为什么deparse（substitute（x））不选择'x'的名称

Question

I was wondering why deparse(substitute(x)) for my xlab does not put the name of x for xlab as expected ( see picture below )?

gg <- function(x, xlab = deparse(substitute(x)), ylab = NA, freq = FALSE, ...) { 
    x <- round(x)
    ylab <- if(is.na(ylab) & freq) {
        "Frequency" 
    } else if(is.na(ylab) & !freq) {
        "Probability" 
    } else ylab
    z <- if(freq) table(x) else table(x)/length(x)
    plot(z, xlab = xlab, ylab = ylab, ...)
}

# Example of use:
gg(mtcars$gear)    # 'mtcars' is a base R built-in dataset

Answer 1

The reason is lazy evaluation. (Don't ask me to explain the details, please. It's complicated and you can study this with the language definition . But basically, x is modified before xlab is evaluated.) You can fix this easily by using force :

gg <- function(x, xlab = deparse(substitute(x)), ylab = NA, freq = FALSE, ...) {
  force(xlab)
  x <- round(x)
  ylab <- if(is.na(ylab) & freq) "Frequency" else if(is.na(ylab) & !freq) "Probability" else ylab
  z <- if(freq) table(x) else table(x)/length(x)
  plot(z, xlab = xlab, ylab = ylab, ...)
}
# Example of use:
gg(mtcars$gear)