similar to: Question regarding analysis of normalised data

Dear Sir/Madam, I could succesfully normalise my microarray data using marrayNorm package. However, i have not been able to get normalised red and green channel intensities through R package. Is there a possibility to write a formula to calculate back the red and green channel intensities after normalisation of the data. Do I need to incorporate this formula in my R script? I am biologist

Hello list! I would like to fit a distribution to each of the peaks in a histogram, such as this: http://photos1.blogger.com/blogger/7029/2724/1600/DU145-Bax3-Bcl-xL.png . The peaks are identified using Petr Pikal peaks function ( http://finzi.psych.upenn.edu/R/Rhelp02a/archive/33097.html), but after that I am quite stuck. Any idea as to how I can: Fit a distribution to each peak Integrate the

Dear List, My data consist of nine columns and about 50,000 rows. It looks like this. -9.0225 3.46464 2.80926 -0.3847 3.73735 1.1058 -2.98936 1.38901 -8.1846 -2.4315 -5.1189 1.8225 3.3798 1.7874 4.693 -3.9286 1.4266 5.7849 -3.4894 -4.0305 3.7879 3.5195 2.9186 2.8685 -6.126 4.978 4.9381 4.5282 3.62558 -3.0455 4.6518 1.39746 0.68652 3.5708 -3.6404 -4.2963 -1.3183 0.6752 -4.0382 -2.5386

Hi all, I have a data frame with some measured values of some animals. Sometimes the measurement failed, resulting in a NA for a measurement and sometimes the animal died, resulting in NA for all measurements. I have several groups of animals. How do I find the size of each group with only alive animals? And how do I find the size of the groups for each measurement? An example: l1 <-

Hello, I have a data frame with different groups (Time, Target, Conc) and each entry has a triplicate value of the measurements OD and ODnorm. How can I merge the triplicates into a single mean value? I tried the following: ``` df = data.frame(Time=rep(1, 9), Well=paste("A", 1:9, sep=""), OD=c(666, 815, 815, 702, 739, 795, 657, 705, 663),

Hi, I have taken one microarray experiment and trying to implement same statistical measures what they have done.I have taken datasets from GEO platform with accession number GSE1557. In the experiment,about half of double transgenic rats (dTGR) over-expressing the human renin and angiotensinogen genes die by age 7 weeks of terminal heart failure (THF); the other (preterminal) half develops

On Mon, 2007-05-14 at 23:41 +0200, vax9000 at gmail.com wrote: > Full_Name: vax, 9000 > Version: 2.4.0, 2.2.1 > OS: 2.4.0: Mac OS X; 2.2.1: Linux > Submission from: (NULL) (192.35.79.70) > > > To reproduce this bug, first go to the website "http://llmpp.nih.gov/DLBCL/" and > download the 14.8M data set "Web Figure 1 Data file". The direct link is >

Hi, I think you're misunderstanding which set of variables go on either side of the formula. Is this what you're looking for? > aggregate(OD ~ Time + Target + Conc, data = df, FUN = "mean") Time Target Conc OD 1 1 BACT 1 765.3333 2 1 BACT 2 745.3333 3 1 BACT 3 675.0000 > aggregate(ODnorm ~ Time + Target + Conc, data = df, FUN =

Thanks, I think I've found the most succinct expression of differences in two data.frames... length(which( rowSums( x1 != x2 ) > 0)) gives a count of the # of records in two data.frames that do not match. // ________________________________________ From: Henrik Bengtsson [henrik.bengtsson at gmail.com] Sent: Sunday, January 28, 2018 11:12 AM To: Ulrik Stervbo Cc: Marsh Hardy ARA/RISK;

The diffobj package (https://cran.r-project.org/package=diffobj) is really helpful here. It provides "diff" functions diffPrint(), diffStr(), and diffChr() to compare two object 'x' and 'y' and provide neat colorized summary output. Example: > iris2 <- iris > iris2[122:125,4] <- iris2[122:125,4] + 0.1 > diffobj::diffPrint(iris2, iris) < iris2 >

Hello Thomas, Ulrik, thanks for your suggestions. Giovanni On Fri, Aug 4, 2017 at 12:13 PM, Thomas Mailund <thomas.mailund at gmail.com> wrote: > Do you mean like this? > > >> f <- function(foo, bar) { > + result <- list(bar) > + names(result) <- foo > + result > + } > >> (x <- f("hello", "world")) > $hello >

Dear R Users, Are there any packages in R which carries out a normalisation to variables as follows: - find the empirical distribution function, using perhaps ecdf - use the empirical distribution function to transform the variables into a series between 0 and 1 - use this series to map the variables into the normal distribution function, using qnorm - perform a regression on the transformed

Also, > aggregate(cbind(OD, ODnorm) ~ Time + Target + Conc, data = df, FUN = "mean") Time Target Conc OD ODnorm 1 1 BACT 1 765.3333 108.33333 2 1 BACT 2 745.3333 88.33333 3 1 BACT 3 675.0000 18.00000 (You might wish for "cbind(OD,ODnorm) ~ . - Well", but aggregate.formula is not smart enough for that.) -pd > On 24 Oct 2023, at

Please keep the list in cc. Sorry, it didn't work as expected. Maybe someone else have an appropriate solution. Best, Ulrik On Sa., 26. Aug. 2017, 12:57 niharika singhal <niharikasinghal1990 at gmail.com> wrote: > Hi > > Thanks for you mail, > I really appreciate your time on my problem > > I have posted this problem on > > >

Hello all, I have some histograms of amount of DNA in some cells (DU145 cells overexpressing Bax and Bcl-xL for those who wish to know). The histograms show not only two peaks as expected, but three, indicating that some cells have more than normal amounts of DNA. I am interested in knowing how much of the cell populations are in each peak as well as between. I am not really sure how to go

The anti_join from the package dplyr might also be handy. install.package("dplyr") library(dplyr) anti_join (x1, x2) You can get help on the different functions by ?function.name(), so ?anti_join() will bring you help - and examples - on the anti_join function. It might be worth testing your approach on a small subset of the data. That makes it easier for you to follow what happens

Cool, looks like that'd do it, almost as if converting an entire record to a character string and comparing strings. -- M. B. Hardy, statistician work: Applied Research Associates, S. E. Div. 8537 Six Forks Rd., # 6000 / Raleigh, NC 27615-2963 (919) 582-3329, fax: 582-3301 home: 1020 W. South St. / Raleigh, NC 27603-2162 (919) 834-1245

The object you load has the same name as the object you saved. In this case datahs0csv and not the name of the file sans .rda On Di., 12. Sep. 2017, 21:26 AbouEl-Makarim Aboueissa < abouelmakarim1962 at gmail.com> wrote: > Dear All: > > > It was saved, but there was a space somewhere. So it works for me now. > > I do have another similar problem. > > I saved an R

You can wrap the list-creating function call (e.g. lapply) in a call to ?setNames, or you can use the ?map function from the purrr package. -- Sent from my phone. Please excuse my brevity. On August 4, 2017 3:14:44 AM PDT, Ulrik Stervbo <ulrik.stervbo at gmail.com> wrote: >Hi Giovani, > >I would create an unnamed list and set the names after. > >Best, >Ulrik > >On

Dear All: It was saved, but there was a space somewhere. So it works for me now. I do have another similar problem. I saved an R data file save(datahs0csv,file=" F:\Fall_2017\5-STA574\2-Notes\1-R\1-R_new\chapter4-Entering_Data/datahs0csv2 .rda") *The new R data file "*datahs0csv2.rda*" is in the directory.* I tried to load the file "" to R, but I got an error