See the Warning section of `?factor`

:

In particular,

`as.numeric`

applied to

a factor is meaningless, and may

happen by implicit coercion. To

transform a factor`f`

to

approximately its original numeric

values,`as.numeric(levels(f))[f]`

is

recommended and slightly more

efficient than

`as.numeric(as.character(f))`

.

The FAQ on R has similar advice.

**Why is as.numeric(levels(f))[f] more efficent than as.numeric(as.character(f))?**

`as.numeric(as.character(f))`

is effectively `as.numeric(levels(f)[f])`

, so you are performing the conversion to numeric on `length(x)`

values, rather than on `nlevels(x)`

values. The speed difference will be most apparent for long vectors with few levels. If the values are mostly unique, there won’t be much difference in speed. However you do the conversion, this operation is unlikely to be the bottleneck in your code, so don’t worry too much about it.

**Some timings**

```
library(microbenchmark)
microbenchmark(
as.numeric(levels(f))[f],
as.numeric(levels(f)[f]),
as.numeric(as.character(f)),
paste0(x),
paste(x),
times = 1e5
)
## Unit: microseconds
## expr min lq mean median uq max neval
## as.numeric(levels(f))[f] 3.982 5.120 6.088624 5.405 5.974 1981.418 1e+05
## as.numeric(levels(f)[f]) 5.973 7.111 8.352032 7.396 8.250 4256.380 1e+05
## as.numeric(as.character(f)) 6.827 8.249 9.628264 8.534 9.671 1983.694 1e+05
## paste0(x) 7.964 9.387 11.026351 9.956 10.810 2911.257 1e+05
## paste(x) 7.965 9.387 11.127308 9.956 11.093 2419.458 1e+05
```