Since you are conditionally indexing df$est
, you also need to conditionally index the replacement vector df$a
:
index <- df$b == 0
df$est[index] <- (df$a[index] - 5)/2.533
Of course, the variable index
is just temporary, and I use it to make the code a bit more readible. You can write it in one step:
df$est[df$b == 0] <- (df$a[df$b == 0] - 5)/2.533
For even better readibility, you can use within
:
df <- within(df, est[b==0] <- (a[b==0]-5)/2.533)
The results, regardless of which method you choose:
df
a b est
1 11.77000 2 0.000000
2 10.90000 3 0.000000
3 10.32000 2 0.000000
4 10.96000 0 2.352941
5 9.90600 0 1.936834
6 10.70000 0 2.250296
7 11.43000 1 0.000000
8 11.41000 2 0.000000
9 10.48512 4 0.000000
10 11.19000 0 2.443743
As others have pointed out, an alternative solution in your example is to use ifelse
.