r - Conditional replacement of values in a data.frame

Question

I am trying to understand how to conditional replace values in a dataframe without using a loop. My data frame is structured as follows:

> df
          a b est
1  11.77000 2   0
2  10.90000 3   0
3  10.32000 2   0
4  10.96000 0   0
5   9.90600 0   0
6  10.70000 0   0
7  11.43000 1   0
8  11.41000 2   0
9  10.48512 4   0
10 11.19000 0   0

and the dput output is this:

structure(list(a = c(11.77, 10.9, 10.32, 10.96, 9.906, 10.7, 
11.43, 11.41, 10.48512, 11.19), b = c(2, 3, 2, 0, 0, 0, 1, 2, 
4, 0), est = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0)), .Names = c("a", 
"b", "est"), row.names = c(NA, -10L), class = "data.frame")

What I want to do, is to check the value of b. If b is 0, I want to set est to a value from a. I understand that df$est[df$b == 0] <- 23 will set all values of est to 23, when b==0. What I don't understand is how to set est to a value of a when that condition is true. For example:

df$est[df$b == 0] <- (df$a - 5)/2.533 

gives the following warning:

Warning message:
In df$est[df$b == 0] <- (df$a - 5)/2.533 :
  number of items to replace is not a multiple of replacement length

Is there a way that I can pass the relevant cell, rather than vector?

Answer 1

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.