[英]Calculate adstock using data.table
I have a dataframe as shown:我有一个数据框,如图所示:
structure(list(ID = c(1, 1, 1, 1, 2, 2, 2, 2), ColA = c(2, 3,
4, 5, 2, 3, 4, 5), ColB = c(1, 2, 3, 4, 1, 2, 3, 4), ColA_0.2 = c(2,
3.4, 4.68, 5.936, 2, 3.4, 4.68, 5.936), ColB_0.2 = c(1, 2.2,
3.44, 4.688, 1, 2.2, 3.44, 4.688)), class = "data.frame", row.names = c(NA,
-8L))
What I need ?我需要的 ? - For each ID, I want to calculate ColA_ad
and ColB_ad
. - 对于每个 ID,我想计算ColA_ad
和ColB_ad
。 User will pass a parameter 'ad'.用户将传递参数“ad”。
For example - if 'ad' is 0.2 then the values will be calculated as:例如 - 如果 'ad' 为 0.2,那么这些值将计算为:
ColA
(ie 2)第一行 - 与ColA
相同(即 2)ColA
to 0.2*First row of ColA_ad
(ie Sum(3,0.2*2)=3.4
)第二排-添加的第二排ColA
为0.2 *第一行ColA_ad
(即Sum(3,0.2*2)=3.4
)ColA
to 0.2*second row of ColA_ad
(ie Sum(4,0.2*3.4)=4.68
) and so on.第三排-添加第三排ColA
为0.2 *第二排ColA_ad
(即Sum(4,0.2*3.4)=4.68
)等。The same will be calculated for all other columns (here ColB), which can be mentioned in separate vector.对于所有其他列(此处为 ColB)计算相同,可以在单独的向量中提及。
Summary - I would take 0.2 times carry over effect of previous calculated row and add to new row.总结 - 我会采用前一个计算行的 0.2 倍结转效果并添加到新行。
The results are displayed in Column ColA_ad
and ColB_ad
.结果显示在列ColA_ad
和ColB_ad
。
As my dataset is very large, I am looking for data.table solution.由于我的数据集非常大,我正在寻找 data.table 解决方案。
Here is a base R solution, where a linear algebra property is applied to speed up your iterative calculation.这是一个基本的 R 解决方案,其中应用了线性代数属性来加速迭代计算。
basic idea (taking id = 1
as example)基本思路(以id = 1
为例)
col
to col_ad
, ie,您首先构造一个低三角矩阵,用于从col
映射到col_ad
,即,l <- 0.2**abs(outer(seq(4),seq(4),"-"))
l[upper.tri(l)] <- 0
which gives这使
> l
[,1] [,2] [,3] [,4]
[1,] 1.000 0.00 0.0 0
[2,] 0.200 1.00 0.0 0
[3,] 0.040 0.20 1.0 0
[4,] 0.008 0.04 0.2 1
l
over columns col
, ie,然后你在列col
使用l
,即,> l %*% as.matrix(subset(df,ID == 1)[-1])
ColA ColB
[1,] 2.000 1.000
[2,] 3.400 2.200
[3,] 4.680 3.440
[4,] 5.936 4.688
ad <- 0.2
col_ad <- do.call(rbind,
c(make.row.names = F,
lapply(split(df,df$ID),
function(x) {
l <- ad**abs(outer(seq(nrow(x)),seq(nrow(x)),"-"))
l[upper.tri(l)]<- 0
`colnames<-`(data.frame(l%*% as.matrix(x[-1])),paste0(names(x[-1]),"_",ad))
}
)
)
)
dfout <- cbind(df,col_ad)
such that以至于
> dfout
ID ColA ColB ColA_0.2 ColB_0.2
1 1 2 1 2.000 1.000
2 1 3 2 3.400 2.200
3 1 4 3 4.680 3.440
4 1 5 4 5.936 4.688
5 2 2 1 2.000 1.000
6 2 3 2 3.400 2.200
7 2 4 3 4.680 3.440
8 2 5 4 5.936 4.688
df <- structure(list(ID = c(1, 1, 1, 1, 2, 2, 2, 2), ColA = c(2, 3,
4, 5, 2, 3, 4, 5), ColB = c(1, 2, 3, 4, 1, 2, 3, 4)), class = "data.frame", row.names = c(NA,
-8L))
A non-recursive option:非递归选项:
setDT(DT)[, paste0(cols,"_",ad) := {
m <- matrix(unlist(shift(ad^(seq_len(.N)-1L), 0L:(.N-1L), fill = 0)), nrow=.N)
lapply(.SD, function(x) c(m%*%x))
}, by = ID, .SDcols = cols]
Another recursive option:另一个递归选项:
library(data.table)
setDT(DT)[, paste0(cols,"_",ad) := {
a <- 0
b <- 0
.SD[, {
a <- ColA + ad*a
b <- ColB + ad*b
.(a, b)
}, seq_len(.N)][, (1) := NULL]
},
by = ID]
output:输出:
ID ColA ColB ColA_0.2 ColB_0.2
1: 1 2 1 2.000 1.000
2: 1 3 2 3.400 2.200
3: 1 4 3 4.680 3.440
4: 1 5 4 5.936 4.688
5: 2 2 1 2.000 1.000
6: 2 3 2 3.400 2.200
7: 2 4 3 4.680 3.440
8: 2 5 4 5.936 4.688
data:数据:
DT <- structure(list(ID = c(1, 1, 1, 1, 2, 2, 2, 2), ColA = c(2, 3,
4, 5, 2, 3, 4, 5), ColB = c(1, 2, 3, 4, 1, 2, 3, 4), ColA_0.2 = c(2,
3.4, 4.68, 5.936, 2, 3.4, 4.68, 5.936), ColB_0.2 = c(1, 2.2,
3.44, 4.688, 1, 2.2, 3.44, 4.688)), class = "data.frame", row.names = c(NA,
-8L))
ad <- 0.2
cols <- c("ColA", "ColB")
Here is one way with data.table
using Reduce
:这是使用Reduce
data.table
一种方法:
#Columns to apply function to
cols <- names(df)[2:3]
#Create a function to apply
apply_fun <- function(col, ad) {
Reduce(function(x, y) sum(y, x * ad), col, accumulate = TRUE)
}
library(data.table)
#Convert dataframe to data.table
setDT(df)
#set ad value
ad <- 0.2
#Apply funnction to each columns of cols
df[, (paste(cols, ad, sep = "_")) := lapply(.SD, apply_fun, ad), .SDcols = cols, by = ID]
df
# ID ColA ColB ColA_0.2 ColB_0.2
#1: 1 2 1 2.000 1.000
#2: 1 3 2 3.400 2.200
#3: 1 4 3 4.680 3.440
#4: 1 5 4 5.936 4.688
#5: 2 2 1 2.000 1.000
#6: 2 3 2 3.400 2.200
#7: 2 4 3 4.680 3.440
#8: 2 5 4 5.936 4.688
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