I am having some trouble replacing values in a dataframe. I would like to replace values based on a separate table. Below is an example of what I am trying to do.
I have a table where every row is a customer and every column is an animal they purchased. Lets call this dataframe table.
> table
# P1 P2 P3
# 1 cat lizard parrot
# 2 lizard parrot cat
# 3 parrot cat lizard
I also have a table that I will reference called lookUp.
> lookUp
# pet class
# 1 cat mammal
# 2 lizard reptile
# 3 parrot bird
What I want to do is create a new table called new with a function replaces all values in table with the class column in lookUp. I tried this myself using an lapply function, but I got the following warnings.
new <- as.data.frame(lapply(table, function(x) {
gsub('.*', lookUp[match(x, lookUp$pet) ,2], x)}), stringsAsFactors = FALSE)
Warning messages:
1: In gsub(".*", lookUp[match(x, lookUp$pet), 2], x) :
argument 'replacement' has length > 1 and only the first element will be used
2: In gsub(".*", lookUp[match(x, lookUp$pet), 2], x) :
argument 'replacement' has length > 1 and only the first element will be used
3: In gsub(".*", lookUp[match(x, lookUp$pet), 2], x) :
argument 'replacement' has length > 1 and only the first element will be used
Any ideas on how to make this work?
You posted an approach in your question which was not bad. Here's a smiliar approach:
new <- df # create a copy of df
# using lapply, loop over columns and match values to the look up table. store in "new".
new[] <- lapply(df, function(x) look$class[match(x, look$pet)])
An alternative approach which will be faster is:
new <- df
new[] <- look$class[match(unlist(df), look$pet)]
Note that I use empty brackets ([]) in both cases to keep the structure of new as it was (a data.frame).
(I'm using df instead of table and look instead of lookup in my answer)