similar to: Combining Data Frames

DateTime,Temp,SpCond,DOConc,Depth,pH,ORP,Turbidity+,Chlorophyll,Battery,Cond,DO%,Salinity,TDS 01/13/2006 17:01,10.87,84,9.36,0.664,7.3,132,28.8,3.1,11.5,0.062,84.6,0.04,0.055 01/13/2006 17:16,10.9,84,9.36,0.66,7.31,133,28.7,2.9,11.5,0.062,84.7,0.04,0.055 01/13/2006 17:31,10.92,84,9.36,0.655,7.3,132,28.4,2.6,11.4,0.062,84.8,0.04,0.055 01/13/2006

I would like to merge zoo objects that are stored in a list into one big zoo object with one index for all of the observations. I have created the list (74 dataframes) with the code below, and have tried the do.call(merge, foo) in the call and the output is not what I expected. Any help would be greatly appreciated. Stephen Sefick ###################################################level logger

# chron library(chron) fmt.chron <- function(x) { chron(sub(" .*", "", x), gsub(".* (.*)", "\\1:00", x)) } z1 <- read.zoo(SC3.csv, sep = ",", header = TRUE, FUN = fmt.chron) z2 <- read.zoo(SC2.csv, sep = ",", header = TRUE, FUN = fmt.chron) z3<-c(z1, z2) write.table(z3, sep="," , "SC.csv") How do you

I have not been able to find anything to do what I want, so I am going to tact to the left. I have twp continuous time series for two years with the same fourteen variables. I would like to simply append the second year to the first. They both have the same column headings etc. Just like tapping two pieces of paper together for a long number series. Thanks Stephen -- Let's not spend our

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,

#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",

#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

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

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

DateTime RM61 11/30/2006 12:31 NA 11/30/2006 12:46 NA 11/30/2006 13:01 2646784125 11/30/2006 13:16 NA 11/30/2006 13:31 NA 11/30/2006 13:46 NA 11/30/2006 14:01 2666435177 11/30/2006 14:16 NA 11/30/2006 14:31 NA 11/30/2006 14:46 NA 11/30/2006 15:01 2653041914 11/30/2006 15:16 NA 11/30/2006 15:31 NA 11/30/2006 15:46 NA 11/30/2006 16:01 2693126189 11/30/2006 16:16 NA 11/30/2006 16:31 NA 11/30/2006

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

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

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

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

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

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 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"

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 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.