similar to: simple graphing question

f <- (structure(list(X = structure(96:97, .Label = c("119DAmm", "119DN", "119DNN", "119DO", "119DOC", "119Flow", "119Nit", "119ON", "119OPhos", "119OrgP", "119Phos", "119TKN", "119TOC", "148DAmm", "148DN", "148DNN", "148DO",

This is what my data looks like DOC TOC TKN RM119mFeb-06 1 2 3 RM61mFeb-06 2 4 6 I have this both in a .csv and .txt I have read this in with read.csv("chemodr.csv", header=T) and this is what I get X dAmon DN.N Nitrite.N DOC OP P TKN TOC 1 RM215mFeb-06 0.000 0.1300 0.0000 2.5

Hi! I am using OpenSSH_3.6.1p2, SSH protocols 1.5/2.0, OpenSSL 0x0090701f on my Mandrake 9.1 machine. I used to run fetchmail to get emails from IRIX 6.3 system which has OpenSSH_3.4p1, SSH protocols 1.5/2.0, OpenSSL 0x0090602f. I had to reinstall my Mandrake, and after that ssh gives me strange error messages. When I run command: ssh -f -L 1234:irixhost.somedomain.fi:110

I ran a function called BoxCox, taken from the book by Venables and Ripley, for checking the need for power transformation. This function works fine using the version 0.50 of R, but gives an error message with version 0.60. The lm function in version 0.60 is different from that in version 0.50. Is there a bug in the new lm function? Kung-Sik Chan >

station month bas 190 5 0.000 190 7 1.563 190 10 0.000 190 11 0.000 202 4 18.750 202 5 18.750 202 7 6.250 202 10 4.800 202 11 3.125 198 4 18.750 198 5 31.250 198 7 3.125 198 10 3.200 198 11 12.500 205 4 0.000 205 5 0.000 205 7 0.000

Hi everyone, > does any one knows how can I calculate mean for different samples > i.e. I have a data like this: > > s1 s2 s3 s4 > 1 0 0 0 1 > 2 1 0 1 0 > 3 0 0 0 0 > 4 0 0 0 0 > 5 0 1 0 1 > 6 1 0 0 0 > 7 0 0 0 0 > 8 0 0 0 0 > 9 0 0 0 0 > 10 0 0 0 1 > > I need to make 5 different sample with 5

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,

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 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"

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",

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 would like to reformat this data frame into something that I can produce some descriptive statistics. I have been playing around with the reshape package and maybe this is not the best way to proceed. I would like to use RiverMile and constituent as the grouping variables to get the summary statistics: 198a 198b mean mean sd sd ... ... etc. for all of these. I have tried

#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 am using the function metaMDS with jaccard distances to ordinate a set of constituent by site matrix. I can post this data if it would be helpful, but it is large to include in an email. I can also provide reproducable code if necessary. I would like to get an R^2 value for the axes of the ordination configuration that I get with metaMDS in the vegan package is there a way to do this- is it

#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))

below is my data frame. I would like to compute summary statistics for mgl for each river mile (mean, median, mode). My apologies in advance- I would like to get something like the SAS print out of PROC Univariate. I have performed an ANOVA and a tukey LSD and I would just like the summary statistics. thanks stephen RM mgl 1 215 0.9285714 2 215 0.7352941 3 215 1.6455696 4 215

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

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

Hello, When I put in the following script line: gap.boxplot(CLI3, CLI4, CLI5, CLI6, CLI7, gap=list(top=c(8000,280000), bottom=c(0,250)), range=50, outline=TRUE, par(ask=FALSE) I get a '+' telling me I am missing something. I have tried adding ')', 'width=NULL', etc and then I get this error: Error: unexpected symbol in: "gap.boxplot(CLI3, CLI4, CLI5, CLI6, CLI7,

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