Error parsing dynamically-built ggplot2 code - r

Question

In the following reproducible example I'm attempting to build a ggplot2 function call dynamically, in order to be able to accommodate unknown number of mixture distribution components. The code produces this error message: Error in parse(text = g) : <text>:8:0: unexpected end of input. What is the problem with the code? (I'm aware of the method of pre-calculating plot data, storing it in a data frame, melting it and supplying it to ggplot2. I would like to explore the option below, as well.) Thank you!
library(ggplot2)
library(scales)
library(RColorBrewer)
library(mixtools)
NUM_COMPONENTS <- 2
set.seed(12345) # for reproducibility
data(diamonds, package='ggplot2') # use built-in data
myData <- diamonds$price
calc.component <- function(x, lambda, mu, sigma) {
lambda * dnorm(x, mean = mu, sd = sigma)
}
overlayHistDensity <- function(data, func) {
# extract 'k' components from mixed distribution 'data'
mix <- normalmixEM(data, k = NUM_COMPONENTS,
maxit = 100, epsilon = 0.01)
summary(mix)
DISTRIB_COLORS <-
suppressWarnings(brewer.pal(NUM_COMPONENTS, "Set1"))
# plot histogram, empirical and fitted densities
g <- "ggplot(data) +\n"
for (i in seq(length(mix$lambda))) {
args <- paste0("args.", i)
assign(args, list(lambda = mix$lambda[i], mu = mix$mu[i],
sigma = mix$sigma[i]))
g <- paste0(g,
"stat_function(fun = func, args = ",
args,
", aes(color = ",
DISTRIB_COLORS[i], ")) +\n")
}
tailStr <-
"geom_line(aes(y = ..density..,colour = 'Empirical'),stat = 'density') +
geom_histogram(aes(y = ..density..), alpha = 0.4) +
scale_colour_manual(name = '', values = c('red', 'blue')) +
theme(legend.position = 'top', legend.direction = 'horizontal')"
g <- paste0(g, tailStr)
gr <- eval(parse(text = g))
return (gr)
}
overlayHistDensity(log10(myData), 'calc.component')

Answer 1

As long as you realize you are going about this a hard way...
If you look at the value of g before it is parsed, it is
ggplot(data) +
stat_function(fun = func, args = args.1, aes(color = #E41A1C)) +
stat_function(fun = func, args = args.2, aes(color = #377EB8)) +
geom_line(aes(y = ..density..,colour = 'Empirical'),stat = 'density') +
geom_histogram(aes(y = ..density..), alpha = 0.4) +
scale_colour_manual(name = '', values = c('red', 'blue')) +
theme(legend.position = 'top', legend.direction = 'horizontal')
Usually the unexpected end of input message is from unbalanced quotes or parentheses, but you've not (obviously) got that problem here. The problem is in the color specification. Literal hex colors should be specified as strings
ggplot(data) +
stat_function(fun = func, args = args.1, aes(color = "#E41A1C")) +
stat_function(fun = func, args = args.2, aes(color = "#377EB8")) +
geom_line(aes(y = ..density..,colour = 'Empirical'),stat = 'density') +
geom_histogram(aes(y = ..density..), alpha = 0.4) +
scale_colour_manual(name = '', values = c('red', 'blue')) +
theme(legend.position = 'top', legend.direction = 'horizontal')
Without the quotes, the hash is a comment character and the rest of the lines (the right parentheses in particular) are not included, and the error you got is given. (Note the syntax highlighting that SO gives on the first code snippet.)
That said, I think you can get what you want without the eval(parse()) approach. In particular, look at aes_string which allows the specification of which variable is used as the aesthetic by the value of a string variable and adding a list of stats or geoms (which can be of un-pre-specified length created using lapply, for example). Also, you seem to be specifying literal colors and then mapping them to just red and blue; possibly you want scale_colour_identity? All this (last paragraph) is more code review and is not what you actually asked about.

Answer 2

You've got several problems:
ggplot's data argument must be a data.frame, not a vector
hex color names starting with # must be quoted, or they'll be interpreted as comments
you must to provide an aes(x = ) mapping
color definitions that are constant do not go in aes
This should work:
overlayHistDensity <- function(data, func) {
# extract 'k' components from mixed distribution 'data'
mix <- normalmixEM(data, k = NUM_COMPONENTS,
maxit = 100, epsilon = 0.01)
summary(mix)
DISTRIB_COLORS <-
suppressWarnings(brewer.pal(NUM_COMPONENTS, "Set1"))
# plot histogram, empirical and fitted densities
g <- "ggplot(as.data.frame(data), aes(x = data)) +\n"
for (i in seq(length(mix$lambda))) {
args <- paste0("args.", i)
assign(args, list(lambda = mix$lambda[i], mu = mix$mu[i],
sigma = mix$sigma[i]))
g <- paste0(g,
"stat_function(fun = func, args = ",
args,
", color = '",
DISTRIB_COLORS[i], "') +\n")
}
tailStr <-
"geom_line(aes(y = ..density..,colour = 'Empirical'),stat = 'density') +
geom_histogram(aes(y = ..density..), alpha = 0.4) +
scale_colour_manual(name = '', values = c('red', 'blue')) +
theme(legend.position = 'top', legend.direction = 'horizontal')"
g <- paste0(g, tailStr)
gr <- eval(parse(text = g))
return (gr)
}
Like Brian, I'll finish with 2 comments:
This is standard debugging and you shouldn't need an SO post for it. It's essentially several syntax errors and a couple little mistakes. I took your code outside of a function and ran it up through the final g <- paste0 line, and put the g output in a code window and looked for problems. Try to write code that works outside of a function first, then put it in a function.
Seconding Brian's comment, a more natural approach is to not use eval(parse()) and all this pasting. Instead, use aes_string, melt your data so that you can use one stat_function call based on a a grouping variable.