similar to: Converting Factor to Vector

Dear all, I tried to find index in repo given a query with this: > repo <- c("AAA", "AAT", "AAC", "AAG", "ATA", "ATT") > qr <- c("AAC", "ATT", "ATT") > which(repo%in%qr) [1] 3 6 Note that the query contain repeating elements, yet the output of which only returns unique. How can I make it

Dear all, I have a repository file (let's call it repo.txt) that contain two columns like this: # tag value AAA 0.2 AAT 0.3 AAC 0.02 AAG 0.02 ATA 0.3 ATT 0.7 Given another query vector > qr <- c("AAC", "ATT") I would like to find the corresponding value for each query above, yielding: 0.02 0.7 However, I want to avoid slurping whole repo.txt

Hi all, I have the following data AAA 1e+06 1222.312 AAC 100.2 0.33 However the following command dat <- read.table("mydat.txt"); print(dat) n1 <- sum(as.matrix(dat$V1)); gives: Error in sum(as.matrix(dat$V1)) : invalid 'type' (character) of argument What's wrong with the data type so that it gives such error? - Gundala Viswanath Jakarta - Indonesia

Hey guys, I have what i think is a really simple problem :( I installed the seqinr library. I want to do an RSCU analysis. But i can't get it to work in even the simplest case. for example, if i have a string read in: > newdata5 $testseq [1] "agtgagatgatagatagatagatagatagatagatagaccccccagata" and then i perform an RSCU analysis on it... >

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

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

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, 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, 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

Hi, I have the following data that looks like this: > names(dat) [1] "(-2329,-2319]" "(-1399,-1389]" "(-669.4,-659.4]" How can I modify those names into just this? [1] -2329 -1399 -669.4 - Gundala Viswanath Jakarta - Indonesia

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, Is there any such package? I am thinking of something equivalent to Perl::Tidy. For example with VI editor one can visual highlight a portion of Perlcode and then issue the command: !perltidy then the code will be automatically arranged. - 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

Hi, Suppose I have a vector of data. Is there a method in R to help us automatically suggest which distributions fits to that data (e.g. normal, gamma, multinomial etc) ? - 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

Dear all, I have the following matrix: > str(mat) Named chr [1:32268] "yQAAA" "jQAAQ" "UQAAg" "FQAAw" "1QABA" ... - attr(*, "names")= chr [1:32268] "CAAAAAAAAA" "CAAAAAAAAC" "CAAAAAAAAG" "CAAAAAAAAT" ... I want to destroy the attribute yielding only this: > str(mat) Named chr [1:32268]

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 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