Menu
You can do that by using the function arrange from dplyr. 2. Use the dplyr filter function to get the first and the last row of each group. This is a combination of duplicates removal that leaves the first and last row at the same time.
The last n rows of the data frame can be accessed by using the in-built tail() method in R. Supposedly, N is the total number of rows in the data frame, then n <=N last rows can be extracted from the structure.
By using bracket notation on R DataFrame (data.name) we can select rows by column value, by index, by name, by condition e.t.c. You can also use the R base function subset() to get the same results. Besides these, R also provides another function dplyr::filter() to get the rows from the DataFrame.
A fast and short data.table solution :
tmp[, .SD[c(1,.N)], by=id]
where .SD represents each (S)ubset of (D)ata, .N is the number of rows in each group and tmp is a data.table; e.g. as provided by fread() by default or by converting a data.frame using setDT().
Note that if a group only contains one row, that row will appear twice in the output because that row is both the first and last row of that group. To avoid the repetition in that case, thanks to @Thell:
tmp[, .SD[unique(c(1,.N))], by=id]
Alternatively, the following makes the logic explicit for the .N==1 special case :
tmp[, if (.N==1) .SD else .SD[c(1,.N)], by=id]
You don't need .SD[1] in the first part of the if because in that case .N is 1 so .SD must be just one row anyway.
You can wrap j in {} and have a whole page of code inside {} if you like. Just as long as the last expression inside {} returns a list- like object to be stacked (such as a plain list, data.table or data.frame).
tmp[, { ...; if (.N==1) .SD else .SD[c(1,.N)] } , by=id]
A plyr solution (tmp is your data frame):
library("plyr")
ddply(tmp, .(id), function(x) x[c(1, nrow(x)), ])
# id d gr mm area
# 1 15 1 2 3.4 1
# 2 15 1 1 5.5 2
# 3 21 1 1 4.0 2
# 4 21 1 2 3.8 2
# 5 22 1 1 4.0 2
# 6 22 1 2 4.6 2
# 7 23 1 1 2.7 2
# 8 23 1 2 3.0 2
# 9 24 1 1 3.0 2
# 10 24 1 2 2.0 3
Or with dplyr (see also here):
library("dplyr")
tmp %>%
group_by(id) %>%
slice(c(1, n())) %>%
ungroup()
# # A tibble: 10 × 5
# id d gr mm area
# <int> <int> <int> <dbl> <int>
# 1 15 1 2 3.4 1
# 2 15 1 1 5.5 2
# 3 21 1 1 4.0 2
# 4 21 1 2 3.8 2
# 5 22 1 1 4.0 2
# 6 22 1 2 4.6 2
# 7 23 1 1 2.7 2
# 8 23 1 2 3.0 2
# 9 24 1 1 3.0 2
# 10 24 1 2 2.0 3
Here is a solution in base R. If there are multiple groups with the same id this code returns the first and last row for each of those individual groups.
EDIT: January 12, 2017
This solution might be a little more intuitive than my other answer farther below:
lmy.df = read.table(text = '
id d gr mm area
15 1 2 3.40 1
15 1 1 4.90 2
15 1 1 4.40 1
15 1 1 5.50 2
21 1 1 4.00 2
21 1 2 3.80 2
22 1 1 4.00 2
23 1 1 2.70 2
23 1 1 4.00 2
23 1 2 3.00 2
24 1 1 3.00 2
24 1 1 2.00 3
24 1 1 4.00 2
24 1 2 2.00 3
', header = TRUE)
head <- aggregate(lmy.df, by=list(lmy.df$id), FUN = function(x) { first = head(x,1) } )
tail <- aggregate(lmy.df, by=list(lmy.df$id), FUN = function(x) { last = tail(x,1) } )
head$order = 'first'
tail$order = 'last'
my.output <- rbind(head, tail)
my.output
# Group.1 id d gr mm area order
#1 15 15 1 2 3.4 1 first
#2 21 21 1 1 4.0 2 first
#3 22 22 1 1 4.0 2 first
#4 23 23 1 1 2.7 2 first
#5 24 24 1 1 3.0 2 first
#6 15 15 1 1 5.5 2 last
#7 21 21 1 2 3.8 2 last
#8 22 22 1 1 4.0 2 last
#9 23 23 1 2 3.0 2 last
#10 24 24 1 2 2.0 3 last
EDIT: June 18, 2016
Since posting my original answer I have learned it is better to use lapply than apply. This is because apply does not work if every group has the same number of rows. See here: Error when numbering rows by group
