Use previous row's calculated value to calculate current

Question

I have this data frame

> df

# rn a b c  d  e  f
# 1  1 2 NA NA NA NA
# 2  5 8 NA 4  5  6 
# 3  8 5 4  2  3  2 
# 4  4 2 5  5  6  2

I am trying to create a new column that's based on these conditions:

• When columns c to f are all NA, then b
• When only c is NA, then return the smallest value in columns b, and f i.e. min(b,f) or pmin
• When no NAs exist in the row, return the least value in b, e, and f + previous calculated value

The desired output is:

> df

# rn a b c  d  e  f  g
# 1  1 2 NA NA NA NA 2
# 2  5 8 NA 4  5  6  6  ### ? [b=8, f=6; least value = 6] 
# 3  8 5 4  2  3  2  3  ### ? [b=5, e=3, 'f + previous calculated value' = 2+6=8; least value = 3]
# 4  4 2 5  5  6  2  2  ### ? [b=2, e=6, 'f + previous calculated value' = 2+3=5; least value = 2]

I have tried this but I have no idea how to access the previously calculated value (using lag(g) as a placeholder) :

df%>%
  mutate(g = case_when(
      is.na(c) & is.na(d) & is.na(e) & is.na(f) ~ b,
      is.na(c) & !is.na(d) & !is.na(e) & !is.na(f) ~ pmin(b,f),
      !is.na(c) & !is.na(d) & !is.na(e) & !is.na(f) ~  pmin(b,e, f+lag(g)),
      TRUE ~ NA)
   )

Maybe I am not thinking about this the right way. Any help is greatly appreciated!

Answer 1

In this case I would say use a simple for-loop. You cannot use lag(g) because you haven't built g column yet.

    res <- rep(0, nrow(df))
    for (i in 1:nrow(df)) {
      row <- df[i, ]
      if (is.na(row["c"]) && is.na(row["f"])) {
        res[i] <- row["b"]
      } else if (is.na(row["c"])) {
        res[i] <- min(row["b"], row["f"])
      } else if (!is.na(row["d"])) {
        res[i] <- min(row["b"], row["e"], row["f") + res[i-1])
      }
    }
    df$g <- res