R根据以前的值填充列

Question

我想填充这样的一个系列。

   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 

下面是我用

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")) 

我没有得到的C右边的代码，你知道我缺少什么。我还附上了实际和预期产出的形象。感谢您的帮助！

Actual and Expected values Image

二图片 - 高附加值产品和商店列

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

列表下面是实际的代码，我也复制了预期与实际输出的图像。谢谢！

Answer 1

你的例子不是我称之为可重复的，代码片段也没有提供关于你想要做什么的很多见解。但是，来自excel的屏幕图像非常有帮助。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 
    } 
}