I have a string variable and I would like to evaluate its value as an expression using expression()
Example:
>expression("Text")
expression("Text")
>a <- "Text"
>expression(a)
expression(a)
I want to do something to make expression(a) evaluate to expression("Text").
maybe you could write why you need it?
Because what you want makes no sense to me.
Type
?expression
and you will see:
expression returns a vector of type "expression" containing its arguments (unevaluated)
You would have to rewrite the function to make it do what you want.
If you wanted to fake it, you would do something like this (not recommended :) ):
paste("expression(", a, ")", sep="")
edit:
From ?expression you can see at the bottom, that as.expression(a) is what you needed
as.expression attempts to coerce its argument into an expression object
Related
I have to use the code bellow but I don't completely understand how it works. Why it won't work if I change du.4 by du.f and then use the f when calling the function? For some reason it only works with numbers and I do not undarstand why.
This is the error that it is giving in the case of du.f
Error in paste("Meth1=", nr, ".ps", sep = "") : object 'f' not found
du.4 <- function(u,v,a){(exp(a)*(-1+exp(a*v)))/(-exp(a)+exp(a+a*u)-exp(a*(u+v))+exp(a+a*v))}
plotmeth1 <- function(data1,data2,alpha,nr) {
psfile <-paste("Meth1=",nr,".ps",sep="")
diffmethod <-paste("du.",nr,sep="")
title=paste("Family",nr)
alphavalue <-paste("alpha=",round(alpha,digits=3),sep="")
#message=c("no message")
postscript(psfile)
data3<-sort(eval(call(diffmethod,data1,data2,alpha)))
diffdata <-data3[!is.na(data3)]
#if(length(data3)>length(diffdata))
#{message=paste("Family ",nr,"contains NA!")}
tq <-((1:length(diffdata))/(length(diffdata)+1))
plot(diffdata,tq,main=title,xlab="C1[F(x),G(y)]",ylab="U(0,1)",type="l")
legend(0.6,0.3,c(alphavalue))
abline(0,1)
#dev.off()
}
In R, a dot is used as just another character in identifiers. It is often used for clarity but doesn't have a formal function in defining the part after the dot as being in a name-space given by the part of the identifier before the dot. In something like du.f you can't refer to the function by f alone, even if your computation is inside of an environment named du. You can of course define a function named du.4 and then use 4 all by itself, but when you do so you are using the number 4 as just a number and not as a reference to the function. For example, if
du.4 <- function(u,v,a){(exp(a)*(-1+exp(a*v)))/(-exp(a)+exp(a+a*u)-exp(a*(u+v))+exp(a+a*v))}
Then du.4(1,2,3) evaluates to 21.08554 but attempting to use 4(1,2,3) throws the error
Error: attempt to apply non-function
In the case of your code, you are using paste to assemble the function name as a string to be passed to eval. It makes sense to paste the literal number 4 onto the string 'du.' (since the paste will convert 4 to the string '4') but it doesn't make sense to paste an undefined f onto 'du.'. It does, however, make sense to paste the literal string 'f' onto 'du.', so that the function call plotmeth1 (data1, data2, alpha, 'f') will work even though plotmeth1 (data1, data2, alpha, f) will fail.
See this question for more about the use of the dot in R identifiers.
I need to have the possibility to fill an expression with the values of the unknown number of variables. The shape of the expression depends on the number of the variables.
Example:
Expression1: "italic(y)==a*italic(x)*b"
to become: "y=1.2 x+4.3"
Expression2: "italic(y)==a*italic(x)*b~c"
to become: "y=1.2 x+4.3 -5.3"
Currently I am using the substitute function, but it does not work along with the expression function:
substitute(expression("italic(y)==a*italic(x)*b"),list(a=1.23,b=2.3))
My expression needs to grow as the number of variables (i.e. length of the list) increases. So, next step would be to add the variable c:
substitute(expression("space1*italic(y)==a*italic(x)*b*c"),list(a=1.23,b=2.3,c=3.2))
But I need to change the expression in the code without any manual interference and these codes do not read the variable values from the list unless I change it to this (in which the expression is not expandable anymore as it is not a string):
substitute(italic(y)==a*italic(x)*b*c,list(a=1.23,b=2.3,c=3.2))
How can I do this?
