Re: [R] Definition of t-value

try this Out - lm(A ~ data$B + data$C + data$D) summary(Out) moreover, by typing 'summary.lm' in your R console you may see how the t-values are computed; check also ?summary.lm. Another way, though less efficient, to obtain the standard errors is the following summ.Out - summary(Out) X -

Re: [R] The most common row in a matrix?

try this: mat - array(1:3, dim = c(4, 5)) # ind - table(apply(mat, 1, paste, collapse = /)) ind - which.max(ind) as.numeric(strsplit(names(ind), /)[[1]]) I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic

Re: [R] Nominal Respose Model in R

AFAIK there exist functions to fit IRT models only for ordinal data. In particular, look at function grm() that fits the Graded Response Model in the 'ltm' package (using Marginal Maximum Likelihood), function PCM() in the 'eRm' package (using Condtional Maximum Likelihood), and function

Re: [R] Selective subsetting

try this: cor.mat - cor(iris[1:4]) cor.mat[lower.tri(cor.mat)] I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax:

Re: [R] Defining factors in GLM

look at ?factor ?formula ?data.frame I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web:

Re: [R] Recursive decreasing sequences

try this: start.val - 2 end.val - 6500 system.time(res - unlist(lapply(start.val:end.val, :, to = end.val))) I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven,

Re: [R] multiline string continuation

maybe you're looking for: cat(datadir) I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web:

Re: [R] Generating bivariate or multivariate data with known parameter values

you can do that using the copula package, e.g., library(copula) bivGamma - mvdc(claytonCopula(2), c(gamma, gamma), list(list(shape = 1, rate = 2), list(shape = 2, rate = 1))) dat - rmvdc(bivGamma, 1000) regarding correlation, the Clayton copula with alpha = 2 gives Kendall's tau =

Re: [R] GLM information matrix

look at summary.glm(), probably you're looking for fit - glm(..., family = binomial) # the inverse Fisher Information matrix summary(fit)$cov.scaled I hope it helps. Best, Dimitris -- Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of

Re: [R] Exponentiate a matrix

Quoting Duncan Murdoch [EMAIL PROTECTED]: On 9/21/2006 10:40 AM, Doran, Harold wrote: Suppose I have a square matrix P P - matrix(c(.3,.7, .7, .3), ncol=2) I know that P * P Returns the element by element product, whereas P%*%P Returns the matrix product. Now, P^2 also returns

Re: [R] Exponentiate a matrix

Quoting Douglas Bates [EMAIL PROTECTED]: On 9/21/06, Dimitrios Rizopoulos [EMAIL PROTECTED] wrote: Quoting Duncan Murdoch [EMAIL PROTECTED]: On 9/21/2006 10:40 AM, Doran, Harold wrote: Suppose I have a square matrix P P - matrix(c(.3,.7, .7, .3), ncol=2) I know that P * P

Re: [R] help on data frame and matrix

look at ?data.matrix() Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://med.kuleuven.be/biostat/

Re: [R] variables in object names

you need something like the following, fit.lis - list(model1, model2, model3, model4, model5) # or if you have many models fit.lis - lapply(paste(model, 1:5, sep = ), get) sapply(fit.lis, function(x) summary(x)$r.squared) I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student

Re: [R] rownumber

maybe the following is what you're looking for: lead.dat[!duplicate(rownames(lead.dat)), ] lead.dat[duplicate(rownames(lead.dat)), ] I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address:

Re: [R] Running wilcox.test function on two lists

try something like the following: lis1 - c(lapply(1:10, rnorm, n = 10)) lis2 - c(lapply(1:10, rnorm, n = 10)) lis1[[5]] - lis2[[8]] - numeric(0) ind - sapply(lis1, length) 0 sapply(lis2, length) 0 lis1 - lis1[ind] lis2 - lis2[ind]

Re: [R] get ylim of current plot()

I don't think you can obtain these values directly since plot.default() does not return something. 'ylim' is in fact calculated from xy.coords() as the range of the finite 'y' abscissa values; check the code of plot.default() for more info. I hope it helps. Best, Dimitris Dimitris

Re: [R] horizontal direct product

maybe something like: %*~% - function(x, y){ n - nrow(x) out - matrix(0, n, ncol(x) * ncol(y)) for(i in 1:n) out[i, ] - c(y[i, ] %o% x[i, ]) out } x - matrix(1:4, 2, 2, TRUE) y - matrix(5:8, 2, 2, TRUE) x %*~% y I hope it helps. Best, Dimitris Dimitris

Re: [R] how to the p-values or t-values from the lm's results

try the following: data - data.frame(matrix(rnorm(900), ncol = 9)) names(data) - c(y, paste(x, 1:8, sep = )) logr - lm(y ~ . - 1, data) a - summary(logr) coef(a) coef(a)[, 3:4] coef(a)[, t value] coef(a)[, Pr(|t|)] I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student

Re: [R] How to order or sort a data.frame

Quoting John Kane [EMAIL PROTECTED]: --- Dimitrios Rizopoulos [EMAIL PROTECTED] wrote: try the following: mdf - data.frame(us.state, count, year, month) mdf[order(mdf$year, mdf$month), ] Thansk to Dimitris and Dieter. This has helped since seems to have shown me a way around

Re: [R] How to order or sort a data.frame

try the following: mdf - data.frame(us.state, count, year, month) mdf[order(mdf$year, mdf$month), ] I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium

