search for: expr1

...to some criteria. In order to do that, what I want to be able to do, is selectively filter the rows, graph some convinient variables, do some further filtering and so on. Let me take a more concrete example to make myself clear. Let's say that I load a given dataset on a dataframe, namely expr1. This dataframe would have the fields expr1$name, expr1$expression, expr1$reliablity, expr1$x, expr1$y and so on, containing, for instance, 26000 rows. Now from these 26000 I'd like to select only those ones satisfying expr1$expression > 2000, expr1$reliability = 100 and plot a graph on expr...

...ilently* recycled, even >> when they are incommensurate lengths, when doing logical indexing? > > It is convenient to use a single `TRUE' in programmatic manipulation of subscripts in the same manner as using an empty subscript interactively: > >> mat<-diag(1:3) >> expr1 <- quote(mat[]) >> expr1[[3]] <- TRUE >> expr1[[4]] <- 2 >> eval(expr1) > [1] 0 2 0 >> mat[,2] > [1] 0 2 0 > > HTH, > > Chuck

...237.7830 0.4114 Total 591 282.5871 If I do with R, I have been trying everything it occurrs to me and looked everywhere and I could not obtain the same results and nothing is clear to me... (I am so sorry... probably it is lack of statistical knowledge): If I do: > anova(lm(Expr1~depth*sector)) Analysis of Variance Table Response: Expr1 Df Sum Sq Mean Sq F value Pr(>F) depth 1 38.2 38.2 92.76 <2e-16 *** sector 6 5.1 0.9 2.07 0.055 . depth:sector 6 1.5 0.3 0.62 0.712 Residuals 578 237.8 0....

...gt; fx2 -(invroot2pi/sigma) * (exp(-0.5 * (((x - mu)/sigma)^2)) * (0.5 * (2 * (1/sigma * (1/sigma)))) - exp(-0.5 * (((x - mu)/sigma)^2)) * (0.5 * (2 * (1/sigma * ((x - mu)/sigma)))) * (0.5 * (2 * (1/sigma * ((x - mu)/sigma))))) So I tried, in the interests of simplification.. > expr1<-expression( ((x-mu)/sigma)^2) > expr1 expression(((x - mu)/sigma)^2) > x [1] 5 15 20 > expr2<-expression(exp(-0.5*expr1)) > expr2 expression(exp(-0.5 * expr1)) > eval(expr2) Error in -0.5 * expr1 : non-numeric argument to binary operator I thought this might mean that I need...

...partial derivatives of my expression? (I suspect it's doing derivatives numerically) Can I specify the partials in advance? I noticed the 'deriv' function, but am baffled by its output: > deriv(~ a0 / (Q*sqrt((1-(x/f0)^2)^2 + ((x/f0) * 1/Q)^2)),"a0") expression({ .expr1 <- x/f0 .expr10 <- Q * sqrt(1 - .expr1^2^2 + .expr1 * 1/Q^2) .value <- a0/.expr10 .grad <- array(0, c(length(.value), 1), list(NULL, c("a0"))) .grad[, "a0"] <- 1/.expr10 attr(.value, "gradient") <- .grad .value }) I think .ex...

...ile="temp.txt") print(prod(x)) file.remove("C:/R-2.5.0/temp.txt") But how to convert the output of the loop to a vector that I can manipulate (by prod or sum etc), without having to write and append to a file? Question 2: > deriv(~gamma(x),"x") expression({ .expr1 <- gamma(x) .value <- .expr1 .grad <- array(0, c(length(.value), 1), list(NULL, c("x"))) .grad[, "x"] <- .expr1 * psigamma(x) attr(.value, "gradient") <- .grad .value }) BUT > deriv3(~gamma(x),"x") Error in deriv3.formu...

Dear list-subscriber, in the process of writing a general code snippet to extract coefficients in an expression (in the example below: 0.5 and -0.7), I stumbled over the following peculiar (at least peculiar to me:-) ) sorting behaviour of the function all.names(): > expr1 <- expression(x3 = 0.5 * x1 - 0.7 * x2) > all.names(expr1) [1] "-" "*" "x1" "*" "x2" > all.vars(expr1) [1] "x1" "x2" > expr2 <- expression(x3 <- 0.5 * x1 - 0.7 * x2) > all.names(expr2) [1] "<-&quo...

hi , this can be done easily if (cond) expr ex:  > for (i in 1: 4)+ {+ if(i==2) print("a")+ if(i==2) print("b")+ } output : [1] "a"[1] "b" but i want this  if (cond) expr1 expr 2 i tried this :  > for (i in 1: 4)+ {+ if(i==2) (print("b") && print("a"))+ } output : [1] "b"Error in print("b") && print("a") : invalid 'x' type in 'x && y' but i expect the same out put as before ....

