Question

My original data set has 62 variables. For variables c(3:56) I would like to loop the function boxplot.with.outlier.label, see

source(“http://www.r-statistics.com/wp-content/uploads/2011/01/boxplot-with-outlier-label-r.txt”)

But I am already stuck on building a function which would allow me to built the loop. Here is some mock up data (which, of course doesn’t show outliers, but to my knowledge this is not part of the problem – proof me wrong!)

x <- rnorm(1000)
y <- rnorm(1000)
z <- sample(letters, 1000)
df <- as.data.frame(cbind(x,y,z))
df$x<- as.numeric(df$x)
df$y<- as.numeric(df$y)
df$z<- as.character(df$z)

This works fine:

boxplot.with.outlier.label(df$x, df$z)

And this does not:

boxplotlabel <- function (data, var, name) {
  datvar <- data[["var"]]
  namevar <- data[["name"]]
  boxplot.with.outlier.label(datvar, namevar)
}
boxplotlabel(df, x, z)
Error in plot.window(xlim = xlim, ylim = ylim, log = log, yaxs = pars$yaxs) :need finite 'ylim' values
In addition: Warning messages:
1: In is.na(x) : is.na() applied to non-(list or vector) of type 'NULL'
2: In is.na(x) : is.na() applied to non-(list or vector) of type 'NULL'
3: In is.na(x) : is.na() applied to non-(list or vector) of type 'NULL'
4: In min(x) : no non-missing arguments to min; returning Inf
5: In max(x) : no non-missing arguments to max; returning -Inf

Where am I going wrong? Or is there a different way of achieving my desired loop for the function boxplot.with.outlier.label?

I appreciate any help! Gerit

Answer 1

The issue lies in the quotes. var and name are variables. But when you call data[["var"]] (with the quotes) you are not using the variable var but rather a string and the value of that string are the characters “var”.

If you remove the quotes you will be halfway there. Var itself should have a string value. so make sure to pass it the name of the column, not the column itself.

eg:

    # If you want to get this: 
    df$x 

    df[["x"]]   # right
    df[[x]]     # wrong

Therefore, if we’re using a variable for x:

    # Wrong
    var <- x
    df[[var]]

    # Right
    var <- "x"
    df[[var]]