similar to: Opening connection to FTP site

Displaying 20 results from an estimated 10000 matches similar to: "Opening connection to FTP site"

2003 Jul 25
5
R won't connect to the internet on SUSE Linux 8.1
Hi Thanks once again for your help, I do appreciate it..... however.... Here is what I get with your test.... (under tcsh - i normally use bash, but I will keep everything the same) users/mwatson> env http_proxy=http://wwwcache.bbsrc.ac.uk:8080/ R >options(internet.info=0) >update.packages() trying URL `http://cran.r-project.org/src/contrib/PACKAGES' unable to connect to
2003 Jul 24
3
R won't connect to the internet on Linux!
OK, I really am struggling with this one! Forgive me if I am being stupid.... I am running R 1.7.1 on Suse Linux 8.1. I connect to the internet through a proxy so I have: IAHC-LINUX03:~ # echo $http_proxy wwwcache.bbsrc.ac.uk:8080 IAHC-LINUX03:~ # echo $HTTP_PROXY wwwcache.bbsrc.ac.uk:8080 just in case ;-) SO, i go into R and I get: >
2004 May 19
7
Help with hclust() and plot()
Hi When I use plot(hclust(dist..)...)...) etc to create a dendrogram of a hierarchial cluster analysis, I end up with a vertical tree. What do I need to do to get a horizontal tree? Also, my users are used to seeing trees who's leaves all "end" at the same place (eg. Like in minitab). Is this possible in R? Thanks Mick Michael Watson Head of Informatics Institute for Animal
2004 Nov 16
3
Simple plot() question
Hi First a simple question to do with plot(). How do I change the x-axis labels on a plot? For example, I am plotting each row of a matrix, and I want each row to be a line on my graph. Simple really. Eg: plot(sg[1,], type="l") When I do this, the x-axis is labelled 1:38, as I have 38 columns in my matrix. When I do: plot(sg[1,order(sg[1,])], type="l") Ideally I would
2003 Jul 18
4
R won't connect to the internet
Hi I can't get R to connect to the internet. I am running R 1.7.1 on Windows XP and whenever I try to download packages etc from within R using the internet, it fails. OK so I am behind a firewall and use a proxy server.... SO, if I go to my MS-DOS prompt and type: RGui.exe --internet2 everything works.... BUT i can't set up a shortcut for this as Windows (I hate Windows) complains
2005 Feb 03
5
How to convert a list to a matrix
Hi Sorry to ask such a basic question. I have a list, each element of which is a vector of two values. What I actually want is a matrix with two columns, and one row per element of the list. Obviously I have tried as.matrix(), and as.vector() but I didn't expect the latter to work. I feel so lame asking this. Any suggestions? Mick
2005 Feb 17
3
A vector or array of data frames
Hi A simple question again, but I can't find it by google-ing R-help. Quite simply, I want to read in the contents of a number of files, using read.table, and assign the results to elements of a vector/array/list/whatever. I want it so that, if my vector/array/whatever is "pos", that pos[1] will give me the first data frame, pos[2] will give me the second etc... Kind of basic
2006 Mar 06
4
Averaging over columns
Hi I've been reading the help for by and aggregate but can't get my head round how to do this. I have a data frame - the first three columns are replicate measurements, then the next 3 are replicates etc up to 36 (so 12 variables with 3 replicate measurements each). I want to compute the mean for each of the 12 variables, so that, for each row, I have 12 means. A grouping variable
2005 Jan 25
2
Rd problems when converting DVI version
Hi Running R v2.0 on SuSe linux 8.2. I'm trying to build a package (which built perfectly on Windows...) on Linux, and I ran: R CMD check mypackage I got: * checking mypackage-maual.tex ... ERROR LaTeX errors when creating DVI version This typically indicates Rd problems OK, there are no problems with my Rd - I got 3 warnings but they were all expected, and the whole package builds find
2005 Feb 17
6
Converting a list to a matrix - I still don't think I have it right
