我有一个包含一些变量的数据集
Plan <- c("A","A","A","B","B","B","B")
Plan_Period <- c(1,2,3,1,2,3,4)
Plan_Elapsed_time <- c(0.5,1,0.25,1,0.5,0.3,0.25)
year <- c(2016,2017,2018,2015,2016,2017,2018)
Inflation <- c(1.014,1.012,1.012,1.013,1.012, 1.080,1.020)
Cost <- c(10,20,30,40,40,50,60)
data <- data.frame(Plan, Plan_Period, Plan_Elapsed_time, year, Inflation, Cost)
计划A的美元价值从名义转换为实际的公式如下：
第1期的实际价值：10*(1.014^0.5)*(1.012^1)*(1.012^0.25)
第2期的实际价值：20*(1.012^1)*(1.012^0.25)
第3期的实际价值：30*(1.012^0.25)
我想使用其他函数在超过1000个不同计划的数据集上执行此操作，而不是编写for循环。
感谢您的帮助！
答案 0 :(得分：2)
使用Base R：我们也可以使用tidyverse
data=data.frame(Plan,Plan_Period ,Plan_Elapsed_time,year, Inflation,Cost
transform(data,m=Cost*ave(Inflation^Plan_Elapsed_time,Plan,
FUN=function(x)rev(cumprod(rev((x))))))
Plan Plan_Period Plan_Elapsed_time year Inflation Cost m
1 A 1 0.50 2016 1.014 10 10.22103
2 A 2 1.00 2017 1.012 20 20.30045
3 A 3 0.25 2018 1.012 30 30.08960
4 B 1 1.00 2015 1.013 40 41.92150
5 B 2 0.50 2016 1.012 40 41.38352
6 B 3 0.30 2017 1.080 50 51.42179
7 B 4 0.25 2018 1.020 60 60.29778
library(tidyverse)
data%>%
group_by(Plan)%>%
mutate(m=Cost*rev(cumprod(rev(Inflation^Plan_Elapsed_time))))
# A tibble: 7 x 7
# Groups: Plan [2]
Plan Plan_Period Plan_Elapsed_time year Inflation Cost m
<fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 A 1.00 0.500 2016 1.01 10.0 10.2
2 A 2.00 1.00 2017 1.01 20.0 20.3
3 A 3.00 0.250 2018 1.01 30.0 30.1
4 B 1.00 1.00 2015 1.01 40.0 41.9
5 B 2.00 0.500 2016 1.01 40.0 41.4
6 B 3.00 0.300 2017 1.08 50.0 51.4
7 B 4.00 0.250 2018 1.02 60.0 60.3
library(data.table)
setDT(data)[,m:=(Cost*rev(cumprod(rev(Inflation^Plan_Elapsed_time)))),by=Plan][]
Plan Plan_Period Plan_Elapsed_time year Inflation Cost m
1: A 1 0.50 2016 1.014 10 10.22103
2: A 2 1.00 2017 1.012 20 20.30045
3: A 3 0.25 2018 1.012 30 30.08960
4: B 1 1.00 2015 1.013 40 41.92150
5: B 2 0.50 2016 1.012 40 41.38352
6: B 3 0.30 2017 1.080 50 51.42179
7: B 4 0.25 2018 1.020 60 60.29778
答案 1 :(得分：2)
不需要循环。使用data.table包，您可以按组计算累计通货膨胀，并将结果乘以成本：
data <- data.frame(
Plan = c("A","A","A","B","B","B","B"),
Plan_Period=c(1,2,3,1,2,3,4),
Plan_Elapsed_time=c(0.5,1,0.25,1,0.5,0.3,0.25),
year=c(2016,2017,2018,2015,2016,2017,2018),
Inflation= c(1.014,1.012,1.012,1.013,1.012, 1.080,1.020),
Cost= c(10,20,30,40,40,50,60)
)
library(data.table)
setDT(data)
data <- data[order(Plan, -Plan_Period)][, Cum_Inflation := cumprod(Inflation^Plan_Elapsed_time), by = Plan][, Real_Cost := Cost * Cum_Inflation]
print(data)
#> Plan Plan_Period Plan_Elapsed_time year Inflation Cost Cum_Inflation Real_Cost
#> 1: A 3 0.25 2018 1.012 30 1.002987 30.08960
#> 2: A 2 1.00 2017 1.012 20 1.015022 20.30045
#> 3: A 1 0.50 2016 1.014 10 1.022103 10.22103
#> 4: B 4 0.25 2018 1.020 60 1.004963 60.29778
#> 5: B 3 0.30 2017 1.080 50 1.028436 51.42179
