search for: xnames

...1,2,2,5,5,5) as.data.frame(cbind(a,b,c))->q21 varnames<-colnames(q21) i<-9999 for (i in 1:3) { round(table(q21[,i])/sum(table(q21[,i])),3)*100->z1 as.data.frame.table(table(q21[,i]))->z1 round(z1[,2]/sum(z1[,2]),3)*100->z1[,2] colnames(z1)<-c(varnames[i],"y") xnames<-levels(z1[,1]) paste(xnames[1]," - low",sep="")->xnames[1] paste(xnames[length(xnames)]," - high",sep="")->xnames[length(xnames)] z1[,1]<-xnames as.factor(z1[,1])->z1[,1] barchart( horizontal=FALSE, z1[,2]~z1[,1], ylab=lis...

(earlier I sent it as html by mistake). Hi, How can I get from C an object name used as a function argument? I have sample code in C that gives me access to the name of the function being called: SEXP xname(SEXP x) { const char *fun_name = CHAR(PRINTNAME(CAR(x))); x = CDR(x); const char *arg_name = isNull(TAG(x)) ? "" : CHAR(PRINTNAME(TAG(x)));

...c 2015-02-11 00:31:24.986198000 -0500 +++ dovecot-2.2.9/src/login-common/ssl-proxy-openssl.c 2015-02-11 00:32:19.262198000 -0500 @@ -951,54 +951,25 @@ return strstr(cert, "PRIVATE KEY---") != NULL; } -static void load_ca(X509_STORE *store, const char *ca, - STACK_OF(X509_NAME) **xnames_r) +static void load_ca(SSL_CTX *ssl_ctx, const char *ca) { - /* mostly just copy&pasted from X509_load_cert_crl_file() */ - STACK_OF(X509_INFO) *inf; - X509_INFO *itmp; - X509_NAME *xname; - BIO *bio; - int i; - - bio = BIO_new_mem_buf(t_strdup_noconst(ca), strlen(ca)); - if (bio == NULL) -...

In an implied call to an S3 print method, deparse(substitute(x)) returns "x", regardless of the name of object in .GlobalEnv, as indicated in the following: > Xnamed <- 1 > class(Xnamed) <- 'name.x' > print.name.x <- function(x, ...){ + namex <- deparse(substitute(x)) + cat('How can I get the name of x in .GlobalEnv?\n', +

...It works, but it prints to screen each time, which is annoying. There must be simple way to suppress this, or to use a slightly different set of commands, which will be obvious to those wiser than I... Here is a simpler mockup which shows the issue: x = data.frame(rbind(c(1,2,3),c(1,2,3))) xnames = c("a", "b", "c") names(x) = xnames for(i in 1:length(x)) { # Create a varying string expression expr = paste("y = x$", xnames[i], "[1]", sep="") # evaluate expression eval(parse(text=print(expr))) # This command prints the expression...

...(dim(y)[2] > 2) stop("Only binomial response matrices (2 columns)") } offset <- model.extract(m, offset) id <- model.extract(m, id) if (is.null(id)) { stop("Id variable not found") } nobs <- nrow(x) p <- ncol(x) xnames <- dimnames(x)[[2]] if (is.null(xnames)) { xnames <- paste("x", 1:p, sep = "") dimnames(x) <- list(NULL, xnames) } if (is.character(family)) family <- get(family) if (is.function(family)) family <- family() if (!...

On 07/05/2017 3:56 PM, Spencer Graves wrote: > In an implied call to an S3 print method, deparse(substitute(x)) returns > "x", regardless of the name of object in .GlobalEnv, as indicated in the > following: > > > > Xnamed <- 1 > > class(Xnamed) <- 'name.x' > > print.name.x <- function(x, ...){ > + namex <-

All, I'm trying to set up a function which calls the partialPlot function but am getting an error that I can't seem to solve. Here's a simplified version of the function and error... > pplot <- function(rf,pred.var){partialPlot(x=rf,pred.data=acoust,x.var=pred.var)} > > attach(acoust) > acoust.rf <-

Hi, A column of my df looks like A 10/20/30 40/20 60/10/10/5 80/10 I want to split it such that the last column has the last composition and if there are not enough the middle columns get the 0s. That way my df would look like A1 A2 A3 A4 10 20 0 30 40 0 0 20 60 10 10 5 80 0 0 10 How can I do that ?? [[alternative HTML version deleted]]

...These looks ok. However, I would like to achieve more. Apart from using a loop, is there are fast way to 'add' the titles to be more informative? that is, in the histograms, I want the titles to be 't distribution with dof=' the degrees of freedom. I have tried apply(Q5,2,hist,xnames=dof) which does not work; apply(Q5,2,hist(,xnames=dof)); does not work either and similarly, how do I add titles to qqnorm plot to make them informative? Thanks! casper -- View this message in context: http://r.789695.n4.nabble.com/use-of-apply-for-hist-tp3093811p3093811.html Sent from the R...

