similar to: plotmath help with expression

Hello! Does someone know how to produce L(y|mu) with plotmath? Some code with unsuccessfull results: plot(dnorm(x = seq(from = -4, to = 4, by = 0.1)), type = "l") ## Not what I want legend(legend = c(expression(L(y:mu))), x = "topright") ## Strange, is this a bug? legend(legend = c(expression(L(y|mu))), x = "top") ## Group produces an error legend(legend =

I have got a data set that is Gross Primary Productivity ~ Total Suspended Solids it is a hyperbola just like: plot(1/c(1:1000)) how do I model this relationship so that I can get all of the neat things that lm gives residuals etc. etc. so that I can see if my eyeball model stands up. Thanks for any help, pointers, or good things to read. -- Stephen Sefick Research Scientist Southeastern

mac osx 10.5.4 R 2.7.1 I have fit a model d<-lm(y~x) with an R^2 of 0.963 but when I issue the command abline(d) the line is below where it ought to be. Looks like the right slope, but not the right intercept. thanks -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so large

I would like to be able to reverse the xlim on qplot this is the code that I am using qplot(a[,"River.Mile"], a[,26] ,ylab=colnames(a)[26], xlab="RiverMile", xlim=rev(c(60, 216)))+geom_smooth()+scale_x_continuous(breaks=c(215,202,198,190,185,179,148,119,61),

Title says it all remember cast() with sum as the aggregation function -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being

Dear all; A simple? question. I'm having a problem with a math expression in the legend of a plot and I haven't found the way to get this to work, so any help will be appreciate. Basically I want to include in the plot is the R-squared and its numerical value, so I tried this: R2c<-0.82879 # R-squared of calibration model plot(1:10,1:10) legend("topleft",

##This is a function that I am trying to write to calculate sunrise and sunset and works "mostly", but returns nonsensical values. What am I #missing? Thanks in advance. ###remember to include maptools as dependence### library(maptools) sunrise.set <- function(lat, long, date, timezone="UTC", num.days=1){ #this needs to be long lat# lat.long <- matrix(c(long, lat),

a <- c(1:26) b <- rnorm(25) e <- rnorm(25) f <- rnorm(25) g <- data.frame(b,e, a,f) I would like to plot a agianst all possibilities and then shoot it out to a pdf one graph per page. I think it would be okay to have this as a lattice plot or a ggplot with many graphs per page. I can figure all of that out I think, but I need something like r <- as.matrix(g) plot(.~a, data=r)

i have a list of 6 each containing a dataframe of 96 observations as a zoo object. Is there a way to plot these in one frame par(mfrow=c(3,2)) this is what I tried lapply(d, FUN=plot) I can provide data, list is large. thanks -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so

a <- c(1:10) b <- c(.5, .6, .9, 10, .4, 3, 4, 9, 0, 11) d <- c(21:30) z <- data.frame(a,b,d) library(fields) results <- c() for(i in 1:(length(rownames(z))-1)){ results[i] <- rdist(z[i,], z[(i+1),]) } results.1 <- data.frame(results) f <- rownames(z) r <- f[-1] rownames(results.1) <- r colnames(results.1) <- f[1] this does what I want it to do - is

library(zoo) d<-(structure(c(1.39981554315924, 0.89196314359498, 0.407816250252697, 0.823496839063978, 1.14429021220358, 1.23971035967413, 0.960868900583432, 0.927685306209829, 1.22072345292821, 0.249842897450642, 1.00879641624694, 0.925372139878243, 0.317259909172362, 0.382677149697482), index = structure(c(11808, 11869, 11961, 11992, 12084, 12173, 12265, 12418, 12600, 12631, 12753, 12996,

#Is there a way to vectorize the for loop #maybe a fancy indexing trick? #thanks d=0.5 L=20 x=seq(20, by=1, length.out=20) reflecting <- function(pre, d, L){ r=L-1 x=rep(0, L) for(j in 2:r){ x[j]=((1-(2*d))*pre[j])+(d*pre[(j+1)])+(d*pre[(j-1)]) } x[1]=((1-d)*pre[1])+(d*pre[2]) x[L]=((1-d)*pre[L])+(d*pre[(L-1)]) y=x } f = reflecting(x, d, L) -- Stephen Sefick Research Scientist

I would like to be able to do the xyplot in ggplot below. I read in the archive that Hadley was working on this for the next release, and I can not find the documentation (Aug. 23rd). River.Mile <- c(215 ,202, 198, 190, 185, 179, 148, 119, 61) Cu <- rnorm(9) Fe <- rnorm(9) Mg <- rnorm(9) Ti <- rnorm(9) Ir <- rnorm(9) r <- data.frame(River.Mile, Cu, Fe, Mg, Ti, Ir) z <-

I have insect data from twelve sites and like most environmental data it is non-normal mostly. I would like to preform an anova and a means seperation like tukey's HSD in a nonparametric sense (on some sort of central tendency measure - median?). I am searching around at this time on the internet. Any suggestions, books, etc. would be greatly appreciated. -- Stephen Sefick Research

I have one hundred and six independent variable that I would like to preform a correlation analysis on. Is there anyway to only get the values that are abolute value 0.6 or greater. thanks -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is

I am was going to look at the as.yearmon function in the zoo package and write a as.day function to aggregate a time series of 96 observations per day into the mean for each day, but I don't know how to look at the code so that I can convert it into something I can use. On top of that I believe that it is probably an S3 method and I haven't quite gotten that far in my programming

Hadley et al., I was using the cast function to reshape some data (aggregate a melted data frame) and I did not put in the fill and for the most part the values that came out were fine, but there were value great than an order of magnitude from the actual value. When I put in the fill argument everything is okay. I don't provide a reproducible example because the data set is to large to post

R version 2.8.0 (2008-10-20) i386-pc-mingw32 locale: LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] StreamMetabolism_0.01 chron_2.3-24 zoo_1.5-4 loaded

h <- structure(list(V1 = c(-0.351714766, 0.188298251, 0.042951816, -0.072490327, -0.691885485, -0.816169763, -0.7066502, -0.856286332, -0.839723411, -0.427242353, -0.372911996, 0.326707494, 0.07847893, 0.687447841, 0.516105863, 0.267076547, 0.727867663, 0.432699191, 0.258610632), V2 = c(0.256636068, -0.824072121, -0.149185618, 0.153280492, 0.01649528, 0.409410528, 0.286015324, 0.366323539,

I have looked at ?as.POSIXct ?POSIXct and many of the references that are on those pages. I am bewildered with timezones. Is there a way to get what would go into tz="" for making a function that uses POSIXct to be able to be used in all of the timezones in just the united states? This is for both windows and mac... this is the function that I am wanting to use it with