Two equal max values in R - r

Question

I have a dataframe with some numbers(score) and repeating ID. I want to get the maximum value for each of the ID.
I used this function
top = aggregate(df$score, list(df$ID),max)
This returned me a top dataframe with maximum values corresponding to each ID.
But it so happens that for one of the ID, we have two EQUAL max value. But this function is ignoring the second value.
Is there any way to retain BOTH the max values.?
For Example:
df
ID score
1 12
1 15
1 1
1 15
2 23
2 12
2 13
The above function gives me this:
top
ID Score
1 15
2 23
I need this:
top
ID Score
1 15
1 15
2 23

Answer 1

I recommend data.table as Chris mentioned (good for speed, but steeper learning curve).
Or if you don't want data.table you could use plyr:
library(plyr)
ddply(df, .(ID), subset, score==max(score))
# same as ddply(df, .(ID), function (x) subset(x, score==max(score)))

Answer 2

You can convert to a data.table:
DT <- as.data.table(df)
DT[, .SD[score == max(score)], by=ID]

Answer 3

Here is a dplyr solution.
library(dplyr)
df %>%
group_by(ID) %>%
filter(score == max(score))
Otherwise, to build on what you have done, we can use a sneaky property of merge on your "top" dataframe, see the following example:
df1 <- data.frame(ID = c(1,1,5,2), score = c(5,5,2,6))
top_df <- data.frame(ID = c(1,2), score = c(5,6))
merge(df1, top_df)
which gives:
ID score
1 1 5
2 1 5
3 2 6

Answer 4

Staying with a data.frame:
df[unlist(by(df, df$ID, FUN=function(D) rownames(D)[D$score == max(D$score)] )),]
# ID score
#2 1 15
#4 1 15
#5 2 23
This works because by splits df into a list of data.frames on the basis of df$ID, but retains the original rownames of df ( see by(df, df$ID, I) ). Therefore, returning the rownames of each D subset corresponding to a max score value in each group can still be used to subset the original df.

Answer 5

A simple base R solution:
df <- data.frame(ID = c(1, 1, 1, 1, 2, 2, 2),
score = c(12, 15, 1, 15, 23, 12, 13))
Several options:
df[df$score %in% tapply(df$score, df$ID, max), ]
df[df$score %in% aggregate(score ~ ID, data = df, max)$score, ]
df[df$score %in% aggregate(df$score, list(df$ID), max)$x, ]
Output:
ID score
2 1 15
4 1 15
5 2 23
Using sqldf:
library(sqldf)
sqldf('SELECT df.ID, score FROM df
JOIN (SELECT ID, MAX(score) AS score FROM df GROUP BY ID)
USING (score)')
Output:
ID score
2 1 15
4 1 15
5 2 23