similar to: formating chron date times for printing

R 2.12.2 on Scientific Linux 6.4 #works chron(times.="15:00:00", format=c(times="h:m:s")) #doesn't work chron(times.="15:00", format=c(times="h:m")) From chron Manual: The times format can be any permutation of "h", "m", and "s" separated by any one non-special character. The default is "h:m:s". what am I

NOTE: I will provide data if necessary, but I didn't want clutter everyones mailbox All: I have a time series with level and temperature data for 11 sites for each of three bases. I will have to do this more than once is what I am saying here. OK, The time series are zoo objects with index values in chron format. The problem is that the date and times should be at even 15 min intervals,

This is the first time that I have tried to update packages with a tinkered around with .Rprofile. I start R with R --vanilla and it does not load my .Rprofile, but when I issue the command update.packages() R downloads the packages as expected, but then seems to load .Rprofile before compiling the packages sources. What am I doing wrong? kindest regards, Stephen Sefick see- Session info

These data represent stream channel cross-sectional surveys. I would like to be able to find the measurement on the tape (measure) where the Bank Full Depth (bkf_depths) is 0. This will happen twice because the channel has two sides. I thought fitting a loess line to these data and then predicting the measurment number would do it. I was wrong. Below is my failed attempt. My naive thought is

I have used the fourthcorner function as suggest by dray and legendre (model 2 and 4 then combine). I plot the combined value with plot(four.comb, type="G"). What do the colors mean? I have both grey and black bars. many thanks, Stephen -- Stephen Sefick ************************************************** Auburn University Biological Sciences 331 Funchess Hall Auburn, Alabama

I know this is a double post, but the subject line was really misleading. Sorry again. This is the first time that I have tried to update packages with a tinkered around with .Rprofile. I start R with R --vanilla and it does not load my .Rprofile, but when I issue the command update.packages() R downloads the packages as expected, but then seems to load .Rprofile before compiling the packages

I am not entirely sure what is going on. I am sure that I am missing something. maptools won't detach... Thanks for all of your help. R 2.11.1 OS Ubuntu 10.04 R --vanilla install.packages("sp", dep=TRUE) Loading required package: reshape Loading required package: plyr Loading required package: grid Loading required package: proto This is vegan 1.17-3 Loading required package:

s <- 1.00 max(s) returns 1 is there anyway that I can get it to return 1.00. I am using the results of this max statement in a grep statement and it returns the wrong numbers, I will provide more information and code if it would make more sense in context. -- Stephen Sefick ____________________________________ | Auburn University? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?? | | Department of

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

I am thinking about submitting a package to CRAN that contains some units conversion functions that I use on a regular basis. Would this be helpful to the community, or would it be better to keep this as a personal package? I don't want to clutter CRAN. many thanks, -- Stephen Sefick ************************************************** Auburn University Biological Sciences 331 Funchess

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",

All, 1. I will try and make this clear and concise. Please let me know any information that would be helpful in figuring out this problem (I don't know the relevant information to post). I am on linux- see below for session information. 2. Problem: working directory: home an old version of a function is sourced into the R session and doesn't work working directory: Desktop the

row <- c("a","b","c","d","e","f","g") #rows from larger data frame row.1 <- c("a","b","c","g") #rows of smaller data frame because d, e, and f don't contain any of the species, but the zeros are important x <- data.frame(sp1=rnorm(4), sp2=rnorm(4), sp3=rnorm(4),

Below is the code that I am using in a much larger function. I would expect a bankfull measure at zero to be between 0.6 and 0.8 approxfun is returning 0.8136986. I am sure that I am missing something. measure_bkf <- (structure(list(measurment_num = c(0, 0.2, 0.4, 0.6, 0.8, 1, 1.2, 1.4, 1.6, 1.8, 2, 2.2, 2.4, 2.6, 2.8, 3, 3.2, 3.4), bankfull_depths_m = c(-0.48, -0.48, -0.42, -0.26, 0.58,

I have just upgraded to R 2.13 and have library(ggplot2) in my .Rprofile (among other things). when i start R I get an error message. Has something in the start up scripts changed? Is there a better way to specify the library calls in .Rprofile? Thanks for all of the help in advance. Error: Loading required package: grid Loading required package: proto Error in rename(x, .base_to_ggplot) :

#Is there a way to break the below zoo object into non-NA data frames algorithmically #this is a small example of a much larger problem. #It is really no even necessary to have the continuous chunks #end up as zoo objects but it is important to have them end #up with the index column. #thanks for all of your help in advance, and #if you need anything else please let me know library(zoo) ind.

I am confused by the behavior of the below piece of code. The NAs are making it past the logical call ==0. I am sure that I am missing something. I just don't understand this behavior. Thanks for your help in advance. ########code####################################################### left <- (structure(list(measurment_num = c(2, 2.2, 2.4, 2.6, 2.8, 2.82, 3, NA, NA, NA),

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.

#This is probably due to my incomplete understanding of grep, but the below code has been working for some time to #search for .R with anything in front of it and return a list of scripts to source. Likely, the syntax for the #grep statement has been wrong all along. scripts2source <- (c("/home/ssefick/R_scripts/Convert_package.R",

I am working on a simple pilot project comparing the capability of SQL, SAS and R to perform a rolling mean per the following instructions. I have completed the SQL and SAS analysis, so now it's R's turn. Calculate mean values of x (x=count) for each date in the dataset where mean = the average count of days [t-9] through day [t-3] for each date/illness combination. Dataset aggpilot