如何根据 R 中现有列的值创建新列？

Question

I have a column for treatment names, unfortunately, the treatment is currently coded so that 1 = 0 kg N/ha, 2 = 40 kg N/ha, 3 = 80 kg N/ha, and so on. I'd like to add a column for the nitrogen rate applied to each treatment.

Here's what I've tried:

dput(head(df))
structure(list(Treatment = c("1", "10", "11", "12", "2", "3"), 
    slope = c(-355.55, -136.125, -137.6625, -96.5, -284.2375, 
    -334.5375)), row.names = 11:16, class = "data.frame")
df$Nrate[which(df$Treatment == 1)] = 0

I get the following error, which I don't understand because there's no reason that I can see why the new column should be any shorter than the old column:

Error in `$<-.data.frame`(`*tmp*`, Nrate, value = c(0, NA, NA, NA, NA,  : 
  replacement has 49 rows, data has 60

Also, I'm not sure how to create a whole column based on all the existing values, where treatment 1 & 7 = 0, treatment 2 & 8 = 40, treatment 3 & 9= 80, treatment 4 & 10 = 120, treatment 5 & 11 = 160, and treatment 6 & 12 = 200. With this code, it looks like it creates a new column with NAs where any number other than 1 is, and I suspect that if I wrote a new line to replace 2 with 40, it would not retain the information that 1 = 0. Any insight into the correct package to use or any sample code that can create a new column based on an existing one would be much appreciated.

Answer 1

You may try this approach.

df$Nrate <- factor(df$Treatment)
levels(df$Nrate)=list("0"=c(1, 7), "40"=c(2, 8), "80"=c(3, 9), 
               "120"=c(4, 10), "160"=c(5, 11), "200"=c(6, 12))
df
#    Treatment     slope Nrate
# 11         1 -355.5500     0
# 12        10 -136.1250   120
# 13        11 -137.6625   160
# 14        12  -96.5000   200
# 15         2 -284.2375    40
# 16         3 -334.5375    80