similar to: Upgrade through the R interface?

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

#I am putting a test together for an introductory biology class and I would like to put different cross hatching inside of each bar for the bar plot below color <- c("Brightly Colored", "Dull", "Neither") lizards <- c(277, 70, 3) liz.col <- data.frame(color, lizards) qplot(color, lizards, data=liz.col, geom="bar", ylab="Observed Matings",

These are date and times in the format YYYYMMDDhhmmss. I would like to take this column and make a chron object form them. I have tried a couple of the split family of functions but they need character input here is the data: date.time <- c(19851001001500, 19851001003000, 19851001004500, 19851001010000, 19851001011500, 19851001013000, 19851001014500, 19851001020000, 19851001021500,

I would like to rotate the axis labels 45 deg. -- Stephen Sefick 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 mammals. -K. Mullis

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),

I have a csv file that I am trying to read in and know that values <0 are erroneous - is there a way to read only value grater than 0. thanks, -- Stephen Sefick 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

Error in firstlib(which.lib.loc, package) : Tcl/Tk support is not available on this system Error : package 'tcltk' could not be loaded Ubuntu 9.10 I have both the 8.5 and 8.4 tcl and tk regular and dev libraries installed. I need both of these because some of my GIS software requires the 8.4 libraries. Should I remove the 8.5 libraries, or ... All help is greatly appreciated!

d <- structure(c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, NA, NA, NA, 14), .Dim = c(14L, 2L ), .Dimnames = list(NULL, c("a", "b"))) plot(d, type="b") This is simplified, but Is there an option I am missing that will force all of the points to be joined by a line? Stephen Sefick -- Let's not spend our time and resources

Is there any easy way to pull out the row indexes for a logical matching statment? #################example code######################################### foo <- data.frame(name=c(rep("A", 25), rep("B", 25), rep("C", 25), rep("A", 25)), stuff=rnorm(100), and=rnorm(100), things=rnorm(100)) #this is what I want but I would like the row indexes

##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)

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

x <- "1041281__2009_08_20_.lev" I would like to split this string up and only extract the leading numbers. 1041281 to use as a label for a data column in a bigger for loop function to read in data. regards, -- Stephen Sefick 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

I have tried to compile this from source. I don't know what Endianess is, but it is probably not debian power pc. Am I would of luck with this package? Stephen Sefick * Installing *source* package ?ifultools? ... ** libs gcc -std=gnu99 -I/usr/local/lib/R/include -I"../inst/include/" -D"MUTIL_STATIC" -D"DEF_TF" -D"INTERRUPT_ENABLE"

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,

has anyone had problems with the upgrade to R 2.8 and chron date classes. I have a large zoo object that has a chron index, and it is taking 5x or so longer to do the same calculation as with 2.7 if it doesn't fail. I will provide anything necessary I am not entirely sure what ya'll would need if anything to try and reproduce the behavior. I am using the package StreamMetabolism. thanks