similar to: ggplot2 45deg axis labels

#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",

I would like to do the following in ggplot: what am I missing? River.Mile <-c(202, 198, 190, 185, 179, 148, 119, 61) TSS <- c(1:8) DOC <- seq(2, by= 0.6, length.out=8) z <- data.frame(River.Mile, TSS, DOC) xyplot(TSS+DOC~River.Mile, data=z, auto.key=TRUE) thanks -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources

plot(1,2, ylab= paste("insects", expression(m^2), sep=" ")) I get insects m^2 I would like m to the 2 what is the problem? -- 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 can't figure this one out- I am the administrator, The file that I want to install is a .tar.gz which is located on my desktop. How do I get it into my packages directory- through the GUI or through brute force? thanks stephen -- 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

I know this is not reproducible, but I don't want to clog up mail boxes with the data frame. I would be happy to send this off list. I am sure that I am missing something simple. The plotting works if I just paste the call to qplot into R and replace the [i] with a number. Thanks for all of your help in advance. #loop to spit out PDFs list.names <-

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

Is there a way to use a continuous variable to pch in qplot? I believe this worked in previous version. I need to specify certain values of a shape for particular points so that multiple graphs all show the same shapes for the same streams. I have gone to the original data and added a pch column that I would like to use to specify the shapes to pch in qplot. Any help would be greatly

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,

#copy and paste this into R f <- (structure(list(TKN = c(0.103011025, 0.018633208, 0.104235702, 0.074537363, 0.138286096), RM = c(215, 198, 148, 119, 61)), .Names = c("TKN", "RM"), class = "data.frame", row.names = 25:29)) plot(f$TKN~f$RM, type="b") I would like to reverse the X-Axis. How do I do this? -- Let's not spend our time and resources

Does anyone know of a test for stationarity of a time series, or like all ordination techniques it is a qualitative assessment of a quantitative result. Books, papers, etc. suggestions welcome. thanks Stephen -- 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

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

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,

library(ggplot2) melt.updn <- (structure(list(date = structure(c(11808, 11869, 11961, 11992, 12084, 12173, 12265, 12418, 12600, 12631, 12753, 12996, 13057, 13149, 11808, 11869, 11961, 11992, 12084, 12173, 12265, 12418, 12600, 12631, 12753, 12996, 13057, 13149), class = "Date"), variable = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,

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

RM215.sp <- SpatialPoints(RM215, proj4string=CRS("+proj=longlat +datum=WGS84")) d060101 <- as.POSIXct("2006-01-01", tz="EST") study_seq <- seq(from=d060101, length.out=761, by="days") up.215 <- sunriset(RM215.sp, study_seq, direction="sunrise", POSIXct.out=TRUE) down.215 <- sunriset(RM215.sp, study_seq, direction="sunset",

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!

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