similar to: R: x-y data

Hello I followed the example in page 59, chapter 11 of the 'Introduction to R' manual. I entered my own x,y data. I used the least squares. My function has 5 parameters: p[1], p[2], p[3], p[4], p[5]. I plotted the x-y data. Then I used lines(spline(xfit,yfit)) to overlay best curves on the data while changing the parameters. My question is how do I calculate the residual sum of squares.

I am trying to fit a curve to a cumulative mortality curve (logistic) where y is the cumulative proportion of mortalities, and t is the time in hours (see below). Asym. at 0 and 1 > y [1] 0.00000000 0.04853859 0.08303777 0.15201970 0.40995074 0.46444992 0.62862069 0.95885057 1.00000000 [10] 1.00000000 1.00000000 > t [1] 0 13 20 24 37 42 48 61 72 86 90 I tried to find starting values for

I want to extra the part of the formula not including the response variable from an lm object. For example if the lm object ABx.lm was created by the call ABx.lm <- lm( y ~ A + B + x, ...) Then ACx.lm is saved as part of a workspace. I wish to extract "~ A + B + x". Later in my code I will fit another linear model of the form z ~ A + B + x for some other response variable z. I

Hello all, So I am using the (formula entry) method for randomForests: randomForest(y~x1+x2+...+x39+x40,data=xxx,...) but the issue is that some of the items in that package dont take a formula entry - you have to explicitly state the y and x vector: randomForest(x=xxx[,c('x1','x2',...,'x40')],y=xxx[,'y'],...) Now my question is whether there is a function/way

Hi All, I have location data set up as degrees: minutes: seconds (for example, 122:45:45) I need to get this data into decimal degree form (eg, 122.7625). I know that I need to use the formula: Decimal degrees = Degrees + (Minutes/60) + (Seconds/3600) Does anyone know how I can get R to do this? The original data is all in one column (called 'lat') and the degrees, mins and secs

A common process when data is obtained in an Excel spreadsheet is to save the spreadsheet as a .csv file then read it into R. Experienced users might have learned to be wary of dates (as I have) but possibly have not experienced what just happened to me. I thought I might just share it with r-help as a cautionary tale. I received an Excel file giving patient details. Each patient had an ID

Hi, I want to run t.test() for several variables among two groups, and I would like to skip the tedious process of collecting information to assemble a table, but I am not sure if the function I want already exists. Any suggestion would be appreciated. I have a working example, as required by the posting guide: my_swiss = swiss[-1,] my_swiss$facto = rep(1:2,nrow(my_swiss)/2)

I am trying to figure out why 'biglm' can handle large data set... According to the R document - "biglm creates a linear model object that uses only p^2 memory for p variables. It can be updated with more data using update. This allows linear regression on data sets larger than memory." After reading the source code below? I still could not figure out how 'update'

I have been using internal snapshots on production qcow2 images for a couple of years, admittedly as infrequently as possible with one exception and that exception has had multiple snapshots taken and removed using virt-manager's GUI. I was unaware of this: > There are some technical downsides to > internal snapshots IIUC, such as inability to free the space used by the > internal

There is a bug in `predict' whereby the order of variables sometimes gets re-arranged compared to the original fit, and then disaster results. Specifically, the 'variables' and 'predvars' attributes of a 'terms' object get out of synch. This only happens when the terms in the original formula get re-ordered during fitting: test> scrunge.data_ data.frame(

Hi all, I've got a problem with the nls function. I have an adjustment which works when I fix one of the argument of my function (Xo=150) : *Xo*=150 f<- function (tt*,Xo*,a,b) ifelse(tt<*Xo*,a*exp(-b**Xo*),a*exp(-b*tt)) ajust<-nls(RER~f(tt,*Xo*,a,b),data=data.frame(tt=Ph2[,2*k],RER=Ph2[,2*k+1]),start=list(a=0.5,b=0.014)) But, when I use it as a "normal" parameter (and

# R for Windows will not send your bug report automatically. # Please copy the bug report (after finishing it) to # your favorite email program and send it to # # r-bugs at r-project.org # ###################################################### This report is joint from Richard Heiberger <rmh at temple.edu> and Burt Holland <bholland at temple.edu>. Burt Holland is the coauthor

Dear R-Help, With the attached data set, I am still getting incorrect bwplots > xyplot(gdf$tt~gdf$OnHour |gdf$Runway, data=gdf) # Is correct > bwplot(gdf$tt~gdf$OnHour |gdf$Runway, data=gdf, horizontal=FALSE) # Puts the boxes on the wrong x-axis values # look especially at 0 and 3. How do I fix this? What is happening? Thanks, Jim Rome

Dear R-users, I would need some advices on the proper way to call a particular function. This function is called scope.char and it is embedded in the step.gam function from the gam package. I am trying to call scope.char directly in a script but I did not find the proper way to do so. Is this even possible? If so, what is the proper syntax? Thank you for your time and help. Sebastien

Dear all, I would like to suggest changing the pairs.formula command such that a command like pairs(GNP ~ . - Year - GNP.deflator, longley) would behave in a similar fashion as lm(GNP ~ . - Year - GNP.deflator, longley) i.e., make a pairwise scatterplot of GNP and all other variables in the (longley) dataframe except for Year and GNP.deflator. The above command, with the

Hello, I am having hard time obtaining a value from a function. "fit" is a survival function that produces some results, such as "median", "confidence intervals" etc. But str() function does not list these values. How can I extract these to be able use them? For example, I need "median" value for the group DrugA which is 48. "Print" function does

Dear list, i am a biologist who needs to do some ttest between disease and non disease, sex, genotype and the serum levels of proteins on a large number of individuals. i have been using excel for a long time but it is very tedious and time consuming. i am posting the data below and ask your help in generating a code to get this analysis done in R. thanks gender disease genotype data M N CC

Hi to all, I have some problems to get the times-scale to the x-axis the times are coming from an excel sheet f. e [1] "0:01:00" "0:02:00" "0:03:00" "0:04:00" "0:05:00" "0:06:00" "0:07:00" [8] "0:08:00" "0:09:00" "0:10:00" "0:11:00" "0:12:00" "0:13:00"

The following code should give me (at least that's what I think) two list boxes with their individual scrollbars. But instead both scrollbars stick to the same listbox even though they work as they should (scroll individual listboxes). When I remove all the widgets on top everything works perfectly. Any help would be highly apprecciated. ps: all subsequent listboxes and scrollbars work

Hi, In R I get the following: > x <- 1:9 > f <- as.factor(1:9) > i <- model.matrix(~x+f) > > attr(i, "assign") [1] 0 1 2 2 2 2 2 2 2 2 whereas in S I get the following: > attr(i, "assign") $"(Intercept)": [1] 1 $x: [1] 2 $f: [1] 3 4 5 6 7 8 9 10 Has anybody written a model.matrix function to get the same result as S?