R: passing column names as variables in custom function - r

Question

I am quite new to R and programming in general and have been struggling with the following for a few hours now.
I am trying to create a function that will take a df and a column name as variables, filter the table based on the column name provided and print the output.
example_function <- function(df=df, col=col){
a <- df[col == 100,]
b <- filter(df, col == 100)
print(a)
print(b)
}
Using example_function(df=example_df, col='percentage') doesn't work, both variables return just the column names but no data rows (despite there being values == 100).
Using example_function(df=df, col=percentage), so percentage isn't surrounded by quotes here, I get:
Error in [.data.frame(df, col == 100, ) : object 'percentage' not
found
However, when I run example_function(df=example_df, col=example_df$percentage) I get the correct result, with my dataframe returning as expected with only those rows where the example_df$percentage is equal to 100.
I really want to be able to pass the df as one variable and the column as another without having to type example_df$percentage each time as I want to be able to re-use the function for many different dataframes and typing that seems redundant.
Based on this I then modified the function thinking that I can just use df$col in the function and it will evaluate to example_df$percentage and work like it did above:
example_function <- function(df=df, col=col){
a <- df[df$col == 100,]
b <- filter(df, df$col == 100)
print(a)
print(b)
}
But now I get another error when using example_function(df=example_df, col=percentage) or when passing col='percentage':
Error in filter_impl(.data, quo) : Result must have length 19, not 0
Would any body be able to help me fix this, or point me in the right direction to understand why what I'm doing isn't working?
Thanks so much
Here is an example of the dataframe I am using (although my real one will have more columns but I hope it won't make a difference for this example.)
name | percentage
-----------------------
tom | 80
john | 100
harry | 99
elizabeth| 100
james | 50
example_df <- structure(list(name = structure(c(5L, 4L, 2L, 1L, 3L), .Label = c("elizabeth",
"harry", "james", "john", "tom"), class = "factor"), percentage = c(80L,
100L, 99L, 100L, 50L)), .Names = c("name", "percentage"), class = "data.frame", row.names = c(NA,
-5L))
as a note, I have updated my col=names to col=percentage in this example to more accurately represent what I am doing. In my attempt to generalise the example I used col=names and now realise that it wasn't a very good example (as you quite rightly asserted that a 'name' is never likely to be numeric). The above problems still persist for me however.
** Update: I managed to get it working with the following:
example_function <- function(df=df, col=col){
a <- df[df[col] == 100,]
print(a)
}
passing example_function(df=example_df, col='percentage')

Answer 1

The first row of example_function should be
a <- df[df[[col]] == 100,]
When you break it down, df[['names']] == 100 will give you a list of logicals corresponding to which rows of df has a names value of 100. But 'names' == 100 is nonsensical: it's always false.