Conditional replacement of values in a data.frame

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

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As others have pointed out, an alternative solution in your example is to use ifelse.

Try data.table's := operator :

DT = as.data.table(df)
DT[b==0, est := (a-5)/2.533]

It's fast and short. See these linked questions for more information on := :

Why has data.table defined :=

When should I use the := operator in data.table

How do you remove columns from a data.frame

R self reference

Here is one approach. ifelse is vectorized and it checks all rows for zero values of b and replaces est with (a - 5)/2.53 if that is the case.

df <- transform(df, est = ifelse(b == 0, (a - 5)/2.53, est))