Replace parts of a variable using numeric indices in dplyr. Do I need to create an index column and use ifelse?

Question

At one stage in longer chain of `dplyr` functions, I need to replace parts of a variable using numeric indices to specify which elements to replace. My data looks like this: ``` df1 <- data.frame(grp = rep(1:2, each = 3), a = 1:6, b = rep(c(10, 20), each = 3)) df1 # grp a b # 1 1 1 10 # 2 1 2 10 # 3 1 3 10 # 4 2 4 20 # 5 2 5 20 # 6 2 6 20 ``` Assume that we, within each group, wish to replace elements in variable `a` with the corresponding elements in `b`, at one or more positions. In this simple example I use a single index (`id`), but this could be a vector of indices. First, here's how I would do it with `ddply`: ``` library(plyr) id <- 2 ddply(.data = df1, .variables = .(grp), function(x){ x$a[id] <- x$b[id] x }) # grp a b # 1 1 1 10 # 2 1 10 10 # 3 1 3 10 # 4 2 4 20 # 5 2 20 20 # 6 2 6 20 ``` In `dplyr` I could think of some different ways to perform the replacement. (1) Use `do` with an anonymous function, similar to the one used in `ddply`. (2) Use `mutate`: concatenate a vecto…

Accepted Answer

Here's a much faster modify-in-place approach:

```
library(data.table)

# select rows we want, then assign b to a for those rows, in place
fun_dt = function(dt) dt[dt[, .I[id], by = grp]$V1, a := b]

# benchmark
df4 = data.frame(grp = rep(1:20000, each = 3),
                 a = rnorm(60000),
                 b = rnorm(60000))
dt4 = as.data.table(df4)

library(microbenchmark)

# using fastest function from OP
microbenchmark(fun_dt(dt4), fun_replace(df4), times = 10)
#Unit: milliseconds
#             expr      min        lq    median        uq       max neval
#      fun_dt(dt4) 15.62325  17.22828  18.42445  20.83768  21.25371    10
# fun_replace(df4) 99.03505 107.31529 116.74830 188.89134 286.50199    10
```

Replace parts of a variable using numeric indices in dplyr. Do I need to create an index column and use ifelse?

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