Is there a faster way to do this? I guess this is unnecessary slow and that a task like this can be accomplished with base functions.
df <- ddply(df, "id", function(x) cbind(x, perc.total = sum(x$cand.perc)))
I'm quite new to R. I have looked at by()
, aggregate()
and tapply()
, but didn't get them to work at all or in the way I wanted. Rather than returning a shorter vector, I want to attach the sum to the original dataframe. What is the best way to do this?
Edit: Here is a speed comparison of the answers applied to my data.
> # My original solution
> system.time( ddply(df, "id", function(x) cbind(x, perc.total = sum(x$cand.perc))) )
user system elapsed
14.405 0.000 14.479
> # Paul Hiemstra
> system.time( ddply(df, "id", transform, perc.total = sum(cand.perc)) )
user system elapsed
15.973 0.000 15.992
> # Richie Cotton
> system.time( with(df, tapply(df$cand.perc, df$id, sum))[df$id] )
user system elapsed
0.048 0.000 0.048
> # John
> system.time( with(df, ave(cand.perc, id, FUN = sum)) )
user system elapsed
0.032 0.000 0.030
> # Christoph_J
> system.time( df[ , list(perc.total = sum(cand.perc)), by="id"][df])
user system elapsed
0.028 0.000 0.028
Since you are quite new to R and speed is apparently an issue for you, I recommend the data.table
package, which is really fast. One way to solve your problem in one line is as follows:
library(data.table)
DT <- data.table(ID = rep(c(1:3), each=3),
cand.perc = 1:9,
key="ID")
DT <- DT[ , perc.total := sum(cand.perc), by = ID]
DT
ID Perc.total cand.perc
[1,] 1 6 1
[2,] 1 6 2
[3,] 1 6 3
[4,] 2 15 4
[5,] 2 15 5
[6,] 2 15 6
[7,] 3 24 7
[8,] 3 24 8
[9,] 3 24 9
Disclaimer: I'm not a data.table expert (yet ;-), so there might faster ways to do that. Check out the package site to get you started if you are interested in using the package: http://datatable.r-forge.r-project.org/
For any kind of aggregation where you want a resulting vector the same length as the input vector with replicates grouped across the grouping vector ave
is what you want.
df$perc.total <- ave(df$cand.perc, df$id, FUN = sum)
Use tapply
to get the group stats, then add them back into your dataset afterwards.
Reproducible example:
means_by_wool <- with(warpbreaks, tapply(breaks, wool, mean))
warpbreaks$means.by.wool <- means_by_wool[warpbreaks$wool]
Untested solution for your scenario:
sum_by_id <- with(df, tapply(cand.perc, id, sum))
df$perc.total <- sum_by_id[df$id]
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