similar to: Is there a neat R trick for this?

I looked at the documentation of source() and summary(), and I could not find the reason why calling something like: > summary(resamps) from the command line, works (it prints the summary) whereas calling summary(resampls) from a file that I source with source("my_file.r") does not print anything. How can I get summary(resamps) to print when I source a file with this command?

Hi all, I've simple code to read a file (verify.txt in the same directory as the script file) but when I run this I get "Error in eval(expr, envir, enclos) : object "y" not found". data_model.df = read.table("./verify.txt", header=TRUE, nrows=10); f <- lm(y ~ x) Could someone pls tell me what's wrong with this code? Sincere thanks!

I am very new to Linux so I probably am doing something stupid but I cannot seem to update to R 2.15 Using Ubuntu 12.02 Precise Penguin I realise that debian packages are not updated regularly so I tried to follow the insructions at the R-site So far, I have modified /etc/apt/sources.list to read ## R CRAN added 2012-06-12 deb http://probability.ca/cran/bin/linux/debian squeeze-cran/ Issued

Hi All I have a vector of about 15 million numbers which I would like to plot. The goal is the see the distribution. I tired the usual steps. 1. Histogram : never gets complete my window freezes w/out log base 10 2. Density : I first calculated the kernel density and then plotted it which worked. It would be nice to superimpose histogram with density but as of now I am not able to get this

What is the simplest way to specify the location of text in a scatter plot (created using the plot function) in relative terms rather than specific x-y coordinates? For example, rather than putting text at (300,49) on a plot, how do I put it 1/10 of the way over from the y axis and 1/2 of the way up from the x axis? Thanks. Tom -- View this message in context:

This may be a really obvious question but I just can't figure out how to do it. I have a small dataset that I am trying to compare to some controls. It is essential that the controls are matched on Cancer Stage (a numerical factor between 1 and 4), and then ideally on Age (integer), Gender (factor), Performance Status(factor). I'm using matchit to try and do this, but it seems to give

Dear Group I have few errors while installing package rattle from CRAN i do the installing from the local zip files... I am using R 2.4.0 do i have to upgrade to R2.4.1 ? ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ utils:::menuInstallLocal() package 'rattle' successfully unpacked and MD5 sums checked updating HTML package descriptions > help(rattle) No

R 2.15.1 OS 10.7 Colleagues I have been an enthusiastic user of the sas7bdat package in R. However, several recent sas7bdat files sent to me from different sources cannot be read by the package. The error message is: Error in read.sas7bdat(FILENAME) : unknown host W32_7PRO please report bugs to sas7bdatRbugs at gmail.com I examined the file with a text editor and it contains

Hello. I am using some dates I read in excel in R. I know the excel origin is supposed to be 1900-1-1. But when I used as.Date with origin=1900-1-1 the dates that R reported me where two days ahead than the ones I read from Excel. I noticed that when I did in R the following: > as.Date("2011-3-4")-as.Date("1900-1-1") Time difference of 40604 days but if I do the same

Hello, How can I apply a function on a vector that refers to actual (n) and previous elements in the vector (e.g. n-1)? For example: I would like to calculate the sum of (n-1) + n for each element of a vector and get a vector as a result. Besides others I tried this: v<-c(3,6,8,1,1,3,9,5,6,3) for (i in 1:NROW(v)){a[i]<-a[i-1]+a[i]} I would like to get this result:

If I have two matrices like x <- matrix(rep(c(1,2,3),3),3) y <- matrix(rep(c(4,5,6),3),3) How can I combine them to get ? 1 1 1 4 4 4 1 1 1 5 5 5 1 1 1 6 6 6 2 2 2 4 4 4 2 2 2 5 5 5 2 2 2 6 6 6 3 3 3 4 4 4 3 3 3 5 5 5 3 3 3 6 6 6 The number of rows and the actual numbers above are unimportant, they are given so as to illustrate how I want to combine the matrices. I.e., I am looking for

Dear All, I am not familiar with R yet I want to use it to perform some task, hence my posting here. I hope someone can help. I have a set of data, genes (rows) and samples (columns). I want to do a Pearson correlation on all the possible pairwise combinations of all the genes (2000). Does anyone have an idea of how to execute this in R? Thanks in advance. -- View this message in context:

Not terribly exciting, but I was happy to think of it. I needed a div with a fixed height that could vary depending on page content. So I did this: <div class="listcntnr" style="height: <%= @users.size / 2 * 125%>px"> Whee! --Al Evans -- Posted via http://www.ruby-forum.com/.

On Wed, 13 Dec 2023 09:04:18 +0100 Hilmar Berger via R-devel <r-devel at r-project.org> wrote: > Still, I feel that default partial matching cripples the functionality > of data.frame for larger tables. Changing the default now would require a long deprecation cycle to give everyone who uses `[.data.frame` and relies on partial matching (whether they know it or not) enough time to

Hi folks, Can anyone suggest an efficient way to do "matching without replacement", or "one-to-one matching"? pmatch() doesn't quite provide what I need... For example, lookupTable <- c("a","b","c","d","e","f") matchSample <- c("a","a","b","d") ##Normal match()

The pmatch help (see also section 4.3.2 in the R Language Definition) claims that pmatch with duplicates.ok=FALSE provides the same functionality as R's argument matching algorithm, modulo how empty strings are matched. Here's an undocumented inconsistency between pmatch and R's argument matching algorithm: > sessionInfo() R version 3.0.1 (2013-05-16) Platform:

Dear Ivan, thanks a lot, that is helpful. Still, I feel that default partial matching cripples the functionality of data.frame for larger tables. Thanks again and best regards Hilmar On 12.12.23 13:55, Ivan Krylov wrote: > ? Mon, 11 Dec 2023 21:11:48 +0100 > Hilmar Berger via R-devel <r-devel at r-project.org> ?????: > >> What was unexpected is that in this case was that

Hi Hilmar and Ivan, I have used your code examples to write a blog post about this topic, which has figures that show the asymptotic time complexity of the various approaches, https://tdhock.github.io/blog/2023/df-partial-match/ The asymptotic complexity of partial matching appears to be quadratic O(N^2) whereas the other approaches are asymptotically faster: linear O(N) or log-linear O(N log N).

Oops, I just erased all my data using this gizmo that I thought would replace -9 with NA. A) Can I get my tcn5 back? B) How do I do it right next time, I learned my lesson, I'll never do it again, I promise! Anders Corr > for(i in 1:dim(tcn5)[2]){ ##for the number of columns + for(n in 1:dim(tcn5)[1]){ ##for the number of rows + tcn5[is.na(tcn5[n,i]) | tcn5[n,i]

Hi, A noobie question: I'm simply trying to run a conditional statement that evaluates if a substring is found within a larger string. I find that if it IS found, my function returns TRUE (great!), but if not, the condition does not evaluate to FALSE. ex): if( grep("hi", "hop", fixed = TRUE) ) print('yes, your substring is in your string') else