search for: savicky

...4. Install the package under Linux and zip the directory library/<package> and unzip it in the library directory on Windows machine. This works. The package behaves correctly. However, I do not think that this is a suggested method. Could you help me? Thank you in advance. Petr Savicky.

Full_Name: Petr Savicky Version: 2.4.0 OS: Fedora Core release 2 Submission from: (NULL) (62.24.91.47) the error is > binom.test(0.56*10000,10000) Error in binom.test(0.56 * 10000, 10000) : 'x' must be nonnegative and integer while > binom.test(5600,10000) yields correct result. The same err...

Full_Name: Petr Savicky Version: 1.5.0 OS: SuSE 7.2 Submission from: (NULL) (147.231.6.1) The function tree() from the tree package crashes for certain input data. All the information needed to reproduce the error (data, script, expected error message) may be found at http://www.cs.cas.cz/~savicky/tree_error/tree_error....

In R-2.10.0, the development version, function as.factor() uses 17 digit precision for conversion of numeric values to character type. This is very good for the consistency of the resulting factor, however, i expect that people will complain about, for example, as.factor(0.3) being [1] 0.29999999999999999 Levels: 0.29999999999999999 I suggest to extend the "Warning" section of

For the package at http://www.cs.cas.cz/~savicky/R-devel/something_0.0.0.tar.gz which is a minor part of some other package only to demonstrate the problem, i get (under R version 2.11.0 Under development 2009-12-12 r50714 and also under R-2.9.2, openSUSE 11.1 (x86_64) and CentOS release 5.2) R CMD check something_0.0.0.tar.gz ... * check...

Dear All, I have two matrices A (40 x 732) and B (40 x 1230) and would like to calculate correlation between them.  I can use: cor(A,B, method="pearson") to calculate correlation between all possible pairs. But the issue is that there is one-many specific mappings between A and B and I just need to calculate correlations for those pairs (not all). Some variables in A (proteins, say p1)

...e function will refuse to modify it if it is not so. The documentation suggests to call "duplicate", if the data are shared, but I cannot afford this due to memory. So, I can only stop if NAMED is not 0. I appreciate, if anybody could give me advice on the above things. Petr Savicky.

Dear List, I will appreciate any advice regarding how to convert the following numbers [I got in return by taxondive()] in numeric integers without the e.g. 6.4836e+01 abbreviations. Thank you very much in advance, Gian > taxa_dive Species Delta Delta* Lambda+ Delta+ S Delta+ Nat1 5.0000e+00 6.4836e+01 9.5412e+01 6.7753e+02 8.7398e+01 436.99 Nat2

Hello All, My algorithm as follows; y <- c(1,1,1,0,0,1,0,1,0,0) x <- c(1,0,0,1,1,0,0,1,1,0) n <- length(x) t <- matrix(cbind(y,x), ncol=2) z = x+y for(j in 1:length(x)) { out <- vector("list", ) for(i in 1:10) { t.s <- t[sample(n,n,replace=T),] y.s <- t.s[,1] x.s <- t.s[,2] z.s <- y.s+x.s out[[i]] <- list(ff <- (z.s), finding=any (y.s==y[j])) kk

List members, Via searches I've seen similar discussion of this topic but have not seen resolution of the particular issue I am experiencing. If my search on this topic failed, I apologize for the redundancy. I am attempting to generate random covariance matrices but would like the corresponding correlations to be uniformly distributed between -1 and 1. The approach I have been using is:

foo <- rnorm(30*34*12) dim(foo) <- c(30, 34, 12) I want to make a data.frame out of this three-dimensional array. Each dimension will be a variabel (column) in the data.frame. I know how this can be done in a very slow way using for loops, like this: x <- rep(seq(from = 1, to = 30), 34) y <- as.vector(sapply(1:34, function(x) {rep(x, 30)})) month <- as.vector(sapply(1:12,

Dear all, I need to generate numbers from multivariate normal with large dimensions (5,000,000). Below is my code and the error I got from R. Sigma in the code is the covariance matrix. Can anyone give some idea on how to take care of this error. Thank you. Hannah > m <- 5000000 > m1 <- 0.5*m > rho <- 0.5 > Sigma <- rho* matrix(1, m, m)+diag(1-rho, m)

Dear all, I have question but cannot explain without providing some context first: I want to calculate how many policy-connected coalitions between 7 parties are possible. I have positions on an one-dimensional scale for each party and I have sorted the parties on the positions (it is sorted from extreme left to extreme right, hence using a left-right scale). A policy-connected coalition

Dear list, I have a big deal concerning the development of a Taylor expansion. require(Matrix) e1 <- as.vector(1:5) e2 <- as.vector(6:10) in order to obtain all the combinations between these two vectors following a Taylor expansion (or more simply through a Maclaurin series) for real numbers. We have f(x) = f(0) + f'(0)(x-0) + f''(0)(x-0)^2/2! + ? + f^(k)(0)(x-0)^k/k!

...pe are equal). However, we have x <- NaN y <- as.double(NA) x # [1] NaN y # [1] NA identical(x,y) # [1] TRUE In my opinion, NaN and as.double(NA) should be distinguished as the help page suggests. Tested under R version 2.5.1 Patched (2007-08-19 r42638) on Linux (CPU Xeon). Petr Savicky.

...oes not contain the directory "R-devel/po". This directory is present in R-devel_2010-01-04.tar.bz2 and contains the file "R-devel/po/Makefile.in.in". I see the directory "R-devel/po" at https://svn.r-project.org/R/trunk/, so the problem could be just temporary. Petr Savicky.

lgamma(x) and lfactorial(x) are defined to return ln|Gamma(x)| {= log(abs(gamma(x)))} or ln|Gamma(x+1)| respectively. Unfortunately, we haven't chosen the analogous definition for lchoose(). So, currently > lchoose(1/2, 1:10) [1] -0.6931472 -2.0794415 NaN -3.2425924 NaN -3.8869494 [7] NaN -4.3357508 NaN -4.6805913 Warning message: In

Hello, I am stuck with selecting the right rows from a data frame. I think the problem is rather how to select them then how to implement the R code. Consider the following data frame: df <- data.frame(ID = c(1,2,3,4,5,6,7,8,9,10), value = c(34,12,23,25,34,42,48,29,30,27)) What I want to achieve is to select 7 rows (values) so that the mean value of those rows are closest to the value of 35

Hello Members, I need to group a data.frame in a a specific way, like shown below, DF raw is like this, id col1 col2 score 1 A B 40 2 B C 55 3 C D 4000 4 D E 100 5 E F 300 I want the out put as List [1] A B C [2] D E [3] F Basically the split should be based on the DF$score > 200, and all the col1 and col2 values should be

Dear all, I am using the ROCR library to compute the AUC and also the Hmisc library to compute the C-index of a predictor and a group variable. The results of AUC and C-index are similar and give a value of about 0.57. The Wilcoxon p-value is <0.001! Why the AUC is showing small value and the p-value is high significant? The AUC is based on Wilcoxon calculation? Many thanks, Lina