similar to: Problem in converting natural numbers to bits and others

Hello all, I have what feels like a simple problem, but I can't find an simple answer. Consider this data frame: > x <- data.frame(sample1=c(35,176,182,193,124), sample2=c(198,176,190,23,15), sample3=c(12,154,21,191,156), class=c('a','a','c','b','c')) > x sample1 sample2 sample3 class 1 35 198 12 a 2 176 176

This is what I am starting with: initial<- matrix(c(1,5,4,8,4,4,8,6,4,2,7,5,4,5,3,2,4,6), nrow=6, ncol=3,dimnames=list(c("1900","1901","1902","1903","1904","1905"), c("sample1","sample2","sample3"))) And I need to apply a filter (in this case, any value <5) to give me one dataframe with only the

Hi Group, I have a data frame below. Within this data frame there are samples (columns) that are measured more than once. Samples are indicated by "idx". So "id1" is present in columns 1, 3, and 5. Not every id is repeated. I would like to create a new data frame so that the repeated ids are averaged. For example, in the new data frame, columns 1, 3, and 5 of the original

Hi all, I am processing 24 samples data and combine them in single table called CombinedSamples using following: CombinedSamples<-rbind(Sample1,Sample2,Sample3) Now variables Sample1, Sample2 and Sample3 have many different columns. To make it more flexible for other samples I'm replacing above code with a for loop: #Sample is a string vector containing all 24 sample names for (k in

Dear experts, Please forgive the puzzled title and the length of this message - I thought it would be best to be as complete as possible and to show the avenues I have explored. I'm trying to fit a linear model to data with a binary dependent variable (i.e. Target.ACC: accuracy of response) using lrm, and thought I would start from the most complex model (of which "sample1.lrm1" is

Hello R list members, I'm a bio informatics student from the Leiden university (netherlands). We were asked to make a program with different clustering methods. The problem we are experiencing is the following. we have a matrix with data like the following research1 research2 research3 enz sample1 0.5 0.2 0.4 sample2 0.4

Hello, I have a data frame with a factor column, which uniquely identifies the observations in the data frame and it looks like this: sample1_condition1_place1 sample2_condition1_place1 sample3_condition1_place1 . . . sample3_condition3_place3 I want to turn it into three separate factor columns "sample", "condition" and "place". This is what I did so far: #

Dear group, I have a matrix like the following: Name Sample1 sample2 sample3 sample4 ..... sample(n) nm1 10.5 13.5 30 31 nm2 8 11 34 29 nm3 9 10.3 27.8 35 nm(j) I want to be able to calculate correlation between all pairs of names. For example (nm1,nm2),

Hi all, As a molecular biologist by training, I'm fairly new to R (and statistics!), and was hoping for some advice. First of all, I'd like to apologise if my question is more methodological rather than relating to a specific R function. I've done my best to search both in the forum and elsewhere but can't seem to find an answer which works in practice. I am carrying out

I am wanting to plot a 95% confidence band using segplot, yet I am wanting to have groups. For example if I have males and females, and then I have them in different races, I want the racial groups in different panels. I have this minor code, completely made up but gets at what I am wanting, 4 random samples and 4 samples of confidence, I know how to get A & B into one panel and C&D in to

Dear Dr.Goslee and anyone may intrested in matrix manipulate, I am using your ecodist to do mantel and partial mantel test, I have locality data and shape variation data, and the two distance matrixs are given as belowings. When I run the analysis, it is always report that the matrix is not square, but I didn't know what's wrong with my data. Would you please help me on this. I am quite

User has to type (input) x. After input....code has to check X with statement if There is a statement IF. If X=0 then Y=5/2 else =7; How to code it please. I tried but my code does not work ;( I wanna see how it looks like( thank u in advance( -- View this message in context: http://r.789695.n4.nabble.com/Help-me-please-to-code-tp4646932.html Sent from the R help mailing list archive at

Hi, I'm looking for some ideas on how to reproduce the attached image in R. There are three samples, each of size n = 10. The first is drawn from a normal distribution with mean 60 and standard deviation 3. The second is drawn from a normal distribution with mean 65 and standard deviation 3. The third is drawn from a normal distribution with mean 70 and standard deviation 3.

Dear All, I am analyzing the miRNA data set in which I have 817 unique probes for each they have 20 features each . I have to group the similar features and take the median across them so that I have a data with no repeats to perform invariant analysis . My data looks something similar format probename sample1 sample2 sample3 A 2.3 2.4 2.5 A

I have to perform ANOVA's on many different data organized in a dataframe. I can run an ANOVA for each sample, but I've got hundreds of data and I would like to avoid manually carrying out each test. in addition, I would like to have the results organized in a simple way, for example in a table, wich could be easy to export. thank you for assistance simone

Hello Everyone, I am writing programs in R from 7 months and I am able to solve most of the errors/issues except for this current post. My Task is to read a Microsoft Excel file(textE_to_affy.csv) which contains the Microarray Expression Values collected from the Illumina Microarray experiment. These collected intensity values need to be normalized(Rank Invariant Normalization) by using the R

## you may need to install HH install.packagess("HH") library(HH) hellisheidi <- read.table(text=" Component Sample1 Sample2 Sample3 CaO 45 52 48 SiO2 25 22 18 Al2O3 15 11 14 TiO2 6 5

Hi group, This is how my test file looks like: Chr start end sample1 sample2 chr2 9896633 9896683 0 0 chr2 9896639 9896690 0 0 chr2 14314039 14314098 0 -0.35 chr2 14404467 14404502 0 -0.35 chr2 14421718 14421777 -0.43 -0.35 chr2 16031710 16031769 -0.43 -0.35 chr2 16036178 16036237 -0.43 -0.35 chr2 16048665 16048724 -0.43 -0.35 chr2 37491676 37491735 0 0 chr2 37702947 37703009 0 0

Dear List Members, I have two small samples (n=20), the distributions are highly skewed. Does it make any sense to do a boostrap test to check for difference in means? And if so, could this be done like this: x <- numeric(10000) for(i in 1:10000) { x[i] <- mean(sample(sample1,replace=TRUE)) - mean(sample(sample2,replace=TRUE)) } (mean(sample1)-mean(sample2))/sd(x) Regards, Erika

Dear Richard,? Thank you very much for your reply. I went through the code and it worked. I was also able to change the colours.? I was wondering if I can change the legend position; instead of being in the bottom to be on the left side.? I tried the following but without any success strip = FALSE strip.right = TRUE likert(t(hellisheidi), ReferenceZero=.5, xlab="X-lab",