search for: sample3

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 2 176 176 154 a 3 182 190 21 c 4 193 23 191 b 5 124 15 156 c Now I...

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 'less than 5' values and with '0' place holders on the values that don't meet this criteria. The second dataframe is the same but for those values >=5. Should look l...

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 1:length(Sample)) { CombinedSamples<-rbind(get(Sample[k]...

Hi, I just started using R for about one week and I have few problems. i)I have a problem in finding right function to convert a table of natural numbers to bitwise. For a simple example; I have the below table:- Column Col1 Col2 Col3 Sample1 5 7 10 Sample2 0 2 1 Sample3 4 0 0 Supposedly i wanted to convert to :- Column Col1 Col2 Col3 Sample1 1 1 1 Sample2 0 1 1 Sample3 1 0 0 ii)Besides, I would like to create a formula of Sum(3*(Element in each column for a row))..does it equivalent to 3*(data[1:3,1:2...

...gram 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 0.4 0.3 sample3 0.7 0.2 0.8 enz now if we use kmeans(matrix,3,20) the clustering method will cluster only on rows by using the columns values as multiple variables. The output from this syntax is for example sample1 = 1 sample2 = 2 sample3 = 3 enz Is there a method that will use...

...r %U sys %S" ./ellbzip2 -1 < sample1.ref > sample1.rb2 echo -n "2:"; /usr/bin/time -f "real %e user %U sys %S" ./ellbzip2 -2 < sample2.ref > sample2.rb2 echo -n "3:"; /usr/bin/time -f "real %e user %U sys %S" ./ellbzip2 -3 < sample3.ref > sample3.rb2 echo -n "4:"; /usr/bin/time -f "real %e user %U sys %S" ./ellbzip2 -d < sample1.bz2 > sample1.tst echo -n "5:"; /usr/bin/time -f "real %e user %U sys %S" ./ellbzip2 -d < sample2.bz2 > sample2.tst echo -n "6...

..., Y, penalty.matrix = penalty.matrix, tol = tol, weights = weights, : NA/NaN/Inf in foreign function call (arg 1) But I cannot find any NA in the data: > table(complete.cases(sample2)) TRUE 500 Some portions of the data don't appear to contain any of the offending "bit": > sample3 <- sample1[12500:13000,] > sample3.lrm = lrm(Target.ACC ~ (Target.RESP + Word.Order)^2, sample3, x=T, y=T) Could one of your shine your light on this puzzle, please? If that includes pointing me towards some background reading, that would be great too. Many thanks in advance. Cecile De Ca...

...[,6] [,7] [,8] [,9] [,10] Sample1 0.7329881 0.76912670 2.45906143 -0.06411602 1.2427801 0.3785717 2.34508664 1.1043552 -0.1883830 0.6503095 Sample2 -2.0446131 1.72783245 -0.40965941 -0.21713655 -0.2386781 -0.1944390 -0.25181541 0.8882743 0.8404783 0.2531209 Sample3 1.7575174 0.03851425 -0.87537424 1.82811160 0.8342636 -0.8155942 -0.90893068 -0.5529098 -1.6586626 0.1761717 Sample4 0.2731749 0.24045167 -0.39913821 -0.48525909 0.1448994 -2.0173360 -0.01073639 -0.3219478 -1.1536431 -0.2521545 Sample5 -0.2014241 -1.04646151 0.28101160 0.74348390 0.17...

...nd C&D in to another but how do I get the x axis to label them properly and have it categorized as two. I am not sure what to have to the left side of the formula. This is the example code: library(lattice) library(latticeExtra) sample1<-rnorm(100,10,2) sample2<-rnorm(100,50,3) sample3<-rnorm(100,20,2) sample4<-rnorm(100,40,1) mu1<-mean(sample1) ci.upper1<-mu1+2*2 ci.lower1<-mu1-2*2 mu2<-mean(sample2) ci.upper2<-mu2+2*3 ci.lower2<-mu2-2*3 mu3<-mean(sample3) ci.upper3<-mu3+2*2 ci.lower3<-mu3-2*2 mu4<-mean(sample4) ci.upper4<-mu4+2...

