similar to: Vectorial analogue of all.equal()?

I am not sure if this question has been asked before - but is there a procedure in R (in lm or glm?) that is equivalent to ENTER and REMOVE regression commands in SPSS? Thanks a lot! -- Dimitri Liakhovitski Ninah.com Dimitri.Liakhovitski at ninah.com

find the position of the first value who equals certain number in a vector: Say a=c(0,0,0,0,0.2, 0.2, 0.4,0.4,0.5) i wish to return the index value in a for which the value in the vector is equal to 0.4 for the first time. in this case, it is 7. [[alternative HTML version deleted]]

Hi! In a function, I may have an instance in which all elements are equal. > x <- rep(1,5) > > x [1] 1 1 1 1 1 > identical(x) Error in .Internal(identical(x, y, num.eq, single.NA, attrib.as.set)) : 'y' is missing > all.equal(x) Error in is.expression(x) : 'x' is missing > I don't care what particular value it is, I just want to know if they are all

I have 2 vectors A and B. For each element of B I'd like to find the index of the next higher or equal element in A. And I'd like to store it effectiv. E.g.: A <- c(1,3,7,8,10) B <- c(5,8) result: 3, 4 I have a possibility but for long vectors it works not very effectiv: ans <- sapply(B, function(x) which.max(A[A < x]) ) as.integer(ans + 1) Is there anyone how has a better

hi ?integer says: Note that on almost all implementations of R the range of representable integers is restricted to about +/-2*10^9: 'double's can hold much larger integers exactly. I am getting very confused as to when to use integers and when not to. In my line I need exact comparisons of large integer-valued arrays, so I often use as.integer(), but the above

Dear All, I am trying to use the do.call command to avoid tedious loops in R when I need to call repetitively a function. I am experiencing a problem when the arguments of a function are given with a vector (I need to express then this way to be able later on to integrate them numerically with the adapt package). Consider the small script: remove(list=ls()) #library(adapt) # first a list of the

Hi, all. I want to write some functions like glm() so i studied it. In glm.fit(), it calls a fortran subroutine named "dqrfit" to compute least squares solutions to the system x * b = y To learn how "dqrfit" works, I just follow how glm() calls "dqrfit" by my own example, my codes are given below: > qr <- >

Just to avoid possible confusion, let me correct a typo (at step [2] in the example below). Apologies! -----FW: <XFMail.111023084327.ted.harding at wlandres.net>----- Date: Sun, 23 Oct 2011 08:43:27 +0100 (BST) Sender: r-help-bounces at r-project.org From: (Ted Harding) <ted.harding at wlandres.net> To: r-help at r-project.org Subject: Re: [R] symmetric matrix multiplication On

Hi Ted, yes this was the problem. Thank you very much. best idaios On Wed, Jan 9, 2013 at 4:51 PM, Ted Harding <Ted.Harding@wlandres.net>wrote: > Ah! You have aqssigned a parameter "equal.var=TRUE", and "equal.var" > is not a listed paramater for t.test() -- see ?t.test : > > t.test(x, y = NULL, > alternative = c("two.sided",

Dear List, Further to my earlier email, I note that, for me at least, R CMD check is *not* reading R_LIBS from ~/.R/check.Renviron on R 2.15.2 patched (r61228) and R Under Development (r61660). The only way I can get R CMD check to look for packages in a user-supplied library is by explicitly exporting R_LIBS set to the relevant directory. R CMD build *does* read R_LIBS from ~/.R/build.Renviron

Hi, I'm a R newbie and want to accelerate the following pre-filtering step of a data set with more than 115,000 rows : #----------------- # Function to filter out constant data columns filter.const<-function(X, vectors=c('column', 'row'), tol=0){ realdata=c() filteredX<-matrix() if( vectors[1] == 'row' ){ for( row in (1:nrow(X)) ){ if(

Greetings! For some reason I am not managing to work out how to do this (in principle) simple task! As a result of applying strsplit() to a vector of character strings, I have a long list L (N elements), where each element is a vector of two character strings, like: L[1] = c("A1","B1") L[2] = c("A2","B2") L[3] = c("A3","B3")

Dear all, I encounter some covariance matrix with quite small eigenvalues (around 1e-18), which are smaller than the machine precision. The dimension of my matrix is 17. Here I just fake some small matrix for illustration. a<-diag(c(rep(3,4),1e-18)) # a matrix with small eigenvalues b<-matrix(1:25,ncol=5) # define b to get an orthogonal matrix b<-b+t(b) bb<-eigen(b,symmetric=T)

Dear all, I need to compute the ACF (autocorrel) of an AR6 process, given the values of its parameters (w1,w2,w3,w4,w5,w6). First, I notice that there is an error as soon as the sum of the wi equals 1 : "Error in drop(.Call("La_dgesv", a, as.matrix(b), tol, PACKAGE = "base")) : system is computationally singular: reciprocal condition number = 1.00757e-18"

I have been told by the CRAN administrators that the following code generated an error on 64-bit Fedora Linux (gcc, clang) and on Solaris machines (sparc, x86), but runs well on all other systems): > fn <- function(x, y) ifelse(x^2 + y^2 <= 1, 1 - x^2 - y^2, 0) > tol <- 1.5e-8 > fy <- function(x) integrate(function(y) fn(x, y), 0, 1,

Hello everyone, this is a very simple question but I could not find the answer. :-( Is there any function in R for calculating the rank of a matrix? (In the sense of the dimension of the vectorial space that spans) Thank you in advance! Manuel --- -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

Greetings All. The function edit.data() allows a convenient spreadsheet-like view of a dataframe with too many rows/columns to fit on the screen (especially when there are many columns). Very useful when scanning through a dataset (row & column are conveniently identified by the labels at the side and above). However, there seens to be no option to set it "read-only" on start-up,

Greetings All. I want to do the following simple thing. I have defined a function med3x3() such that, given vectors X,Y, med3x3(X,Y) returns a 3x3 table where: Row 1: X > median(X) Row 2: X = median(X) Row 3: X < median(X) Col 1: Y < median(Y) Col 2: Y = median(Y) Col 3: Y > median(Y) (with intersections of these conditions for the individual cells). I can easily define fixed

Hi, I was trying to draw an empirical distribution function with uniform confidence bands. So I tried to find a way to calculate values of the Kolmogorov-Smirnov Distribution but failed. I guess it must be hidden somewhere (since the ks-test is implemented), but I was unable to find it. Is there any way to do this? Thanks Leif Boysen

A quick question: Is there anywhere a listing of the RGB components of the named colours listed by colors()? For example, where would I find the RGB for "orange1" or "salmon"? When I look at an EPS file from R where I have used these colours, it seems that for: "salmon": 0.9804 0.5020 0.4471 rgb "orange1": 1 0.6471 0 rgb However, this is a tedious way