I have a data frame with 3 columns. a,b,c. There are multiple rows corresponding to each unique value of column a. I want to select top 5 rows corresponding to each unique value of column a. column c is some value and the data frame is already sorted by it in descending order, so that would not be a problem. Can anyone please suggest how can I do this in R.
Asked
Active
Viewed 998 times
0
-
2a combination of `plyr::ddply` and `head` or `data.table`. There are plenty of examples on SO – mnel Aug 30 '12 at 07:00
-
1and if, after you look for SO posts that are similar to your question, you still can't find a helpful answer, a [reproducible example](http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example) would help a lot – BenBarnes Aug 30 '12 at 07:16
2 Answers
4
Stealing @ptocquin's example, here's how you can use base function by
. You can flatten the result using do.call
(see below).
> by(data = data, INDICES = data$a, FUN = function(x) head(x, 5))
# or by(data = data, INDICES = data$a, FUN = head, 5)
data$a: 1
a b c
21 1 0.1188552 1.6389895
41 1 1.0182033 1.4811359
61 1 -0.8795879 0.7784072
81 1 0.6485745 0.7734652
31 1 1.5102255 0.7107957
------------------------------------------------------------
data$a: 2
a b c
15 2 -1.09704040 1.1710693
85 2 0.42914795 0.8826820
65 2 -1.01480957 0.6736782
45 2 -0.07982711 0.3693384
35 2 -0.67643885 -0.2170767
------------------------------------------------------------
A similar thing could be achieved by splitting your data.frame based on a
and then using lapply
to step through each element subsetting first n
rows.
split.data <- split(data, data$a)
subsetted.data <- lapply(split.data, FUN = function(x) head(x, 5)) # or ..., FUN = head, 5) like above
flatten.data <- do.call("rbind", subsetted.data)
head(flatten.data)
a b c
1.21 1 0.11885516 1.63898947
1.41 1 1.01820329 1.48113594
1.61 1 -0.87958790 0.77840718
1.81 1 0.64857445 0.77346517
1.31 1 1.51022545 0.71079568
2.15 2 -1.09704040 1.17106930
2.85 2 0.42914795 0.88268205
2.65 2 -1.01480957 0.67367823
2.45 2 -0.07982711 0.36933837
2.35 2 -0.67643885 -0.21707668

Roman Luštrik
- 69,533
- 24
- 154
- 197
1
Here is my try :
library(plyr)
data <- data.frame(a=rep(sample(1:20,10),10),b=rnorm(100),c=rnorm(100))
data <- data[rev(order(data$c)),]
head(data, 15)
a b c
28 6 1.69611039 1.720081
91 11 1.62656460 1.651574
70 9 -1.17808386 1.641954
6 15 1.23420550 1.603140
23 7 0.70854914 1.588352
51 11 -1.41234359 1.540738
19 10 2.83730734 1.522825
49 10 0.39313579 1.370831
80 9 -0.59445323 1.327825
59 10 -0.55538404 1.214901
18 6 0.08445888 1.152266
86 15 0.53027267 1.066034
69 10 -1.89077464 1.037447
62 1 -0.43599566 1.026505
3 7 0.78544009 1.014770
result <- ddply(data, .(a), "head", 5)
head(result, 15)
a b c
1 1 -0.43599566 1.02650544
2 1 -1.55113486 0.36380251
3 1 0.68608364 0.30911430
4 1 -0.85406406 0.05555500
5 1 -1.83894595 -0.11850847
6 5 -1.79715809 0.77760033
7 5 0.82814909 0.22401278
8 5 -1.52726859 0.06745849
9 5 0.51655092 -0.02737905
10 5 -0.44004646 -0.28106808
11 6 1.69611039 1.72008079
12 6 0.08445888 1.15226601
13 6 -1.99465060 0.82214319
14 6 0.43855489 0.76221979
15 6 -2.15251353 0.64417757

ptocquin
- 325
- 3
- 9
-
Note that the solutions by @RomanLuštrik is far more efficient than mine : 10 times quicker ! – ptocquin Aug 30 '12 at 12:33