Reputation: 89
Im tryng to avoid a time consuming for loop by using an aggregate on a data.frame. But I need that the values of one of the columns enters in the final computation.
dat <- data.frame(key = c('a', 'b', 'a','b'),
rate = c(0.5,0.4,1,0.6),
v1 = c(4,0,3,1),
v2 = c(2,0,9,4))
>dat
key rate v1 v2
1 a 0.5 4 2
2 b 0.4 0 0
3 a 1.0 3 9
4 b 0.6 1 4
aggregate(dat[,-1], list(key=dat$key),
function(x, y=dat$rate){
rates <- as.numeric(y)
values <- as.numeric(x)
return(sum(values*rates)/sum(rates))
})
Note: The function is just an example!
The problem of this implementation is that y=dat$rate
gives all 4 rates on dat, when what I want is just the 2 aggregated rates!
Anny sugestion on how I could do this?
Thanks!
Upvotes: 8
Views: 9306
Reputation: 193687
Here's what I managed to achieve, using the "data.table
" package:
DT <- data.table(dat, key = "key")
DT[, list(v1 = sum(rate * v1)/sum(rate), v2 = sum(rate * v2)/sum(rate)), by = "key"]
# key v1 v2
# 1: a 3.333333 6.666667
# 2: b 0.600000 2.400000
OK. So that's easy to write out for just two variables, but what about when we have a lot more columns. Use lapply(.SD,...)
in conjunction with your function:
First, some data:
set.seed(1)
dat <- data.frame(key = rep(c("a", "b"), times = 10),
rate = runif(20, min = 0, max = 1),
v1 = sample(10, 20, replace = TRUE),
v2 = sample(20, 20, replace = TRUE),
v3 = sample(30, 20, replace = TRUE),
x1 = sample(5, 20, replace = TRUE),
x2 = sample(6:10, 20, replace = TRUE),
x3 = sample(11:15, 20, replace = TRUE))
library(data.table)
datDT <- data.table(dat, key = "key")
datDT
# key rate v1 v2 v3 x1 x2 x3
# 1: a 0.26550866 10 17 28 3 9 15
# 2: a 0.57285336 7 16 14 2 7 13
# 3: a 0.20168193 3 11 20 4 9 14
# 4: a 0.94467527 1 1 15 4 6 13
# 5: a 0.62911404 9 15 3 2 10 12
# 6: a 0.20597457 5 10 11 2 10 13
# 7: a 0.68702285 5 9 11 4 7 11
# 8: a 0.76984142 9 2 15 4 6 15
# 9: a 0.71761851 8 7 26 3 9 13
# 10: a 0.38003518 8 14 24 5 8 15
# 11: b 0.37212390 3 13 9 4 7 13
# 12: b 0.90820779 2 12 10 2 10 11
# 13: b 0.89838968 4 16 8 2 7 13
# 14: b 0.66079779 4 10 23 1 8 12
# 15: b 0.06178627 4 14 27 1 8 13
# 16: b 0.17655675 6 18 26 1 9 11
# 17: b 0.38410372 2 5 11 5 8 14
# 18: b 0.49769924 7 2 27 4 6 13
# 19: b 0.99190609 2 11 12 3 6 13
# 20: b 0.77744522 5 9 29 4 9 13
Second, aggregate:
datDT[, lapply(.SD, function(x, y = rate) sum(y * x)/sum(y)), by = "key"]
# key rate v1 v2 v3 x1 x2 x3
# 1: a 0.6501303 6.335976 8.634691 15.75915 3.363832 7.658762 13.19152
# 2: b 0.7375793 3.595585 10.749705 16.26582 2.792390 7.741787 12.57301
If you have a really large dataset, you might want to explore data.table
in general.
For what it is worth, I was also successful in base R, but I'm not sure how efficient this would be, particularly because of the transposing and so on.
t(sapply(split(dat, dat[1]),
function(x, y = 3:ncol(dat)) {
V1 <- vector()
for (i in 1:length(y)) {
V1[i] <- sum(x[2] * x[y[i]])/sum(x[2])
}
V1
}))
# [,1] [,2] [,3] [,4] [,5] [,6]
# a 6.335976 8.634691 15.75915 3.363832 7.658762 13.19152
# b 3.595585 10.749705 16.26582 2.792390 7.741787 12.57301
Upvotes: 5
Reputation: 60984
One solution is to use ddply
from the plyr
package:
res = ddply(dat, .(key), summarise, result = sum(v1 * rate) / sum(rate))
> res
key result
1 a 3.333333
2 b 0.600000
If you want to apply this to all the v
columns, I would recommend first changing your data structure a bit:
dat = melt(dat, id.vars = c("key", "rate"))
> dat
key rate variable value
1 a 0.5 v1 4
2 b 0.4 v1 0
3 a 1.0 v1 3
4 b 0.6 v1 1
5 a 0.5 v2 2
6 b 0.4 v2 0
7 a 1.0 v2 9
8 b 0.6 v2 4
and then using ddply
again:
res = ddply(dat, .(key, variable), summarise, result = sum(value * rate) / sum(rate))
> res
key variable result
1 a v1 3.333333
2 a v2 6.666667
3 b v1 0.600000
4 b v2 2.400000
...or is you need a standard R solution, you can use by
:
res = by(dat, list(dat$key), function(x) sum(x$v1 * x$rate) / sum(x$rate))
> res
: a
[1] 3.333333
------------------------------------------------------------
: b
[1] 0.6
Upvotes: 3