我正在尝试在R中进行库存计算,这需要对每种Mat-Plant组合进行逐行计算。这是一个测试数据集-
df <- structure(list(Mat = c("A", "A", "A", "A", "A", "A", "B", "B"
), Plant = c("P1", "P1", "P1", "P2", "P2", "P2", "P1", "P1"),
Day = c(1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L), UU = c(0L, 10L,
0L, 0L, 0L, 120L, 10L, 0L), CumDailyFcst = c(11L, 22L, 33L,
0L, 5L, 10L, 20L, 50L)), .Names = c("Mat", "Plant", "Day",
"UU", "CumDailyFcst"), class = "data.frame", row.names = c(NA,
-8L))
Mat Plant Day UU CumDailyFcst
1 A P1 1 0 11
2 A P1 2 10 22
3 A P1 3 0 33
4 A P2 1 0 0
5 A P2 2 0 5
6 A P2 3 120 10
7 B P1 1 10 20
8 B P1 2 0 50
我需要一个新字段“ EffectiveFcst”,使得when Day = 1 then EffectiveFcst = CumDailyFcst
及其后的几天-
这是所需的输出-
Mat Plant Day UU CumDailyFcst EffectiveFcst
1 A P1 1 0 11 11
2 A P1 2 10 22 22
3 A P1 3 0 33 23
4 A P2 1 0 0 0
5 A P2 2 0 5 5
6 A P2 3 120 10 10
7 B P1 1 10 20 20
8 B P1 2 0 50 40
我当前正在使用for循环,但实际表超过30万行,因此希望使用tidyverse
进行此操作,以实现更优雅,更快速的方法。尝试了以下方法,但没有解决-
group_by(df, Mat, Plant) %>%
mutate(EffectiveFcst = ifelse(row_number()==1, CumDailyFcst, 0)) %>%
mutate(EffectiveFcst = ifelse(row_number() > 1, CumDailyFcst - lag(CumDailyFcst, default = 0) + max(lag(EffectiveFcst, default = 0) - lag(UU, default = 0), 0), EffectiveFcst)) %>%
print(n = nrow(.))
答案 0 :(得分:5)
我们可以使用accumulate
中的purrr
library(tidyverse)
df %>%
group_by(Mat, Plant) %>%
mutate(EffectiveFcst = accumulate(CumDailyFcst - lag(UU, default = 0), ~
.y , .init = first(CumDailyFcst))[-1] )
# A tibble: 8 x 6
# Groups: Mat, Plant [3]
# Mat Plant Day UU CumDailyFcst EffectiveFcst
# <chr> <chr> <int> <int> <int> <dbl>
#1 A P1 1 0 11 11
#2 A P1 2 10 22 22
#3 A P1 3 0 33 23
#4 A P2 1 0 0 0
#5 A P2 2 0 5 5
#6 A P2 3 120 10 10
#7 B P1 1 10 20 20
#8 B P1 2 0 50 40