Dear R users, I am looking for the source code of the implementation of ppr (Projection Pursuit Regression) in R. It will be great if citations of the source papers on which the implementation is based, are also provided. Thank you, Arvind Iyer, Grad student, Deptt. of Biomedical Engineering Viterbi School of Engineering University of Southern California, Los Angeles [[alternative HTML

I'm not understanding something. I'm trying to add xlab & ylab to a balloon plot of a table object. From docs I thought following should work: require(gplots) # From balloonplot example: # Create an example using table xnames <- sample( letters[1:3], 50, replace=2) ynames <- sample( 1:5, 50, replace=2) tab <- table(xnames, ynames) balloonplot(tab) # Try xlab, ylab: balloonplot(tab, xlab = "MyX", ylab = "MyY") But second plot is no different from first. R.version.string:...

Dear list, can someone explain to me why deparse(substitute(x)) does not seem to work when x is of a user-defined S3 class? In my actual problem, my print method is part of a package, and the method is registered in the NAMESPACE, if that should make a difference. > print.testclass <- function(x,...){ xname <- deparse(substitute(x)) cat("Your object name

Hi all, In the histogram created with the `hist` command, the label of the y axis reads "Frequency ?. Finding that the ylab key is used to change (localize) that word may seem difficult for people who are beginners either in R or in english. I suggest very minor modifications of the file hist.Rd at https://svn.r-project.org/R/trunk/src/library/graphics/man/hist.Rd. 1) make the default value

R2.1.1 Win 2k I have a function, B, within a function, A. I would like to have B print the name of the argument passed to it (not the value of the arguments). i.e., A<-function() { B<-function(x,y) { fit1<-lm(y~x,data=jo) print(summary(fit1) I want B to print the string "age" and the string "height". } B(age,height) }

...included in Redhat Enterprise Linux 6.2: *** src/login-common/ssl-proxy-openssl.c 2010-12-30 10:42:54.000000000 +0100 --- src/login-common/ssl-proxy-openssl.c_1 2012-03-20 09:48:28.359508087 +0100 *************** *** 924,930 **** X509_STORE *store; STACK_OF(X509_NAME) *xnames = NULL; ! SSL_CTX_set_options(ssl_ctx, SSL_OP_ALL | SSL_OP_NO_SSLv2); if (*set->ssl_ca != '\0') { /* set trusted CA certs */ store = SSL_CTX_get_cert_store(ssl_ctx); --- 924,930 ---- X509_STORE *store; STACK_OF(X509_NAME) *xn...

...glm.fit ========================= 1 glm.fit <- 2 function (x, y, weights = rep(1, nobs), start = NULL, etastart = NULL, 3 mustart = NULL, offset = rep(0, nobs), family = gaussian(), 4 control = glm.control(), intercept = TRUE) 5 { 6 x <- as.matrix(x) 7 xnames <- dimnames(x)[[2]] 8 ynames <- names(y) 9 conv <- FALSE 10 nobs <- NROW(y) 11 nvars <- NCOL(x) 12 if (nvars == 0) { 13 cc <- match.call() 14 cc[[1]] <- as.name("glm.fit.null") 15 return(eval(cc, parent.fra...

Is there a way to create a 'bubble plot' in R? For example, if we define the following data frame containing the level of y observed for 5 patients at three time points: time<-c(rep('time 1',5),rep('time 2',5),rep('time 3',5))

...exported function called updateList for a while, which looks like > lattice:::updateList function (x, val) { if (is.null(x)) x <- list() if (!is.list(x)) stop("x must be NULL or a list") if (!is.list(val)) stop("val must be a list") xnames <- names(x) for (v in names(val)) { existing <- v %in% xnames if (existing && is.list(x[[v]]) && is.list(val[[v]])) x[[v]] <- updateList(x[[v]], val[[v]]) else x[[v]] <- val[[v]] } x } Basically, it recursively replaces el...

Hola! I am producing postscript output from R (windows 98), both using firts the windows devicd and then " copy to postscript", or using the postscript device directly. The problem shows up equally both ways. The function producing the plot is given below. The only difference I can see with this function and other I have producing plots, is that this uses split.screen(). I include the