R聚合在函数中有多个参数 [英] R aggregate with multiple arguments in function
问题描述
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
密钥率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 )
函数(x,y = dat $ rate){
rate < - as.numeric(y)
values < - as.numeric(x)
return (sum(value * rate)/ sum(rates))
})
注意:这个功能只是一个例子!
这个实现的问题是, y = dat $ rate
给出了dat上的所有4个费率,当我想要只是2个汇总率!
Anny sugestion我该怎么做?
谢谢!
这是我设法实现的,使用 data.table
包:
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
好的。所以这很容易写出两个变量,但是当我们有更多的列时呢。使用 lapply(.SD,...)
结合您的功能:
首先,一些数据: / p>
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))
库(data.table)
datDT< - data.table(dat,key =key)
datDT
#密钥速率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
二,聚合: / p>
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
如果您有一个非常大的数据集,您可能需要探索 data.table
一般来说
对于什么是值得的,我也是成功的在基地R,但我不知道这会有多高效,特别是因为转置等等。
t (i(i,i))中的(i,i,b,b) 1:length(y)){
V1 [i]< - sum(x [2] * x [y [i]])/ sum(x [2])
}
}))
#[,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
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!
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
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