Reduce dataset based on value - r

Question

I have a dataset
dtf<-data.frame(id=c("A","A","A","A","B","B","B","B"), value=c(2,4,6,8,4,6,8,10))
for every id the values are sorted with ascending order
i want to reduce the dtf to include only the first row for every id that the value exceeds a specified limit. Only one row per id, and that should be the one that the value first exceed a specified limit.
For this example and for the limit of 5 the dtf should reduce to :
A 6
B 6
Is the a nice way to do this?
Thanks a lot

Answer 1

It could be done with aggregate:
dtf<-data.frame(id=c("A","A","A","A","B","B","B","B"), value=c(2,4,6,8,4,6,8,10))
limit <- 5
aggregate(value ~ id, dtf, function(x) x[x > limit][1])
The result:
id value
1 A 6
2 B 6
Update: A solution for multiple columns:
An example data frame, dtf2:
dtf2 <- data.frame(id=c("A","A","A","A","B","B","B","B"),
value=c(2,4,6,8,4,6,8,10),
col3 = letters[1:8],
col4 = 1:8)
A solution including ave:
with(dtf2, dtf2[ave(value, id, FUN = function(x) cumsum(x > limit)) == 1, ])
The result:
id value col3 col4
3 A 6 c 3
6 B 6 f 6

Answer 2

Here is a "nice" option using data.table:
library(data.table)
DT <- data.table(dft, key = "id")
DT[value > 5, head(.SD, 1), by = key(DT)]
# id value
# 1: A 6
# 2: B 6
And, in the spirit of sharing, an option using sqldf which might be nice depending on whether you feel more comfortable with SQL.
sqldf("select id, min(value) as value from dtf where value > 5 group by id")
# id value
# 1 A 6
# 2 B 6
Update: Unordered source data, and a data.frame with multiple columns
Based on your comments to some of the answers, it seems like there might be a chance that your "value" column might not be ordered like it is in your example, and that there are other columns present in your data.frame.
Here are two alternatives for those scenarios, one with data.table, which I find easiest to read and is most likely the fastest, and one with a typical "split-apply-combine" approach that is commonly needed for such tasks.
First, some sample data:
dtf2 <- data.frame(id = c("A","A","A","A","B","B","B","B"),
value = c(6,4,2,8,4,10,8,6),
col3 = letters[1:8],
col4 = 1:8)
dtf2 # Notice that the value column is not ordered
# id value col3 col4
# 1 A 6 a 1
# 2 A 4 b 2
# 3 A 2 c 3
# 4 A 8 d 4
# 5 B 4 e 5
# 6 B 10 f 6
# 7 B 8 g 7
# 8 B 6 h 8
Second, the data.table approach:
library(data.table)
DT <- data.table(dtf2)
DT # Verify that the data are not ordered
# id value col3 col4
# 1: A 6 a 1
# 2: A 4 b 2
# 3: A 2 c 3
# 4: A 8 d 4
# 5: B 4 e 5
# 6: B 10 f 6
# 7: B 8 g 7
# 8: B 6 h 8
DT[order(value)][value > 5, head(.SD, 1), by = "id"]
# id value col3 col4
# 1: A 6 a 1
# 2: B 6 h 8
Second, base R's common "split-apply-combine" approach:
do.call(rbind,
lapply(split(dtf2, dtf2$id),
function(x) x[x$value > 5, ][which.min(x$value[x$value > 5]), ]))
# id value col3 col4
# A A 6 a 1
# B B 6 h 8

Answer 3

Another approach with aggregate:
> aggregate(value~id, dtf[dtf[,'value'] > 5,], min)
id value
1 A 6
2 B 6
This does depend on the elements being sorted, as that will be the entry returned by min

Answer 4

might aswell, an alternative with plyr and head :
library(plyr)
dtf<-data.frame(id=c("A","A","A","A","B","B","B","B"), value=c(2,4,6,8,4,6,8,10))
limit <- 5
result <- ddply(dtf, "id", function(x) head(x[x$value > limit ,],1) )
> result
id value
1 A 6
2 B 6

Answer 5

This depends on your data.frame being sorted:
threshold <- 5
foo <- dtf[dtf$value>=threshold,]
foo[c(1,which(diff(as.numeric(as.factor(foo$id)))>0)),]