Passing a column as a function argument

Question

My df looks like this:

library(data.table)
df <- data.table(id = c(1,1,1,2,2), 
             starts = c(0,0,6,0,9), 
             ends = c(0,6,10,9,20))

I want to create a function which serves to create a data table with a new column containing the mean:

mean <- function(df, col_name){
library(data.table)

a <- data.table(

mean_df = df[col_name != 0, mean(col_name)]

)
return(a)
}

Then when I run the function

mean(df, "starts")

This gives me the error:

 Error in data.table(mean = df[col_name != 0, mean(col_name)]) : 
    argument "col_name" is missing, with no default

I'm not sure what I'm doing wrong - I tried get() and substitute() to pass the column name but I'm not sure if I'm doing it correctly.

Any help is appreciated!

EDIT1: Editted name of column to be created in function

you are overwriting the mean function with a new function called mean, then trying to call the original mean function inside your function...its best not to create functions with the same name as base R stuff — morgan121, Mar 01 '19 at 02:55
maybe `meanfun <- function(df, col_name){ df[abs(get(col_name)) > 0, mean(get(col_name))] } meanfun(df, "starts")` — chinsoon12, Mar 01 '19 at 02:56

score 1 · Accepted Answer · answered Mar 01 '19 at 11:37

I think using get() like so is pretty standard:

my_mean <- function(data, col_name) {
  data[get(col_name) != 0, mean(get(col_name))]
} 

my_mean(data = df, col_name = "starts")
[1] 7.5

BTW. it's a very bad idea to name your new function mean() since it will replace the base function.

1 Answers1