similar to: Create NA

Hi all, May I request for your help if you have time and if you have an idea on how to do this. I want to add three vectors... And my goal is to obtain the sum of the vectors, ignoring the vector of "na"... Here is what i did in R.. I'm adding the three vectors, e,z,k, and my objective is to get an answer = -23. I tried putting the na.omit but it did not work. Thanks. > z [1]

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

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

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 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",

There is a series of data contains time in fixed step and energy varying with time, how to test its periodicity?In R, it seems there is no direct tools since I have search the R manual with periodic and I have not found any related topic. Thanks a lot

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,

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

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

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

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

#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

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 a model Q=K*A*(R^r)*(S^s) A, R, and S are data I have and K is a curve fitting parameter. I have linearized as log(Q)=log(K)+log(A)+r*log(R)+s*log(S) I have taken the log of the data that I have and this is the model formula without the K part lm(Q~offset(A)+R+S, data=x) What is the formula that I should use? Thanks for all of your help. I can provide a subset of data if necessary.

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 would like to be able to grab x and y columns out of a dataframe and then grab all of the columns that are not equal to x or y. I am sure that I am missing something easy. ftbr_UTM_downstream <- (structure(list(site = c("Jennie_Creek_Main_Stem", "Wolf_Pit_Creek_Main_Stem", "Little_Rockfish_Main_Stem_North", "Big_Muddy_Creek_Main_Stem",

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

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),