similar to: Alternative and more efficient data manipulation

## Hello there, ## I have an issue where I need to use the value of column names to multiply with the individual values in a column and I have many columns to do this over. I have data like this where the column names are numbers: mydf <- data.frame(`2.72`=runif(20, 0, 125), `3.2`=runif(20, 50, 75), `3.78`=runif(20, 0, 100), yy=

I know that this has been asked before in other variations but I just can't seem to figure out my particular application from previous posts. My apologies if I have missed the answer to this question somewhere in the archives. I have indeed looked. I am running Ubuntu 11.04, with R 2.12.1 and ESS+Emacs. For journal formatting requirements, I need to italicize all the greek letters in any

Dear all, I am stumped at what should be a painfully easy task: predicting from an lm object. A toy example would be this: XX <- matrix(runif(8),ncol=2) yy <- runif(4) model <- lm(yy~XX) XX.pred <- data.frame(matrix(runif(6),ncol=2)) colnames(XX.pred) <- c("XX1","XX2") predict(model,newdata=XX.pred) I would have expected the last line to give me the

Dear R users, I try to compute this summation, http://www.nabble.com/file/p25054272/dd.jpg where f(y|x) = Negative Binomial(y, mu=exp(x' beta), size=1/alp) http://www.nabble.com/file/p25054272/aa.jpg http://www.nabble.com/file/p25054272/cc.jpg In fact, I tried to use "do.call" function to compute each u(y,x) before the summation, but I got an error, "Error in X[i, ]

Hi, I ran into a problem where I actually need rank(, ties.method="last"). It would be great to have this feature in base and it's also simple to get (see below). Thanks & cheers, Marius rank2 <- function (x, na.last = TRUE, ties.method = c("average", "first", "last", # new "last" "random", "max",

On 15 Jun 2004 at 13:52, Vito Muggeo wrote: > Dear Petr, > Probably I don't understand exactly what you are looking for. > > However your "plot(x,c(y,z))" suggests a broken-line model for the > response "c(y,x)" versus the variables x. Therefore you could estimate > a segmented model to obtain (different) slope (and breakpoint) > estimates. See

Dear all Suppose I have teeth like data similar like x <- 1:200 y <- 0.03*x[1:100]+rnorm(100, mean=.001, sd=.03) z <- 3-rep(seq(1,100,10),each=10)*.03+rnorm(100,mean=.001, sd=.03) plot(x,c(y,z)) and I want to have a gradient estimations for some values from increasing part of data like y.agg <- aggregate(diff(c(y,z)), list(rep(seq(1,200,10),each=10)[1:199]), mean) y.agg[1:10,]

Dear All, I have got the limits for removing extreme values for each variables using following function . f=function(x){quantile(x, c(0.25, 0.75),na.rm = TRUE) - matrix(IQR(x,na.rm = TRUE) * c(1.5), nrow = 1) %*% c(-1, 1)} #Example: n <- 100 x1 <- runif(n) x2 <- runif(n) x3 <- x1 + x2 + runif(n)/10 x4 <- x1 + x2 + x3 + runif(n)/10 x5 <-

Dear Rich, Assuming that I understand what you want to do, try adding the following to your script (which, by the way, is more complicated that it needs to be): xx <- 10:50 m <- lm(y ~ x) yy <- predict(m, data.frame(x=xx)) lines(spline(xx, yy), col="blue") m <- lm(y ~ log(x)) yy <- predict(m, data.frame(x=xx)) points(xx, yy, col="magenta") The first set of

Hi, I have a matrix of say 1024x1024 and I want to look at it in chunks. That is I'd like to divide into a series of submatrices of order 2x2. | 1 2 3 4 5 6 7 8 ... | | 1 2 3 4 5 6 7 8 ... | | 1 2 3 4 5 6 7 8 ... | | 1 2 3 4 5 6 7 8 ... | ... So the first submatrix would be | 1 2 | | 1 2 | the second one would be | 3 4 | | 3 4 | and so on. That is I want the matrix to be evenly divided

Hi Rich, If I understand your comment about "uniformly distributed along the log=x axis" then I think John's (second) set of commands needs a change to the definition of xx, as follows: xx <- exp(seq(from=log(min(x)),to=log(max(x)),length=50)) m <- lm(y ~ log(x)) yy <- predict(m, data.frame(x=xx)) points(xx, yy, col="red") HTH, Eric On Mon, Sep 25, 2017 at

Hi, I'm using panel.polygon inside a custom panel function to generate filled polygons for an xyplot. Everything is as expected until I specify a value < 1 for alpha to fill with a semi-transparent color and output to pdf. The plot symbols appear malformed. Am I doing something wrong here? Thanks, Mark Here is an example (pdfs are attached): library(lattice) ## example data

Hello, I am trying to create a plot often seen in hydrodynamic work than includes a contour plot representing the water speed with arrows pointing in the direction of flow. Does anyone have any idea how I might add arrows based on wf$angle (in the example below) to the plot below? Thanks in advance! Sam library(lattice) speed <- runif(100, 0, 20) wf <- data.frame(speed) wf$width <-

Hello all, I have a fairly simple data manipulation question. Say I have a dataframe like this: dat <- as.data.frame(runif(7, 3, 5)) dat$cat <- factor(c("1","4","13","1","4","13","13A")) dat runif(7, 3, 5) cat 1 3.880020 1 2 4.062800 4 3 4.828950 13 4 4.761850 1 5 4.716962 4 6

Hi, I try to fit a non linear regression by minimising the sum of the sum of squares. The model is number[2]-(x/number[1])^number[3] Number [2] and number [1] change as the data changes but for all the set of data number[3] must be identical. I have 3 set of data (x1,y1), (x2,y2), (x3,y3). x_a<-c(0,0.5,1,1.5,2,3,4,6) y_a<-c(5.4,5,4.84,4.3,4,2,1.56,1.3)

Hi, I'm trying to follow the suggestions given by Deepayan Sarkar in this message: http://tolstoy.newcastle.edu.au/R/help/05/11/16135.html to plot 3-D points on a wireframe plot. The problem is that I keep getting a partly formed plot- with the colored lattice visible but no axis labels or additional points- with the error message "error using packet 1, 'x' and 'units'

Full_Name: Hallgeir Grinde Version: 2.1.1 OS: Windows XP Submission from: (NULL) (144.127.1.1) While using lm(y~(x*z*c*...*v)^2) R crashes/closes if the numbers of variables are at least 8.

Hello I'm having trouble figuring out how to use the output of "segmented()" with a new set of predictor values. Using the example of the help file: ??set.seed(12) xx<-1:100 zz<-runif(100) yy<-2+1.5*pmax(xx-35,0)-1.5*pmax(xx-70,0)+15*pmax(zz-.5,0)+rnorm(100,0,2) dati<-data.frame(x=xx,y=yy,z=zz) out.lm<-lm(y~x,data=dati) o<-## S3

Hello all, This is one of those "Is there a better way to do this questions". Say I have a dataframe (df) with a grouping variable (z). This is my base data. Now I know that there is a higher order level of grouping that exist for my group variable. So what I want to do is create a new column that express that higher order level of grouping based on values in the sub-group (z in this

Dear R-users I have a probelm running stepAIC in R1.7.1 I wrote a program which used stepAIC as a part of it, and it worked fine while I was using the previous version of R1.7.0. However, I found the program did not work any more. Now, R produces a message which tells "Error in as.data.frame.default(data) : can't coerce function into a data.frame" every time I run the part of