Automatically Convert Polynomials to Expression for ggplot Legend in R

Question

I'm currently trying to convert polynomial variable strings into expressions in R and have seen this similar link. However, I'm struggling to conceptualize a method that would allow me to provide a list of variables and develop a string that would appear correct for a ggplot legend. For example, I have the below string:

variable.names <- c("x", "y", "yy", "xxx", "xxy", "yyy", 
                    "xxxxx", "xxxxy", "xxxyy", "xxyyy", "yyyyy")

and would like to convert it automatically to:

new.variable.names <- c("x", "y", expression(y^2),
                    expression(x^3), expression(paste(x^2, y)), expression(y^3),
                    expression(x^5), expression(paste(x^4, y)), expression(paste(x^3, y)),
                    expression(paste(x^2, y^3)), expression(y^5))

This is one example, but I am hoping to write a function that can do this with possibly more variables in variable.names. I was thinking there was possibly a way to use regular expressions, but don't know how to develop a function that would determine the pattern of letters, and put them in the right position for the expression and paste functions to create the names automatically.

Thanks in advance.

How are your variable names created? Can you use that process to create expressions instead? — Roland, Nov 24 '20 at 14:33
The variable names are created using a similar method with ```strsplit```, they were created and reordered during an initial expansion of polynomials. the Method you've provided below seems to work how I'd want it. — AW27, Nov 24 '20 at 14:38
"expansion of polynomials" I would first create the expressions and than the variable names from the expressions, not the other way around. I dislike the need for parsing. — Roland, Nov 24 '20 at 14:41
The expansion was used based on the `polym` function in `R`. It generated the variable names that based on the numbers provided by the resulting columns. This was a way to keep the columns in order based on polynomial degree rather than alphabetical. — AW27, Nov 24 '20 at 14:53
Probably. I have never felt the need for using the tidyverse. Thus, I can't help you there. — Roland, Nov 24 '20 at 15:05
Let us [continue this discussion in chat](https://chat.stackoverflow.com/rooms/225038/discussion-between-aw27-and-roland). — AW27, Nov 24 '20 at 15:06

score 2 · Accepted Answer · answered Nov 24 '20 at 14:29

variable.names <- c("x", "y", "yy", "xxx", "xxy", "yyy", 
                    "xxxxx", "xxxxy", "xxxyy", "xxyyy", "yyyyy")

foo <- function(x) {
  runlength <- rle(x)
  string <- gsub("^1", "", 
                 paste0(paste(runlength$values, 
                              runlength$lengths, sep = "^"), 
                        collapse = "*"), 
                 fixed = TRUE)
  parse(text = string)
}

res <- lapply(strsplit(variable.names, ""), foo)



plot.new()

plot.window(c(0, 2), c(0, 11))
text(1, 10:0, do.call(c, res))

Synchronicity · Answer 2 · 2020-11-25T01:20:01.023

1

Here's a tidy solution UPDATED 2020-11-24T13:35:33 to work for any number of variables and to not include the ^ and output expression:

purrr::map(strsplit(variable.names, ""), ~{
    
    .str <- .x
    .vars <- purrr::map(unique(.str), ~{
        .instances <- sum(.str %in% .x)
        if (.instances == 0) .v <- NULL else if (.instances == 1) .v <- .x else if  (.instances > 1) .v <- paste0(.x,"^",.instances)
    })
    
    parse(text = paste0(.vars, collapse = "*"))  
    
})

edited Nov 25 '20 at 01:20

answered Nov 24 '20 at 14:29

Synchronicity

133
8

The only problem I'm facing with the tidy solution is that the variables don't look right in the legend. For example, I wouldn't want the ```^``` symbol in the legend. – AW27 Nov 24 '20 at 14:39
So you would want x2y2 for x^2y^2? – Synchronicity Nov 24 '20 at 16:03
@AW27 I updated the above to do what I think you want it to do? – Synchronicity Nov 24 '20 at 16:15
Based on Roland's answer, I want the result as an ```expression```. The updated tidy solution gives a string still, but I need it in an ```expression``` such as ```expression(y^5)```. Thanks – AW27 Nov 24 '20 at 16:33
1

@AW27 You just need to update the output line to give an expression: `rlang::parse_expr(do.call(paste0,.vars))` as i've done so above. – Synchronicity Nov 24 '20 at 18:27
I get the error ```Error: `x` must be a character vector or an R connection``` when attempting to change ```paste0``` with ```expression``` – AW27 Nov 24 '20 at 21:30
@AW27 The objects that output from the above code are already expressions... – Synchronicity Nov 24 '20 at 21:55
I'm having a hard time getting the same results as Roland did with his method. With the updated code for tidy, I get `y5` rather than `expression(y^5)`. The latter allows me to pass into `ggplot` and return what Roland shows. – AW27 Nov 24 '20 at 23:41
1

@AW27 Ok, your first comment said >"I wouldn't want the `^` in the legend so I thought you didnt want that. It's updated to give you the `expression` call – Synchronicity Nov 25 '20 at 01:19

Automatically Convert Polynomials to Expression for ggplot Legend in R

2 Answers2