similar to: par(mfrow=c(4,2)) vs. lattice

Hi All, I'm going to draw 8 plots in one page. I want the plots to be arranged in two rows, each row has 4 plots. So, I set the it par(mfrow=c(2,4)). However, there might be too many plots in a page, all the 8 plots were drawn in triangular shape, which makes the x-y coordinate scale not in same length, the x-axis length is much shorter than the y-axis length. Could anybody let me know how

#this is my little function that I would like to use the column names of the x and y arguments in the function. I would like it to read # site1-site2 how would I do this diff.temp <- function(x, y ,use="pairwise.complete.obs") { na.method <- pmatch(use, c("all.obs", "complete.obs", "pairwise.complete.obs")) par(mfrow=c(2,1))

I am having a hard time interpreting the phase and coherency plots. x is two timeseries that occur at the same time i.e. a b 1 11.2 12.3 16 11.3 12.4 31 11.4 12.5 46 11.5 12.6 ...etc even though my example is does not show this they are oscillating at more or less the same frequency just shifted by t=x (imagine two sine waves offset with the 2nd sine

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

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 would like to use the data below where the plots are close to what I want. Instead of color I would like to use different symbols, and have the symbols in the legend match the graphs. I am also going to add a regression line to these I know about the type="r" (which is fine for these particular graphs) argument, but it fits the subsets instead of the entire data set-- should I use a

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 <-

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",

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

#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 package to make a map from GIS data, and/or would it be easier in one of the free GIS programs. I would like to make a map of the savannah river area with our sampling locations. 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

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

#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",

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

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

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,

have a data frame composed of many columns. I would like to hold one column constant and scatter plot it agianst all of the other columns? how do I do this? I have only been able to plot(x) and use indexes to create an entire scatterplot matrix, but not hold one constant. -- Let's not spend our time and resources thinking about things that are so little or so large that all they really do

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),