lmy.df = read.table(text = '
id d gr mm area
15 1 2 3.40 1
15 1 1 4.90 2
15 1 1 4.40 1
15 1 1 5.50 2
21 1 1 4.00 2
21 1 2 3.80 2
22 1 1 4.00 2
23 1 1 2.70 2
23 1 1 4.00 2
23 1 2 3.00 2
24 1 1 3.00 2
24 1 1 2.00 3
24 1 1 4.00 2
24 1 2 2.00 3
', header = TRUE)
lmy.seq <- rle(lmy.df$id)$lengths
lmy.df$first <- unlist(lapply(lmy.seq, function(x) seq(1,x)))
lmy.df$last <- unlist(lapply(lmy.seq, function(x) seq(x,1,-1)))
lmy.df
lmy.df2 <- lmy.df[lmy.df$first==1 | lmy.df$last == 1,]
lmy.df2
# id d gr mm area first last
#1 15 1 2 3.4 1 1 4
#4 15 1 1 5.5 2 4 1
#5 21 1 1 4.0 2 1 2
#6 21 1 2 3.8 2 2 1
#7 22 1 1 4.0 2 1 1
#8 23 1 1 2.7 2 1 3
#10 23 1 2 3.0 2 3 1
#11 24 1 1 3.0 2 1 4
#14 24 1 2 2.0 3 4 1
Here is an example in which each group has two rows:
lmy.df = read.table(text = '
id d gr mm area
15 1 2 3.40 1
15 1 1 4.90 2
21 1 1 4.00 2
21 1 2 3.80 2
22 1 1 4.00 2
22 1 1 6.00 2
23 1 1 2.70 2
23 1 2 3.00 2
24 1 1 3.00 2
24 1 2 2.00 3
', header = TRUE)
lmy.seq <- rle(lmy.df$id)$lengths
lmy.df$first <- unlist(lapply(lmy.seq, function(x) seq(1,x)))
lmy.df$last <- unlist(lapply(lmy.seq, function(x) seq(x,1,-1)))
lmy.df
lmy.df2 <- lmy.df[lmy.df$first==1 | lmy.df$last == 1,]
lmy.df2
# id d gr mm area first last
#1 15 1 2 3.4 1 1 2
#2 15 1 1 4.9 2 2 1
#3 21 1 1 4.0 2 1 2
#4 21 1 2 3.8 2 2 1
#5 22 1 1 4.0 2 1 2
#6 22 1 1 6.0 2 2 1
#7 23 1 1 2.7 2 1 2
#8 23 1 2 3.0 2 2 1
#9 24 1 1 3.0 2 1 2
#10 24 1 2 2.0 3 2 1
Original answer:
my.seq <- data.frame(rle(my.df$id)$lengths)
my.df$first <- unlist(apply(my.seq, 1, function(x) seq(1,x)))
my.df$last <- unlist(apply(my.seq, 1, function(x) seq(x,1,-1)))
my.df2 <- my.df[my.df$first==1 | my.df$last == 1,]
my.df2
id d gr mm area first last
1 15 1 2 3.4 1 1 4
4 15 1 1 5.5 2 4 1
5 21 1 1 4.0 2 1 2
6 21 1 2 3.8 2 2 1
7 22 1 1 4.0 2 1 3
9 22 1 2 4.6 2 3 1
10 23 1 1 2.7 2 1 3
12 23 1 2 3.0 2 3 1
13 24 1 1 3.0 2 1 4
16 24 1 2 2.0 3 4 1
Another approach utilizing dplyr could be:
tmp %>%
group_by(id) %>%
filter(1:n() %in% range(1:n()))
id d gr mm area
<int> <int> <int> <dbl> <int>
1 15 1 2 3.4 1
2 15 1 1 5.5 2
3 21 1 1 4 2
4 21 1 2 3.8 2
5 22 1 1 4 2
6 22 1 2 4.6 2
7 23 1 1 2.7 2
8 23 1 2 3 2
9 24 1 1 3 2
10 24 1 2 2 3
Or the same idea with using row_number():
tmp %>%
group_by(id) %>%
filter(row_number() %in% range(row_number()))
Or performing the operation with slice():
tmp %>%
group_by(id) %>%
slice(c(which.min(1:n()), which.max(1:n())))
slice_head() and slice_tail()
library(tidyverse)
tmp <- structure(list(id = c(15L, 15L, 15L, 15L, 21L, 21L, 22L, 22L,
22L, 23L, 23L, 23L, 24L, 24L, 24L, 24L), d = c(1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), gr = c(2L, 1L,
1L, 1L, 1L, 2L, 1L, 1L, 2L, 1L, 1L, 2L, 1L, 1L, 1L, 2L), mm = c(3.4,
4.9, 4.4, 5.5, 4, 3.8, 4, 4.9, 4.6, 2.7, 4, 3, 3, 2, 4, 2), area = c(1L,
2L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 2L, 3L)), class = "data.frame", row.names = c(NA,
-16L))
tmp %>%
group_by(id) %>%
slice_head()
# A tibble: 5 x 5
# Groups: id [5]
id d gr mm area
<int> <int> <int> <dbl> <int>
1 15 1 2 3.4 1
2 21 1 1 4 2
3 22 1 1 4 2
4 23 1 1 2.7 2
5 24 1 1 3 2
tmp %>%
group_by(id) %>%
slice_tail()
# A tibble: 5 x 5
# Groups: id [5]
id d gr mm area
<int> <int> <int> <dbl> <int>
1 15 1 1 5.5 2
2 21 1 2 3.8 2
3 22 1 2 4.6 2
4 23 1 2 3 2
5 24 1 2 2 3
By default, slice_head() and slice_tail() return 1 row, but you can also specify the arguments n and prop with slice a number of rows or a proportion of rows respectively. See ?slice for more details.
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With