CODEWITHSUNDEEP
rexpressionsubstitution
Ben
Ben

Reputation: 103

How to use variable character strings with 'substitute' function in R

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?

Upvotes: 2

Views: 704

Answers (1)

Tim Biegeleisen
Tim Biegeleisen

Reputation: 520878

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.

Upvotes: 1

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