#> 6: B 2 0.50 2016 1.012 40 1.034588 41.38352
#> 7: B 1 1.00 2015 1.013 40 1.048038 41.92150
基于@Sathish评论的优化版本：
data <- data.frame(
Plan = c("A","A","A","B","B","B","B"),
Plan_Period=c(1,2,3,1,2,3,4),
Plan_Elapsed_time=c(0.5,1,0.25,1,0.5,0.3,0.25),
year=c(2016,2017,2018,2015,2016,2017,2018),
Inflation= c(1.014,1.012,1.012,1.013,1.012, 1.080,1.020),
Cost= c(10,20,30,40,40,50,60)
)
library(data.table)
setDT(data)[order(Plan, -Plan_Period), real_val := Cost * cumprod( Inflation ^ Plan_Elapsed_time ), by = .(Plan)]
data
#> Plan Plan_Period Plan_Elapsed_time year Inflation Cost real_val
#> 1: A 1 0.50 2016 1.014 10 10.22103
#> 2: A 2 1.00 2017 1.012 20 20.30045
#> 3: A 3 0.25 2018 1.012 30 30.08960
#> 4: B 1 1.00 2015 1.013 40 41.92150
#> 5: B 2 0.50 2016 1.012 40 41.38352
#> 6: B 3 0.30 2017 1.080 50 51.42179
#> 7: B 4 0.25 2018 1.020 60 60.29778
答案 2 :(得分：1)
这是一个tidyverse解决方案。我认为我理解所解释的逻辑，即更大的计划周期更新，因此inflation ^ plan_elapsed1的术语越少越好。在这里，我们arrange获取按plan然后plan_period排序的行，然后使用cumprod制作正确的术语，将cost乘以。 / p>
library(tidyverse)
data <- tibble(
Plan = c("A","A","A","B","B","B","B"),
Plan_Period = c(1,2,3,1,2,3,4),
Plan_Elapsed_time = c(0.5,1,0.25,1,0.5,0.3,0.25),
year = c(2016,2017,2018,2015,2016,2017,2018),
Inflation = c(1.014,1.012,1.012,1.013,1.012, 1.080,1.020),
Cost = c(10,20,30,40,40,50,60)
)
data %>%
`colnames<-`(str_to_lower(colnames(.))) %>%
mutate(deflate = inflation ^ plan_elapsed_time) %>%
group_by(plan) %>%
arrange(plan, desc(plan_period)) %>%
mutate(
cum_deflate = cumprod(deflate),
real_cost = cost * cum_deflate
) %>%
select(plan:cost, real_cost)
#> # A tibble: 7 x 7
#> # Groups: plan [2]
#> plan plan_period plan_elapsed_time year inflation cost real_cost
#> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 A 3. 0.250 2018. 1.01 30. 30.1
#> 2 A 2. 1.00 2017. 1.01 20. 20.3
#> 3 A 1. 0.500 2016. 1.01 10. 10.2
#> 4 B 4. 0.250 2018. 1.02 60. 60.3
#> 5 B 3. 0.300 2017. 1.08 50. 51.4
#> 6 B 2. 0.500 2016. 1.01 40. 41.4
#> 7 B 1. 1.00 2015. 1.01 40. 41.9
由reprex package（v0.2.0）创建于2018-04-09。
答案 3 :(得分：1)
考虑by内部sapply调用运行条件产品：
by_list <- by(data, data$Plan, function(sub){
sub$RealValue <- sapply(sub$Plan_Period, function(i)
sub$Cost[sub$Plan_Period == i] * prod((sub$Inflation[sub$Plan_Period >= i])^(sub$Plan_Elapsed_time[sub$Plan_Period >= i]))
)
return(sub)
})
finaldata <- do.call(rbind, unname(by_list))
finaldata
# Plan Plan_Period Plan_Elapsed_time year Inflation Cost RealValue
# 1 A 1 0.50 2016 1.014 10 10.22103
# 2 A 2 1.00 2017 1.012 20 20.30045
# 3 A 3 0.25 2018 1.012 30 30.08960
# 4 B 1 1.00 2015 1.013 40 41.92150
# 5 B 2 0.50 2016 1.012 40 41.38352
# 6 B 3 0.30 2017 1.080 50 51.42179
# 7 B 4 0.25 2018 1.020 60 60.29778
答案 4 :(得分：0)
不确定我是否误解了这个问题，但为这个数据写一个for循环似乎相当简单。
chkconfig NetworkManager off
service NetworkManager stop
希望这有用！