In R, how can I access the first element of each l

Question

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

Answer 1

Surprised not to see this classic in the answer stream yet.

> do.call(rbind, lapply(split(df, df$n), function(x) x[1,]))
##   n s
## 2 2 a
## 3 3 c
## 4 4 e

Answer 2

You could also use data.table

library(data.table)
dt = as.data.table(df)
dt[, list(firstelement = s[1]), by=n]

which would get you:

   n firstelement
1: 2            a
2: 3            c
3: 4            e

The by=n bit groups everything by each value of n so s[1] is getting the first element of each of those groups.

To get this as an extra column you could do:

dt[, newcol := s[1], by=n]
dt
#   n s newcol
#1: 2 a      a
#2: 2 b      a
#3: 3 c      c
#4: 3 d      c
#5: 4 e      e
#6: 4 f      e

So this just takes the value of s from the first row of each group and assigns it to a new column.

Answer 3

Here is an approach using match:

 df$s[match(levels(n), df$n)]

EDIT: Maybe this looks a bit confusing ...

To get a column which lists the first elements you could use match twice (but with x and table arguments swapped):

 df$firstelement <- df$s[match(levels(n), df$n)[match(df$n, levels(n))]]
 df$firstelement
 # [1] a a c c e e
 # Levels: a b c d e f

Lets look at this in detail:

 ## this returns the first matching elements
 match(levels(n), df$n)
 # [1] 1 3 5

 ## when we swap the x and table argument in match we get the level index
 ## for each df$n (the duplicated indices are important)
 match(df$n, levels(n))
 # [1] 1 1 2 2 3 3

 ## results in
 c(1, 3, 5)[c(1, 1, 2, 2, 3, 3)]
 # [1] 1 1 3 3 5 5
 df$s[c(1, 1, 3, 3, 5, 5)]
 # [1] a a c c e e
 # Levels: a b c d e f

Answer 4

the function ave is useful in these cases:

df$firstelement = ave(df$s, df$n, FUN = function(x) x[1])
df
  n s firstelement
1 2 a            a
2 2 b            a
3 3 c            c
4 3 d            c
5 4 e            e
6 4 f            e

Answer 5

df$s[sapply(levels(n), function(particular.level) { which(df$n == particular.level)[1]})]

I believe your problem is that you are comparing two vectors df$n is a vector and levels(n) is a vector. vector == vector only happens to work for you since df$n is a multiple length of levels(n)

Answer 6

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