Displaying 20 results from an estimated 20434 matches for "expressibility".
2005 Jun 06
2
Bug in new() or validObject() in methods package (PR#7922)
Something in new() or validObject() in the methods package is messing
up. This happens in both 2.1.0 and R-devel in Windows.
I'd like to have an empty expression for a slot in a class. An empty
expression is an expression:
> is.expression(expression())
[1] TRUE
>
> is(expression(), "expression")
[1] TRUE
> class(expression())
[1] "expression"
2012 Mar 15
3
single, double or no quotes in expression
Dear all,
I am confused about how to create an expression. I use a package (rsbml)
which uses expressions and seems to make a difference if there is a quote
around the expression or not.
For example, package works with expressions such as
> expression(A + B)
but not with
> expression("A + B")
I now have a set of math expressions represented as strings, something like
this:
>
2010 Nov 26
3
Calling substitute(expr, list(a=1)) when expr <- expression(a+b+c)
# The result I am after is the result after a substitution in an expression, such as
substitute(expression(a+b+c), list(a=1))
expression(1 + b + c)
# However, the way I want to do it is for a an expression "stored as a variable" as
(expr <- expression(a+b+c))
expression(a + b + c)
# a) The following does not work
(expr2 <- substitute(expr, list(a=1)))
expr
# b) - whereas this
2004 Jul 27
2
Incorrect display of b[hat((a))] expression in plots
Hi, I am not sure if this is a bug or a non-implement feature of
text-drawing functions with TeX-style expression, but hat() and some of its
equals does not get the right "bounding boxes" if they are put in sub- or
superscripts. For instance, for the expression 'b[hat(a)]' the hat() seems
to shift 'hat(a)' too much to the right of 'b'. Try the below example and
2007 Mar 27
7
Replacement in an expression - can't use parse()
Dear all,
Suppose I have a very long expression e. Lets assume, for simplicity, that it is
e = expression(u1+u2+u3)
Now I wish to replace u2 with x and u3 with 1. I.e. the 'new'
expression, after replacement, should be:
> e
expression(u1+x+1)
My question is how to do the replacement?
I have tried using:
> e = parse(text=gsub("u2","x",e))
> e =
2007 Feb 04
4
Reading expressions from character vectors
Greetings,
I have a problem that I am sure is very straightforward, but I just
can't wrap my head around it. I've read the help pages on text,
plotmath, expression, substitute, but somehow I can't find the answer
to this simple question.
Basically consider the following example:
plot( NULL, xlim = c(0,2), ylim = c(0,2) )
expressions <- expression( -infinity, infinity )
2010 Jan 29
2
evaluating expressions with sub expressions
Hallo
I'm having trouble figuring out how to evaluate an expression when one of
the variables in the expression is defined separately as a sub expression.
Here's a simplified example
mat <- expression(0, f1*s1*g1) # vector of formulae
g1 <- expression(1/Tm) # expansion of the definition of g1
vals <- data.frame(f1=1, s1=.5, Tm=2) # one set of possible values for
2013 Aug 23
3
[LLVMdev] defining symbols with lld
On Aug 22, 2013, at 6:42 PM, Shankar Easwaran <shankare at codeaurora.org> wrote:
> On 8/22/2013 3:44 PM, Nick Kledzik wrote:
>>
>> Linker scripts have the same need. My idea for this was to allow atoms to have an associated expression tree and would have references to all symbols in that tree. The resolver wouldn't need any special handling for this, and the backend
2009 Jul 23
3
How to pass a character argument which contains expressions to arg.names in barplot?
Hi all
Can anybody help me with this? I am trying to include in an automatic way
the argument in arg.names in a barplot. I generate the labels I want to
appear below the bars with a for loop, and they contain subscripts, so I
need to use expression
anch<-0.05
esp<-4
for (i in 1:dim(Ntot)[1])
{
naux<-Ntot[i,]
naux2<-naux[naux>0]
nind<-which(naux>0)
2001 Apr 30
2
plotting an expression
I am sure it is just me not understanding how R works, but could somebody
explain why
curve(cos(x))
works and
curve(expression(cos(x))
does not?
