R根据以前的值填充列

Question

我正在尝试填充这样的系列。

                My result                 ACTUAL         Expected       
FWK_SEQ_NBR  a  initial_d   initial_c   b   c   d       b   c   d
914        9.161    131       62        0   62  69      0   62  69
915        9.087    131       0         0   53  78      0   53  78
916        8.772    131       0         0   44  140     0   44  87
917        8.698    131       0         0   0   140     0   35  96
918        7.985    131       0        69   52  139    69   96  35
919        6.985    131       0        78   63  138    78  168   0
920        7.077    131       0       140   126 138    87  247   0
921        6.651    131       0       140   126 138    96  336   0
922        6.707    131       0       139   125 138    35  364   0

逻辑

a     given
b     lag of d by 4
c     initial c for first week thereafter (c previous row + b current - a current)
d     initial d - c current

这是我使用的代码

DS1 = DS %>% 
mutate(c    = ifelse(FWK_SEQ_NBR == min(FWK_SEQ_NBR), intial_c, 0)   ) %>%
mutate(c    = lag(c) + b - a)) %>% 
mutate(d    = initial_d - c) %>% 
mutate(d    = ifelse(d<0,0,d)) %>%
mutate(b    = shift(d, n=4, fill=0, type="lag"))

我没有得到正确的，你知道我错过了什么吗？我还附上了实际和预期输出的图像。谢谢你的帮助！

Actual and Expected values Image

第二张图片 - 添加了产品并存储到列列表

Image - Product and Store as the first two columns- please help

下面是实际代码，我还复制了预期和实际输出的图像。谢谢！

Answer 1

您的示例不是我称之为可重复的示例，而且代码段也没有提供有关您尝试执行的操作的详细信息。然而，来自excel的屏幕图像非常有用。这是我的解决方案

df <- as.data.frame(cbind(a = c(1:9), b = 0, c = 0, d = NA))
c_init = 62
d_init = 131
df$d <- d_init
df$c[1] <- c_init # initial data frame is ready at this stage

iter <- dim(df)[1] # for the loop to run item times

for(i in 1:iter){
  if(i>4){
    df[i, "b"] = df[i-4,"d"] # Calculate b with the lag
  }
 if(i>1){
    df[i, "c"] = df[i-1, "c"] + df[i, "b"] - df[i, "a"] # calc c
  }
  df[i, "d"] <- d_init - df[i, "c"] # calc d
  if(df[i, "d"] < 0) {
    df[i, "d"] <- 0 # reset negative d values
  }
}