similar to: ggplot2 and lattice

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,

updn.gg <- (structure(list(date = structure(c(11808, 11869, 11961, 11992, 12084, 12173, 12265, 12418, 12600, 12631, 12753, 12996, 13057, 13149), class = "Date"), unrestored = c(1.13789418691602, 0.704948049842955, 0.276777348238899, 0.417586861554189, 0.504870337754768, 0.673201771716216, 0.560704221510771, 0.835737007551542, 1.10773858390693, 0.197070828834836, 0.942350681588179,

I don't know if there is a way to use the scale relation free argument in ggplot2 like in lattice. I have a feeling that there is not, but I would like to make a plea for this feature. It would be nice to be able to plot Total Inorganic Nitrogen Total Phosphorus and the ratio of the two- the numbers on the axis are not related, but the previous two are surely related to the last (this ratio

r <-(structure(list(TSS = c(2.8, 8.4, 11, 1.3, 4.2, 2, 3.4, 14, 8.2, 3.1, 1.4, 0.9, 0.5, 6.1, 9.2, 0.6, 1, 11, 2.4, 1.2, 1.3, 1.3, 0, 1.8, 8, 11, 11, 8.5, 8.5, 1.8, 13, 4.4, 1.4, 2.1, 0.5, 25, 25, 9.3, 6.1, 1.6, 1.5, 19, 19, 24, 9.6, 1.8, 1.4, 1), GPP = c(1.213695235, 3.817313822, 1.267930498, 10.45692825, 3.268295623, 3.505286001, 4.468225245, 0.915653726, 1.635617261, 3.726133898,

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

##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),

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

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 <-

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)

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

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

?for doesn't return anything help.search("for") doesn't return anything- Is the for loop so prevelant in computer programing that the documentation is implicit or is R paradigm to discourage the use of the for loop. I will post data probably tonight, but here is my problem. I have preformed an MDS on a set of data. I have the scores of the four axes that are the optimal

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

#how do I break these up into first two letters (RM), number, and then the last part #is there an easily accessible regex tutorial on the internet? v = (structure(1:122, .Label = c("RM215Temp", "RM215SpCond", "RM215DO.Conc", "RM215Depth", "RM215pH", "RM215ORP", "RM215Turbidity.", "RM215Battery", "RM215DO.",

I have a data set of mean velocity, discharge, and mean depth. I need to find out which model best fits them out of log linear, linear, some other kind of model... Using excel I have found that linear is not that bad and log10(discharge) vs. the other two variables (I am trying to predict velocity and depth from discharge) is not that bad either. How do I test and see which one of these models