Here is a script which might be along the lines of what you want. We can iterate the list of replacements using a for loop, and then make a regex replacement of the placeholder in the expression with the corresponding value from the list.
lst <- list(a=1.23,b=2.3)
expression <- "italic(y)==a*italic(x)*b"
for (name in names(lst)) {
expression <- gsub(paste0("\\b", name, "\\b"), lst[[name]], expression)
}
print(expression)
[1] "italic(y)==1.23*italic(x)*2.3"
Note carefully that I search for the variable name surrounded by word boundaries on both sides. If your placeholder would ever be surrounded by other word characters, then my solution would fail, and we would need to change the replacement logic.
I wanted to answer a question regarding plotmath but I failed to get my desired substitute output.
My desired output:paste("Hi", paste(italic(yes),"why not?"))
and what I get: paste("Hi", "paste(italic(yes),\"why not?\")")
text<-'paste(italic(yes),"why not?")'
text
[1] "paste(italic(yes),\"why not?\")"
noqoute_text<-noquote(text)
noqoute_text
[1] paste(italic(yes),"why not?")
sub<-substitute(paste("Hi",noqoute_text),
env=list(noqoute_text=noqoute_text))
sub
paste("Hi", "paste(italic(yes),\"why not?\")")
You're using the wrong function, use parse instead of noquote :
text<-'paste(italic(yes),"why not?")'
noquote_text <- parse(text=text)[[1]]
sub<- substitute(paste("Hi",noquote_text),env=list(noquote_text= noquote_text))
# paste("Hi", paste(italic(yes), "why not?"))
noquote just applies a class to an object of type character, with a specific print method not to show the quotes.
str(noquote("a"))
Class 'noquote' chr "a"
unclass(noquote("a"))
[1] "a"
Would you please elaborate on your answer?
In R you ought to be careful about the difference between what's in an object, and what is printed.
What noquote does is :
add "noquote" to the class attribute of the object
That's it
The code is :
function (obj)
{
if (!inherits(obj, "noquote"))
class(obj) <- c(attr(obj, "class"), "noquote")
obj
}
Then when you print it, the methods print.noquote :
Removes the class "noquote" from the object if it's there
calls print with the argument quote = FALSE
that's it
You can actually call print.noquote on a string too :
print.noquote("a")
[1] a
It does print in a similar fashion as quote(a) or substitute(a) would but it's a totally different beast.
In the code you tried, you've been substituting a string instead of a call.
For solving the question I think Moody_Mudskipperss answer works fine, but as you asked for some elaboration...
You need to be careful about different ways similar-looking things are actually stored in R, which means they behave differently.
Especially with the way plotmath handles labels, as they try to emulate the way character-strings are normally handled, but then applies its own rules. The 3 things you are mixing I think:
character() is the most familiar: just a string. Printing can be confusing when quotes etc. are escaped. The function noquote basically tells R to mark it's argument, so that quotes are not escaped.
calls are "unevaluated function-calls": it's an instruction as to what R should do, but it's not yet executed. Any errors in this call don't come up yet, and you can inspect it.
Note that a call does not have its own evironment given with it, which means a call can give different results if evaluated e.g. from within a function.
Expressions are like calls, but applied more generally, i.e. not always a function that needs to be executed. An expression can be a variable-name, but also a simple value such as "why not?". Also, expressions can consist of multiple units, like you would have with {
Different functions can convert between these classes, but sometimes functions (such as paste!) also convert unexpectedly:
noquote does not do that much useful, as Moody_Mudskipper already pointed out: it only changes the printing. But the object basically remains a character
substitute not only substitutes variables, but also converts its first argument into (most often) a call. Here, the print bites you, for when printing a call, there is no provision for special classes of its members. Try it: sub[[3]] from the question gives[1] paste(italic(yes),"why not?")
without any backslashes! Only when printing the full call the noquote-part is lost.
parse is used to transform a character to an expression. Nothing is evaluated yet, but some structure is introduced, so that you could manipulate the expression.
paste is often behaving annoyingly (although as documented), as it can only paste together character-strings. Therefore, if you feed it anything but a character, it firs calls as.character. So if you give it a call, you just get a text-line again. So in your question, even if you'd use parse, as soon as you start pasting thing together, you get the quotes again.