Dear R experts, code: >m<-read.table("test.txt",sep='\t',header=TRUE,colClasses=c('character','integer','integer','rep('numeric',150)) > s<-data.frame(c(rle(m$Sample1)[[2]],rle(m$Sample2)[[2]],rle(m$Sample3)[[2]]),c(rle(m$Sample1)[[1]],rle(m$Sample2)[[1]],rle(m$Sample3)[[1]])) > names(s)=c("Values","Probes") #Suppose the test file looks like with ofcourse more samples with values: Chr Start End Sample1 Sample2 Sample3 chr2 9896633 9896683 0 0 0 chr2 9896639 9896690 0 0 0 chr2...

...8614 Loc11 3180.844 2407.1921 2818.8222 10046.113 1983.5246 2092.9254 8568.467 3518.9508 1182.826 6572.264 0.000 3647.8307 Loc12 3112.680 1314.6927 1063.8092 10253.957 1670.4862 1558.8962 8654.392 631.5406 3397.490 7497.861 3647.831 0.0000 > distancematrix sample1 sample2 sample3 sample4 sample5 sample6 sample7 sample8 sample9 sample10 sample11 sample12 sample1 0.0000 0.0229 0.0258 0.0394 0.0295 0.0337 0.0269 0.0345 0.0314 0.0418 0.0577 0.0853 sample2 0.0229 0.0000 0.0219 0.0373 0.0337 0.0352 0.0320 0.0310 0.0275 0.0508 0.0533 0.0739 sample3...

..., and 5 of the original will be replaced by 1 new column that is the mean of these three. Thanks for any suggestions. Juliet myData <- data.frame("sample1.id1" =rep(1,10), "sample1.id2"=rep(2,10), "sample2.id1" = rep(2,10), "sample1.id3" = 1:10, "sample3.id1" = rep(1,10), "sample1.id4" = 1:10, "sample2.id2" = rep(1,10)) repeat_ids <- c("id1","id2")

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...

...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 8.9 3.6 7.9 A 4.7 4.4 2.2 C 2.2 2.1 2.3 C 1.3 1.3 2.9 B 2.2 2.1 2.7 B 2.2...

...ute) > example(impute.knn) > str(khan.imputed) num [1:2308, 1:63] 0.773 -2.438 -0.483 -2.721 -1.217 ... - attr(*, "dimnames")=List of 2 ..$ : chr [1:2308] "2" "3" "4" "5" ... ..$ : chr [1:63] "sample1" "sample2" "sample3" "sample4" ... - attr(*, "rng.seed")= num 3.62e+08 - attr(*, "rng.state")= int [1:626] 403 624 1638542565 108172386 -1884566405 -1 838154368 -250773631 919185230 -1001918601 -1002779316 ... impute v1.14.0: > library(impute) > example(impute.knn) > str(...

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: # generate a factor column for the example fctr<- factor(c("sample1_condition1_place1",...

## 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 6 Na2O 5 4 5...

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.

...convert the intensity values present in the (textE_to_affy.csv) file into an AffyBatch Object as follows: > testFile <- tempfile() >textAffy<-read.table("textE_to_affy.csv",header=TRUE,sep=",",row.names=1) > textAffy Sample1 Sample2 Sample3 GI_10047089-S -6.100 -5.12500 -5.61250 GI_10047091-S 10.725 9.70625 10.21562 GI_10047093-S 1392.100 1378.70000 1385.40000 GI_10047099-S 264.925 260.98125 262.95312 GI_10047103-S 5315.675 5412.01875 5363.84688 GI_10047105-S 21.750 22.53750 22.14375 > write.table(TA,testFile,...

...d regards, Maria ???? ????? 2 ????? 2023 ???? 08:51:16 ?.?. GMT+1, ? ??????? Richard M. Heiberger <rmh at temple.edu> ??????: ## 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? ? ? ? ? ? ? ? 6 Na2O? ? ? ? ? ? ? ? ? ? 5? ? ? ? ? ? ? ? ? 4? ? ? ? ? ? ?...