similar to: How to get a proportion of a Vector Member

Hi all, Suppose I have 2 matrices A and B. And I want to measure how good each of this matrix is. So I intend to compare A and B with another "gold standard" matrix X. Meaning the more similar a matrix to X the better it is. What is the common way in R to measure matrix similarity (ie. A vs X, and B vs X) ? - Gundala Viswanath Jakarta - Indonesia

Hi, I have the following vector which is created from 3 distinct distribution (three components) of gamma: x=c(rgamma(30,shape=.2,scale=14),rgamma(30,shape=12,scale=10),rgamma(30,shape=5,scale=6)) I want to plot the density curve of X, in a way that it shows a distinct 3 curves that represent each component. How can I do that? I tried this but doesn't work: lines(density(x)) Please

Hi Dieter, Sorry my mistake. I wanted to convert them into Decimal (not Hexadecimal). Given this string, the desired answer follows: > ascii_str <- "ORQ>IK" 79 82 81 62 73 75 > ascii_str2 <- "FDC" 70 68 67 - Gundala Viswanath Jakarta - Indonesia On Mon, Dec 22, 2008 at 5:49 PM, Dieter Menne <dieter.menne at menne-biomed.de> wrote: > Gundala

Hi, I have the following (M x N) matrix, where M = 10 and N =2 What I intend to do is to group index of (M) based on this condition of "x_mn" , namely For each M, If x_m1 > x_m2, assign index of M to Group1 otherwise assign index of M into Group 2 > x [,1] [,2] [1,] 4.482909e-01 0.55170907 [2,] 9.479594e-01 0.05204063 [3,] 8.923553e-01 0.10764474

Dear all, I have the following matrix. > dat A A A A A A A A A A [1,] 0 0 0 0 0 0 0 0 0 0 [2,] 0 0 0 0 0 0 0 0 0 1 [3,] 0 0 0 0 0 0 0 0 0 2 How can I change it into: [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] 0 0 0 0 0 0 0 0 0 0 [2,] 0 0 0 0 0 0 0 0 0 1

Hi, How can I extract the label only from a given data frame. Fore example from this data frame. > print(dataf) V1 V2 V3 V4 V5 V6 V7 V8 V9 11145 14.3 17.1 31.2 41.7 45.8 49.8 68.6 70.6 72.9 3545 10.2 15.6 20.9 23.2 31.4 31.7 36.2 48.4 51.9 8951 15.2 17.5 20.0 21.4 32.4

Dear all, Suppose I have this data frame: > data_main V1 V2 foo 13.1 bar 12.0 qux 10.4 cho 20.33 pox 8.21 And I want to split the data into two parts first part are the one contain in the source array: > src [1] "bar" "pox" and the other one the complement. In the end we hope to get this two dataframes: > data_child1 V1 V2 bar 13.1 pox

I have the following data frame: > foo w x y z n 1.51550092 1.4337572 1.2791624 1.1771230 q 0.09977303 0.8173761 1.6123402 0.1510737 r 1.17083866 1.2469347 0.8712135 0.8488029 What I want to do is to change it into : > newdf 1 n w 1.51550092 2 q w 0.09977303 3 r w 1.17083866 4 n x 1.43375725 5 q x 0.81737606 6 r x

Dear all, For example I want to process set of files. Typically Perl's idiom would be: __BEGIN__ @files = glob("/mydir/*.txt"); foreach my $file (@files) { # process the file } __END__ What's the R's way to do that? - Gundala Viswanath Jakarta - Indonesia

Dear all, The basename() function returns the extension also: > myfile <- "path1/path2/myoutput.txt" > basename(myfile) [1] "myoutput.txt" Is there any other function where it just returns plain base: "myoutput" i.e. without 'txt' - Gundala Viswanath Jakarta - Indonesia

Dear all, What's the R way to compress the string into smaller 2~3 char/digit length. In particular I want to compress string of length >=30 characters, e.g. ACGATACGGCGACCACCGAGATCTACACTCTTCC The reason I want to do that is because, there are billions of such string I want to print out. And I need to save disk space. - Gundala Viswanath Jakarta - Indonesia

Hi, How can I partitioned an example vector like this > print(myvector) [1] 30.9 60.1 70.0 73.0 75.0 83.9 93.1 97.6 98.8 113.9 into the following pairwise partition: PAIR1 part1 = 30.9 part2 = 60.1 70.0 73.0 75.0 83.9 93.1 97.6 98.8 113.9 PAIR2 part1 = 30.9 60.1 part2 = 70.0 73.0 75.0 83.9 93.1 97.6 98.8 113.9 .... PAIR9 part1 = 30.9

Hi all, Given a matrix: > mat [,1] [,2] [,3] [1,] 0 0 0 [2,] 3 3 3 [3,] 1 1 1 [4,] 2 1 1 How can I convert it to a list of strings: > desired_output [1] "aaa" "ttt" "ccc" "gcc" In principle: 1. Number of Column in matrix = length of string (= 3) 2. Number of Row in matrix = length of vector ( = 4). 3.

Dear all, Suppose I have a hash created with this x <- list() for (i in c('test', 'some', 'more')){ x[[i]] <- runif(1) } then I want to extract the elem of that hash with a vector > q <- c("some", "more", "not_there") But why this failed? > extracted <- x[[q]] Error in x[[q]] : subscript out of bounds we expect the

Hi all, I have the following named list: > print(y) $`200052_s_at` [1] -1066.975 -1063.893 -1062.815 -1062.121 -1059.004 $`200071_at` [1] -959.823 -953.980 -953.886 -948.781 -974.890 $`200084_at` [1] -1135.804 -1132.863 -1128.197 -1128.633 -1125.890 What I want to do is to iterate this name list and process its members. To do that I attempt the following code (but failed): __BEGIN__ ny

Dear all, I found that printing with 'cat' is very slow. For example in my machine this snippet __BEGIN__ # I need to resolve to use this type of loop. # because using write(), I need to create a matrix which # consumes so much memory. Note that "foo, bar, qux" object # is already very large (>2Gb) for ( s in 1:length(x) ) {

Hi, I have the following data frame: > print(mydatframe) __DATAFRAME__ V1 V2 V3 1 1007_s_at DDR1 discoidin domain receptor tyrosine kinase 1 2 1053_at RFC2 replication factor C (activator 1) 2, 40kDa 3 117_at HSPA6 heat shock 70kDa protein 6 (HSP70B') __END__ Is there a way to create a hash with V2 as Key and V3 as its value? - Gundala Viswanath Jakarta - Indonesia

Hi, Suppose I have the following data frame: V1 V2 V3 var 1 100 200 400 2.3 2 40.5 1.2 20.3 ... In this example the maximum V column is 3. In my code there the column number can be varied. My question is how can I access data frame from column V1 up to Vk (some "k" and excluding'var' column)? - Gundala Viswanath Jakarta - Indonesia

Hi, With the following codes, I attempt to convert the data.frame into a matrix. However I notice that data.matrix function doesn't seem to work. __ BEGIN__ dat <- read.table("mydata", comment.char = "!" , na.strings = "null"); # Select n-genes by random sample # n = 1 nosamp <- 1 geneid <- sequence(nrow(dat)) geneid.samp <- sample(geneid,nosamp)

Dear all, I have no problem with this individual grep command: > datk <- grep("XM_528056", source$V1) > dat2 <- source[datk,] > print(dat2) V1 V2 V3 V4 V5 V6 V7 35995 XM_528056 panTro2 chr8 + 1775569 1896107 Chimpanzee BUT, when I run them under the loop it gives this error: > hm_acc <-