na [[alternative HTML version deleted]]

I have some data measured with a coarsely-quantized clock. Let's say the real data are q<- sort(rexp(100,.5)) The quantized form is floor(q), so a simple quantile plot of one against the other can be calculated using: plot(q,type="l"); points(floor(q),col="red") which of course shows the characteristic stair-step. I would like to smooth the quantized form back into an app...

...rather, confirm an existing one (bug #751). I have had some difficulty using the polyroot() function. For example, in Win 98, R 1.1.1, > polyroot(c(2,1,1)) correctly (per the help index) gives the roots of 1 + (1*x) + (2*x^2) as [1] -0.5+1.322876i -0.5-1.322876i However, > polyroot(c(-100,0,1)) gives the roots of [1] 10+0i -10+0i which corresponds to -100 + x^2, not 1 - (100*x^2). Also, > polyroot(c(0,0,0,1)) gives [1] 0+0i 0+0i -10+0i which aren't all roots of x^3. Thanks for all your efforts. I apologize if I'm missing something. Kristian Markon...

...t, but I am looking for something in the spirit of mtext. That is, putting the legend anywhere I want on the device. In my situation I have: par(mfrow=c(2,2), ann = FALSE) p.names <- c("1","2","3","4") lapply(p.names, function(x) { d1 <- density(rnorm(100,0,1)) d2 <- density(rnorm(100,.3,1.2)) plot(range(d1$x, d2$x), range(d1$y, d2$y), type = "n", xlab = NULL, ylab = NULL) lines(d1, col = "black", lty = 2) lines(d2, col = "black", lty = 1) }) mtext("Learning R the hard way", side = 1, line=-1.5, outer...

...plot command does not bring up a plot (e.g. plot(), boxplot(), hist()). I found a thread concerning X11 windows under Mac OS X but I feel these users will most likely be just using the native quartz device. Invoking a call to quartz() first does not seem to help, e.g. >quartz() >plot(rnorm(100,0,1)) produces no output and no error message (Nothing happens). A call to dev.cur() seems to indicate a device is active. >quartz() >dev.cur() quartz 2 but again a plot command produces no figure. Sorry am I not a Mac OS user and I did check the archives but found mostly discussions on X11...

Dear R users, Can anyone please tell me how to generate a large number of samples in R, given certain distribution and size. For example, if I want to generate 1000 samples of size n=100, with a N(0,1) distribution, how should I proceed? (Since I dont want to do "rnorm(100,0,1)" in R for 1000 times) Thanks for help Debbie _________________________________________________________________ Looking to change your car this year? Find car news,...

hi - if I sample from a normal distribution with something like n100<-rnorm(100,0,1) and add an outlier with n100[10]<-4 then qqnorm(n100) visually shows the point 4 as an outlier and calculating the probablity of a value of 4 or bigger in 100 samples of norm(0,1) gives > 1-exp(log(pnorm(4,0,1))*100) [1] 0.003162164 If I have more than 1 sample above out...

Hi there, I am working on a project which needs the value of the interaction area under two distributions( eatimated by kernel density estimators). For example: x<-rnorm(100,0,1) y<-rnorm(100,0.2,1) density(x) # This produces the summary of dependent variable and independent variable. How can I get the individual values of variables and reform a curve to calculate the area under curve? Is there a way to get aroud this and give me the solution? Thanks a lot....

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Dear members, I was trying to simulate W which is iid and depends on X and Y. Here are 2 methods: Method 1: x<-rnorm(100) y<-rnorm(100) w<-rnorm(100, 2x+y,1) Method 2: x<-rnorm(100) y<-rnorm(100) w<-2x+y+rnorm(100,0,1) Are these methods comparable? Since x and y are vectors, the term 2x+y would return a vector. It seems that, given Method 1, each element of W would come from a different populat...

Dear R-list, the following code had been running well over the last months: exam <- matrix(rnorm(100,0,1), 10, 10) gg <- factor(c(rep("A", 5), rep("B", 5))) mlmfit <- lm(exam ~ 1); mlmfitG <- lm(exam ~ gg) result <- anova(mlmfitG, mlmfit, X=~0, M=~1) Until, all of a sudden the following error occured: Fehler in apply(abs(sapply(deltassd, funct...

...) is introduced at the extremes of the x and y axes of my lattice plots. I've tried defining the scales using xlim, scale and num.limits in x.scale.components, but haven't succeded in getting overriding the introduction of extra space. Here's the problem: test.df<-data.frame(a=runif(100,0,1),b=c(runif(50,0,1),runif(50,0,2)),c=c(rep(c("A","B"),c(50,50)))) 1) Space appears when I plot multiple panels: xyplot(a~b|c,data=test.df) 2) I can get rid of the space by explicitly defining scales (shown for x axis)... xyplot(a~b|c,data=test.df...

Hi, I'd like to create a function that accepts as arguments a string that is to be substituted within a variable name. For instance, suppose I have a data frame df: df<-data.frame(x_narrow=c(rnorm(100,0,1)),x_wide=c(rnorm(100,0,10))) What I have in mind is something like: f <- function(string){ var = paste("x_",string,sep = "") df$var } which does not work. Any suggestion for modifications? Thanks in advance, Matteo -- View this message in context: http://r.789695.n4.n...