Re: [R] Unique rows

if you want the first row for the unique 'aa' entries, try the following: cc[!duplicated(cc$aa), ] I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium

Re: [R] locating intervals (corrected version)

?findInterval() could be of help in this case. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web:

Re: [R] workaround for numeric problems

I'd compute this in the log-scale (taking also advantage of the 'log' and 'log.p' arguments of dnorm() and pnorm(), respectively), and then transform back, e.g., fn1 - function(B){ -(pnorm(B) * dnorm(B) * B + dnorm(B)^2)/pnorm(B)^2 } fn2 - function(B){ p1 - dnorm(B, log = TRUE) +

Re: [R] how to recode in my dataset?

probably ?cut() is what you're looking for, e.g., something like: ind - cut(mydata$soiltem, seq(0, 60, 0.2), labels = FALSE) seq(0.1, 60, 0.2)[ind] I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of

Re: [R] Interpreting as.factor

probably you want to look at the help page of ?contr.treatment I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax:

Re: [R] about the analysis of strings, thanks

I'm not aware of package for amino acid sequence analysis, but it'd better to search yourself using, e.g., RSiteSearch() and help.search(). Regarding your question consider the following: strg1 - c(AAA, AA, ) strg2 - c(BBB, , BBB) strg - paste(strg1, strg2, sep = ) nA - nchar(gsub(B, ,

Re: [R] Vector Manipulation

try the following: x - c(B, F, N, Y) y - numeric(length(x)) y[x != N] - 1 y I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel:

Re: [R] select the last row by id group

one approach is the following: math - c(80, 75, 70, 65, 65, 70) reading - c(65, 70, 88, NA, 90, NA) id - c('001', '001', '001', '002', '003', '003') score - data.frame(id, reading, math) # ind - cumsum(tapply(score$id, score$id, length)) score[ind, ] I hope it helps. Best, Dimitris

Re: [R] How to call a value labels attribute?

maybe you could consider something like the following: varlabs - function(x){ if (is.null(names(x))) NULL else x[!duplicated(x)] } varlabs- - function(x, value){ names(x) - names(value[x]) x } ### x - c(1, 2, 3, 3, 2, 3, 1) x varlabs(x) varlabs(x) - c(apple=1, banana=2,

Re: [R] slanted ends of horizontal lines for certain line widths

probably you want to look at the `lend' argument of ?par, e.g., op - par(lend = 2) plot(c(1,2), c(1,1), xlim = c(0,3), lwd = 20, type = l) par(op) I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven

Re: [R] evaluation of the alternative expression in ifelse

Quoting Renaud Lancelot [EMAIL PROTECTED]: Dear all, I am trying to avoid the warnings produced by: x - -2:2 log(x) [1] NaN NaN -Inf 0.000 0.6931472 Warning message: production de NaN in: log(x) I thought that using ifelse would be a solution, but it is not

Re: [R] Random number generation

according to ?.Random.seed, `Note' section, different R sessions give differen simulation results; maybe each time you start R you load a previously saved workspace, with an existing .Random.seed, i.e., check if the .Random.seed value is the same each time you start R. I hope it helps. Best,

Re: [R] Joining variables

look at ?paste(). Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://med.kuleuven.be/biostat/

Re: [R] conditional replacement

you could try something like: ifelse(df 30, 30, ifelse(df 60, 60, df)) I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax:

Re: [R] intervals from cut() as numerics?

as an alternative, you can have a look inside cut.default and use the part that produces the breaks, i.e., breaks - 10 groups - cut(x, breaks = breaks) max.bias - as.vector(tapply(error, groups, mean)) # from cut.default() nb - as.integer(breaks + 1) dx - diff(rx - range(x, na.rm = TRUE))

Re: [R] extracting values from data.frame given arbitrary keys

maybe merge() is what you're looking for, e.g., dat - data.frame(x = rnorm(10), v1 = gl(5, 2), v2 = rep(1:2, 5)) ids - data.frame(v1 = c(2, 3, 4), v2 = c(1, 2, 1)) # dat merge(dat, ids) I hope it helps. Best, Dimitris -- Dimitris Rizopoulos Ph.D. Student

Re: [R] matrix transformation into 3 columns

probably ?reshape() could be used in this case; I hope it helps. Best, Dimitris -- Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web:

Re: [R] Converting character strings to numeric

your code works for me, e.g., x1 - 1159 1129 1124 -5 -0.44 -1.52 x2 - c(1159, 1129, 1124, -5, -0.44, -1.52) x3 - unlist(strsplit(x1, )) all.equal(x2, x3) [1] TRUE as.numeric(x2) [1] 1159.00 1129.00 1124.00 -5.00 -0.44 -1.52 as.numeric(x3) [1] 1159.00 1129.00 1124.00

Re: [R] What does it mean to be masked from data when attaching? (Newbie question)

the problem you get comes from the fact that you attach the data.frame, e.g., it can be re-produced by: x1 - rnorm(30) dat - data.frame(x1 = rnorm(30)) attach(dat) look at ?attach for more details; you could consider with() as an alternative. I hope it helps. Best, Dimitris -- Dimitris

Re: [R] integer vector to a string

probably you need ?paste, e.g., paste(1:500, collapse = ,) I hope it helps. Best, Dimitris Quoting Søren Merser [EMAIL PROTECTED]: hi there is there a way that i can coerce a vector of integers to ONE string with the numbers comma separated like: 1:500 -1,2,3, ..., 500 i've