I have done some investigating and here is what I found. If I comment out
the line of curve indicated below, both calls work fine.
function (expr, from, to, n = 101, add = FALSE, type = "l", ylab = NULL,
log = NULL, xlim =
2006 Nov 21
2
Symbolic derivation using D in package stats - how do I properly convert the returned call into a character string?
Dear all,
I am using the function 'D' in the 'stats' package to perform symbolic
derivation.
This works very well and it is much faster than e.g. Mathematica (at
least for my purposes).
First, I would like to thank the development team for this excellent
function.
However, I run into trouble in some cases, particularly when I am to do
some operations on long expressions
1997 Nov 03
1
R-alpha: expression(..) objects -- c(ex1, ex2) `fails'
When investigating the
legend(x,y, expression ( <math>, <math> ), ....)
[as suggested by Jim Lindsey],
I once again stumbled over the fact that expression objects behave somewhat
``wierdly'', IMHO.
We can have expressions of length(.) > 1, and access elements of these like
list elements.
However, other natural things don't work with expressions; some example
2004 Sep 06
4
substitution in expression
I have been struggling with this problem for a while and I hope someone
could help me. Or if someone could point me to a section in the manual I
would be grateful.
x <- "my"
plot(1:10, main=expression(paste( x, Delta, "values" )))
Q : How do I get the title to say "my (triangle symbol) values" ?
The following trial-and-error produced mainly errors :
2013 Aug 22
3
[LLVMdev] defining symbols with lld
On Aug 22, 2013, at 1:32 PM, Michael Spencer <bigcheesegs at gmail.com> wrote:
> On Thu, Aug 22, 2013 at 12:54 PM, Shankar Easwaran <shankare at codeaurora.org> wrote:
> Hi Nick,
>
> I am planning to work on adding support for definining expressions for the Gnu flavor.
>
> Currently Gnu ld supports an option --defsym symbol=expression. The expression may be
2013 Nov 04
2
[LLVMdev] compile error when using overloaded = operator of DenseMap
Hi,
I am trying to implement Available Expressions data flow analysis. I
created the following class (I am giving here code snippet.):
namespace {
typedef DenseMap<Expression, uint32_t> DMTy; //Expression is a class I
defined.
struct DataFlowValue {
DMTy ExprMap;
llvm::BitVector* DFV;
// Functions operating on the data //
bool operator==(const DataFlowValue V) const;
2006 Mar 06
3
how to make plotmath expression work together with paste
Recent questions about using plotmath have renewed my interest in this question
I want to have expressions take values of variables from the
environment. I am able to use expressions, and I am able to use paste
to put text and values of variables into
plots. But the two things just won't work together.
Here is some example code that shows what I mean.
plot(NA,xlim=c(0,100),ylim=c(0,100))
2024 Apr 27
1
Should c(..., recursive = TRUE) and unlist(x, recursive = TRUE) recurse into expression vectors?
Reading the body of function 'AnswerType' in bind.c, called from 'do_c'
and 'do_unlist', I notice that EXPRSXP and VECSXP are handled identically
in the recurse = TRUE case.
A corollary is that c(recursive = TRUE) and unlist(recursive = TRUE)
treat expression vectors like expression(a, b) as lists of symbols and
calls. And since they treat symbols and calls as
2013 Oct 10
2
Help with expression()
Hi everyone,
I am hoping someone can help with my attempted use of the expression
function. I have a long series of text and variable to paste together
including a degree symbol. The text is to be placed on my scatter plot
using the mtext function.
Using expression like this:
changetext = expression(paste("Change from ",mini," to ", maxi, ":",
diff
2006 Sep 27
2
How to pass expression as an argument
Hi,
I am writing a function and need to pass a function expression as an argument, for instance,
myfun <- function( express) {
x<- c(1,2,3)
y<-express
}
if I call the above function by myfun( x*2 ), I get 2 as the result, instead of 2,4,6 , could someone help me to
fix this problem ?
Furthermore, is that
2001 Jul 03
2
How do one modify an expression?
I am using expressions to annotate x and y labels in plots. I start of with
a label as
xlab <- expression(X)
Note that it is the user that sets this. It could also be something more
complicated as expression(sqrt(R*G)). Can I easily add a log[2] "around"
this expression? I would like to get
xlab <- expression(log[2](X))
I basically know nothing about expressions.
Thanks