Finally, your problem is harder because it's using plotmaths internal logic.
That means that as soon as you try to evaluate your text, you'll probably get an error "could not find function italic" (or a more confusing error if there is a function italic defined elsewhere). When providing it in plotmath, it works because the call is only evaluated by plotmath, which will give it a nice environment, where italic works as expected.
This all means you need to treat it all as an expression or call. As long as evaluation cannot be done (as long as it's you that handles the expression, instead of plotmath) it all needs to remain an expression or call. Giving substitute a call works, but you can also emulate more closely what happens in R, with
call('paste', 'Hi', parse(text=text)[[1]])
Trying to get into Julia after learning python, and I'm stumbling over some seemingly easy things. I'd like to have a function that takes strings as arguments, but uses one of those arguments as a regular expression to go searching for something. So:
function patterncount(string::ASCIIString, kmer::ASCIIString)
numpatterns = eachmatch(kmer, string, true)
count(numpatterns)
end
There are a couple of problems with this. First, eachmatch expects a Regex object as the first argument and I can't seem to figure out how to convert a string. In python I'd do r"{0}".format(kmer) - is there something similar?
Second, I clearly don't understand how the count function works (from the docs):
count(p, itr) → Integer
Count the number of elements in itr for which predicate p returns true.
But I can't seem to figure out what the predicate is for just counting how many things are in an iterator. I can make a simple counter loop, but I figure that has to be built in. I just can't find it (tried the docs, tried searching SO... no luck).
Edit: I also tried numpatterns = eachmatch(r"$kmer", string, true) - no go.
To convert a string to a regex, call the Regex function on the string.
Typically, to get the length of an iterator you an use the length function. However, in this case that won't really work. The eachmatch function returns an object of type Base.RegexMatchIterator, which doesn't have a length method. So, you can use count, as you thought. The first argument (the predicate) should be a one argument function that returns true or false depending on whether you would like to count a particular item in your iterator. In this case that function can simply be the anonymous function x->true, because for all x in the RegexMatchIterator, we want to count it.
So, given that info, I would write your function like this:
patterncount(s::ASCIIString, kmer::ASCIIString) =
count(x->true, eachmatch(Regex(kmer), s, true))
EDIT: I also changed the name of the first argument to be s instead of string, because string is a Julia function. Nothing terrible would have happened if we would have left that argument name the same in this example, but it is usually good practice not to give variable names the same as a built-in function name.
I hope to get some help on the use of quotation marks within a string for get().
Say, I want to retrieve an element from a list
some_list <- list(element1=11,element2=22,element3=33)
naturally, I can simply reference this element through
some_list[['element1']]
However, once I use this as a string within get(), R throws this error message
get("some_list[['element1']]")
> Error in get("some_list[['element1']]") :
object 'some_list[['element1']]' not found
I cannot figure out why this is the case. get() works fine when used with strings that do not have quotation marks within them, e.g.
get("some_list")
I also tried escaping the quotation marks within the string (although I don't this I would need to since they are single quotation marks) but it does not work either.
some_list[["\'"element1"\'"]]
What am I missing.
get won't do that.
some_list[['element1']] isn't the name of an object in an R environment (in a technical sense). When you type some_list[['element1']] at the console, R parses the expression, looks up the symbol some_list and then calls the function [[. get is intended just for the symbol lookup piece of that.
(Technically, my sequence of events there probably isn't right, but I listed them that way to help make the issue clear. Really, R is just parsing the expression, and then calling [[ with arguments some_list and 'element1', and those symbols are subsequently looked up.)
The quotes have nothing to do with it. Run:
get("some_list")[['element1']]