similar to: .Rprofile interfering with update.packages()

Displaying 20 results from an estimated 700 matches similar to: ".Rprofile interfering with update.packages()"

2010 Jul 14
2
qplot in ggplot2 not working any longer - (what did I do?)
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
2012 Nov 16
2
source file on startup question - why does an old version of a function show up? ggplot or R?
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
2011 Feb 15
0
RPostgreSQL [Expired connection at startup]
I have set up a postgresql database that I connect to each R startup in my .Rprofile file. ##############relevant lines from .Rprofile############################ library(RPostgreSQL) m <- PostgreSQL() con <- dbConnect(m, user="xxx", password="yyy", dbname="zzz") ####################################################################### in R con <Expired
2011 Nov 03
2
grep fixed (?) in 2.14
#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",
2024 Mar 22
1
Problem with new version of R: Mutated vocals
Dear ladies and gentlemen, I have recently installed the latest version of R (4.3.3) for windows. Now I have the following problems with mutated vowels like ?, ?, etc. Here is an example: If I type the command: Dir <- "C/Users/macho/Documents/_LVn/Experimentelle ?bungen" in the R console there is no problem. However, if I put the same command into a source file (e.g. Test.r) and call
2010 Nov 03
1
rgl.snapshot() : no longer works?
Hi all, > library(rgl) > plot3d(1,1,1) > snapshot3d("somefile.png") Error in rgl.snapshot(...) : pixmap save format not supported in this build Why does this no longer work? thanks, Remko > sessionInfo() R version 2.12.0 (2010-10-15) Platform: i386-pc-mingw32/i386 (32-bit) locale: [1] LC_COLLATE=English_Australia.1252 LC_CTYPE=English_Australia.1252 [3]
2012 Jan 21
2
4th corner analysis ade4 - what do the colors mean
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
2010 Sep 08
1
problem with max in a function
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
2012 Apr 25
1
fill a dataframe with zeros where the rows are a smaller subset of a larger dataframe (species by site)
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),
2012 Jul 24
2
Convert Package Interest?
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
2012 Jul 18
1
ggplot2 qplot pch not working anymore
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
2010 Jul 08
0
0 exit status on packages update try
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:
2010 Sep 15
1
approxfun returning value higher than I would expect
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,
2011 Jan 06
1
[zoo] - Individual zoo or data frames from non-continuous zoo series
#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.
2010 Jul 18
2
NA preserved in logical call - I don't understand this behavior because NA is not equal to 0
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),
2010 Jun 28
1
Zoo series to a date time stamp that is regular
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,
2013 May 01
3
Chron format question h:m not working
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
2010 Jul 15
1
loess line predicting number where the line crosses zero twice
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
2010 Jun 29
3
formating chron date times for printing
the date were created with chron with this argument format=c(dates="Y/m/d", times="H:M:S")) so I have the dates being displayed as (10/06/22 12:00:00) I would like to have them displayed as "2010-06-22 12:00:00" or "%Y-%m-%d %H:%M:%S" and then I can convert these for mergeing with another data frame x <- (structure(c(14464, 14464.0104166667,
2010 Aug 03
2
subset based on column names and then subset based on the inverse (grep?, or...)
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",