...onfint(). I thought that getting derivatives with deriv might help but found that deriv does not automatically handle the indexing of the beta parameter. I modified the output of deriv from the case when the term beta is not indexed to give: dd.deriv2 <- function (Blev, beta, gamm, GL) { .expr1 <- GL^gamm .value <- Blev + rep(beta, each = 17) * .expr1 .grad <- array(0, c(length(.value), 5), list(NULL, c("Blev", "beta.rouge", "beta.vert", "beta.bleu", "gamm" ))) .grad[, "Blev"] <- 1 .grad[1:17,...

Sorry if this has been covered here somewhere in the past, but ... Does anyone know why logical vectors are *silently* recycled, even when they are incommensurate lengths, when doing logical indexing? This is as documented: For ?[?-indexing only: ?i?, ?j?, ?...? can be logical vectors, indicating elements/slices to select. Such vectors are recycled if necessary to match

Hello! I work with: R : Copyright 2006, The R Foundation for Statistical Computing Version 2.3.1 (2006-06-01) On Windows XP Professional (Version 2002) SP2 I think there is a bug in the conditional execution if (expr1) {expr2} else {expr3} If I try: "if (expr1) expr2 else expr3" it works well but when I put the expression expr2 and expr3 between {} I receive an error message like this one: "Erreur dans parse(file, n = -1, NULL, "?") : erreur de syntaxe ? la ligne 4: } 5:...

Hi, Is there any equivalent for ifelse (except if (cond) expr1 else expr2) which takes an atomic element as argument but returns vector since ifelse returns an object of the same length as its argument? x = c(1,2,3) y = c(4,5,6,7) z = 3 ifelse(z <= 3,x,y) would return x and not 1 thanks

...valuated) expression or not, *without* evaluating it if not? Currently, I do various rather ad hoc eval()+substitute() tricks for this that most likely only work under certain circumstances. Ideally, I'm looking for a isParseTree() function such that I can call: expr0 <- foo({ x <- 1 }) expr1 <- foo(expr0) stopifnot(identical(expr1, expr0)) where foo() is: foo <- function(expr) { if (!isParseTree(expr)) expr <- substitute(expr) expr } I also want to be able to do: expr2 <- foo(foo(foo(foo(expr0)))) stopifnot(identical(expr2, expr0)) and calling foo() from within...

Hey, everybody--- Ignorance CAN be bliss, at least for a while, but, .... Just so you know... A week or two ago, some upgrades to the expression parser (you know, the expressions you put in $[ ... ] in your extensions.conf file) that I submitted, have been merged into the CVS HEAD of the source. Hopefully, for around 99.9% of you, it won't make any difference to you. The Makefile has also

Hi, I am using R a lot to make plots relating to radioactivity, I am often using expression() to label the plots with nuclide names written with superscripts, e.g. expression(paste("Releases of ", { }^{99},Tc," (TBq/year)"))->ywtext But, is there any simple way to change the number and name of the nuclide through a variable? I tried nuccode=expression({ }^{99},Tc)

I have a situation where this is fine: > if (length(x)>15) { clever <- rr.ATM(x, maxtrim=7) } else { clever <- rr.ATM(x) } > clever $ATM [1] 1848.929 $sigma [1] 1.613415 $trim [1] 0 $lo [1] 1845.714 $hi [1] 1852.143 But this variant, using ifelse(), breaks: > clever <- ifelse(length(x)>15, rr.ATM(x, maxtrim=7), rr.ATM(x))

Quick question: Which function do you use to calculate partial derivatives from a model equation? I've looked at deriv(), but think it gives derivatives, not partial derivatives. Of course my equation isn't this simple, but as an example, I'm looking for something that let's you control whether it's a partial or not, such as: somefunction(y~a+bx, with respect to x,

Using an example provided by "The Hitchhiker's Guide to Asterisk", I made the following addition to my extensions.conf file: [inbound-analog] exten => s,1,Wait(1) exten => s,2,SetVar(counter=0) exten => s,3,Answer() exten => s,4,Wait(1) exten => s,5,DigitTimeout(15) exten => s,6,ResponseTimeout(10)

...the last line, which reads the file nevertheless, and I also don't know the number of lines in the file. Perhaps one could catch an error, when the first invocation of read.table fails, and try the second one. However tryCatch doesn't seem to make it simple to write something like E = try(expr1 otherwise expr2) (if expr1 fails, evaluate expr2 instead) ? Oliver

Dear list # I have a function ff <- function(a,b=2,c=4){a+b+c} # which I programmatically want to modify to a more specialized function in which a is replaced by 1 ff1 <- function(b=2,c=4){1+b+c} # I do as follows: vals <- list(a=1) (expr1 <- as.expression(body(ff))) expression({ a + b + c }) (expr2 <- do.call("substitute", list(expr1[[1]], vals))) { 1 + b + c } # This "works", eval(expr2, list(b=10,c=12)) [1] 23 # - but I would like a function. Hence I do: ll <- alist(b=, c=, eval(expr...