Hi We have touched on this before, but I don't think I quite got it right. So I have a list, each element of which is a a vector of 2 numbers: > l2 $cat000_a01 [1] 0.3429944 4.5138244 $cat000_a02 [1] 0.1929336 4.3064944 $cat000_a03 [1] -0.2607796 4.1551591 What I actually want to convert this into is a matrix with the names (cat000_a01 etc) as row names, the first element of each of
2006 Nov 28
2
Sorting a data frame when you don't know the columns
Hi Sorry to ask such a well oiled question, but even with multiple google hits I don't think this has been answered very well. It's all well and good doing a sort of a data frame on multiple columns when you know in advance which columns you want to sort on, but what about when the names of the columns you wish to sort on are in a vector? At the minute I'm messing about with
2006 Mar 06
2
Problems with heatmap.2 in the gregmisc package
Hi Sorry to revisit an old problem, I seemed to solve this in 2004, only for it to resurface :-S I am trying to plot a heatmap, and I don't want the columns of my matrix re-ordered. The function doesn't seem to behave as the help would have you believe: a <- matrix(rnorm(100),nr=20) a.d <- dist(a) a.hc <- hclust(a.d) a.de <- as.dendrogram(a.hc) # columns are re-ordered
2004 Nov 16
5
Difference between two correlation matrices
Hi Now a more theoretical question. I have two correlation matrices - one of a set of variables under a particular condition, the other of the same set of variables under a different condition. Is there a statistical test I can use to see if these correlation matrices are "different"? Thanks Mick
2005 Jan 14
5
Replacing NAs in a data frame using is.na() fails if there are no NAs
Hi This is a difference between the way matrices and data frames work I guess. I want to replace the NA values in a data frame by 0, and the code works as long as the data frame in question actually includes an NA value. If it doesn't, there is an error: df <- data.frame(c1=c(1,1,1),c2=c(2,2,NA)) df[is.na(df)] <- 0 df df <- data.frame(c1=c(1,1,1),c2=c(2,2,2)) df[is.na(df)] <-
2006 Feb 23
4
Changing the x-axis labels in plot()
Hi Hopefully this one isn't in the manual or I am about to get shot :-S One of my colleagues wants a slightly strange graph. We basically have a data matrix, and she wants to plot, for each row, the values in the row as points on the graph. The following code draws the graph just fine: plot(row(d)[,3:9],d[,3:9]) So as there are 12 rows in my matrix, there are 12 columns of points, which
2008 Aug 05
1
Greek characters in plots
Hi I am running an R script that creates 100s of graphs, and I need to use the greek CAPITAL letter delta in the mtext() function. I got as far as expression(delta) but this gives me the lowercase version. Can anyone help? I'm using R 2.7 on Windows XP Mick Head of Informatics Institute for Animal Health Compton Berks RG20 7NN 01635 578411
2006 Mar 10
2
lapply and list attributes
Hi I have a list that has attributes: attributes(lis[2]) $names [1] "150096_at" I want to use those attributes in a function and then use lapply to apply that function to every element of the list, eg for simplicity's sake: my.fun <- function(x) { attributes(x) } Then l2 <- lapply(lis, my.fun) It seems that "attributes(x)" within the function is not the
2006 Oct 17
2
Finding out about objects and classes
When R help simply states something like: Value: An object of class '"loess"'. How do I find out more about that class? Shouldn't there be a link in the help file or something? ATB Mick
2005 Feb 10
3
Using a number as a name to access a list
Hi Dumb question time again, for which I apologise. I have a variable that contains the following numerical text "04010". This is the name to access a list: > as.list(KEGGPATHID2NAME)$"04010" [1] "MAPK signaling pathway" Marvellous! Except I want to do that when "04010" is assigned to a variable called path and I can't figure out how to do it!
2004 Dec 10
4
cbind() and factors.
Hi I'm seeing some "odd" behaviour with cbind(). My code is: > cat <- read.table("cogs_category.txt", sep="\t", header=TRUE, quote=NULL, colClasses="character") > colnames(cat) [1] "Code" "Description" > is.factor(cat$Code) [1] FALSE > is.factor(cat$Description) [1] FALSE > is.factor(rainbow(nrow(cat))) [1]