Question

I am looking to use a function to speed up a data cleaning process. In the example shown I am looking to remove values reported in the am and pm columns if the “.no” column for that day has a value of 1.

df1 = data.frame (identifier = c(1:4),
mon.no = c(1,NA,NA,NA),mon.am = c(2,1,NA,3),mon.pm = c(3,4,NA,5),
tues.no = c(NA,NA,1,NA),tues.am = c(2,3,1,4),tues.pm = c(3,3,2,3))

I envisage using a function uses the day to clean the data:

clean1 = function (day) {
df1$day.am[df1$day.no==1] = NA
df1$day.pm[df1$day.no==1] = NA
return (df1)}
df2 = clean1(mon)

However this returns the following error.

Error in `$<-.data.frame`(`*tmp*`, "day.am", value = logical(0)) : 
replacement has 0 rows, data has 4

I assume that this is because the function expects a full column name and cannot fill in the gaps around a text input? Is it possible to use a function in that way?

Having read these notes I think that it would be better practice to have my data in a tidy format and am working on a solution which involves reorganising my data. However it would also be handy to be able to do this while the data is in it’s original format.

Thanks.

Answer 1

You’re really close. @Tyler Rinker in comments has explained why it doesn’t work. Here’s a fix:

clean1 = function (day) {
  day.am = paste(day, "am", sep=".") # make a string from the variable day and the suffixes
  day.pm = paste(day, "pm", sep=".")
  day.no = paste(day, "no", sep=".")
  df1[day.am][df1[day.no]==1] = NA 
  df1[day.pm][df1[day.no]==1] = NA
  return (df1)}

df2 = clean1("mon")   # "mon" should be a string

Somebody else might offer more efficient ways of doing this. Note that you’re only ever working from your original df1 here. If you now run

df3 = clean1("tues")

you won’t get a dataframe with both days cleaned. You could fix this by supplying the dataframe to be acted on to the function too:

clean2 = function(df, day){...