I have a data frame like this:
n = c(2, 2, 3, 3, 4, 4)
n <- as.factor(n)
s = c("a", "b", "c", "d", "e", "f")
df = data.frame(n, s)
df
n s
1 2 a
2 2 b
3 3 c
4 3 d
5 4 e
6 4 f
and I want to access the first element of each level of my factor (and have in this example a vector containing a, c, e
).
It is possible to reach the first element of one level, with
df$s[df$n == 2][1]
but it does not work for all levels:
df$s[df$n == levels(n)]
[1] a f
How would you do that?
And to go further, I’d like to modify my data frame to see which is the first element for each level at every occurrence. In my example, a new column should be:
n s rep firstelement
1 2 a a a
2 2 b c a
3 3 c e c
4 3 d a c
5 4 e c e
6 4 f e e
Edit. The first part of my answer addresses the original question, i.e. before "And to go further" (which was added by OP in an edit).
Another possibility, using duplicated
. From ?duplicated
: "duplicated()
determines which elements of a vector or data frame are duplicates of elements with smaller subscripts."
Here we use !
, the logical negation (NOT), to select not duplicated elements of 'n', i.e. first elements of each level of 'n'.
df[!duplicated(df$n), ]
# n s
# 1 2 a
# 3 3 c
# 5 4 e
Update Didn't see your "And to go further" edit until now. My first suggestion would definitely be to use ave
, as already proposed by @thelatemail and @sparrow. But just to dig around in the R toolbox and show you an alternative, here's a dplyr
way:
Group the data by n
, use the mutate
function to create a new variable 'first', with the value 'first element of s' (s[1]
),
library(dplyr)
df %.%
group_by(n) %.%
mutate(
first = s[1])
# n s first
# 1 2 a a
# 2 2 b a
# 3 3 c c
# 4 3 d c
# 5 4 e e
# 6 4 f e
Or go all in with dplyr
convenience functions and use first
instead of [1]
:
df %.%
group_by(n) %.%
mutate(
first = first(s))
A dplyr
solution for your original question would be to use summarise
:
df %.%
group_by(n) %.%
summarise(
first = first(s))
# n first
# 1 2 a
# 2 3 c
# 3 4 e