I am trying to rank counts, conditioned by two factors in a dataframe. However I would like to have a special treatment of the ties. If two values are equaly, I want them to have an equal tie value. However the next value in the rank should have the next integer rank.
Where I'm stuck is when I have the get the dataframe of unique values, conditional on the factor species. (In my actual data set it is conditional on three factors).
species <- c(rep("a", 3), rep("b", 4))
df <- data.frame(species, count = c("1", "1", "5", "1", "3", "3", "4"))
df$rank <- ave(df$count, df$species, FUN = rank)#doesnt get the output i'd like
#desired output
df$rank.good <- c("1", "1", "2", "1", "2", "2", "3")
df
With your data in its current form you have two problems, one of which is an R syntactic concern and the other is a "semantic" concern. The syntactic concern has been raised by @ARobertson who is really suggesting that you convert the "count" column to character. That will prevent the creation of spurious <NA>'s but won't solve the semantic problem of what to do if this is more than just a toy problem. If those count values are coming in as character values then sorting as characters will make the ordering: 1,10,11,12,...,19,2,20,21, .... So immediately after converting factors with as.character, you also need an as.numeric step, even if you resort to using dplyr::dense_rank:
dense_rank <- # copied from pkg::dplyr
function (x)
{ r <- rank(x)
match(r, sort(unique(r)))
}
df$rank.good <- ave(as.numeric(as.character(df$count)), df$species, FUN = dense_rank)
If you really want these to be character class you can wrap an outer as.character(.) around the ave function-call.
Try this:
# added more tests that are not sequential and fixed up data.frame
species <- c(rep("a", 3), rep("b", 4),rep("c",10))
df <- data.frame(species, count = c("1", "1", "5", "1", "3", "3", "4",'1','7','3','3','7','2','10','3','11','2'),stringsAsFactors = F)
df$count <- as.numeric(df$count)
# solution
df$rank <- ave(df$count, df$species, FUN = function(x){
r <- rank(x,ties.method = 'min')
as.numeric(factor(rank(sort(r))))[r]
})
