how to search with number of occurrences in factor

Question

Find the date on which three movies had launched on the same day and store it in the variable date_three

releasedate<-count(bollywood$Rdate)

> releasedate

            x freq
1  01-05-2015    1
2  02-10-2015    2
3  03-07-2015    1
4  04-09-2015    1
5  04-12-2015    1
6  05-06-2015    1
7  06-02-2015    1
8  06-03-2015    1
9  07-08-2015    1
10 08-05-2015    2
11 09-01-2015    1
12 09-10-2015    1
13 10-04-2015    1
14 11-09-2015    1
15 12-06-2015    1
16 12-11-2015    1
17 13-02-2015    1
18 13-03-2015    1
19 14-08-2015    1
20 15-05-2015    1
21 16-01-2015    1
22 16-10-2015    1
23 17-04-2015    1
24 17-07-2015    1
25 18-09-2015    1
26 18-12-2015    2
27 19-06-2015    1
28 20-02-2015    1
29 20-03-2015    1
30 21-08-2015    2
31 22-05-2015    1
32 22-10-2015    1
33 23-01-2015    2
34 25-09-2015    2
35 26-06-2015    1
36 27-02-2015    2
37 27-11-2015    1
38 28-05-2015    1
39 28-08-2015    1
40 30-01-2015    2
41 30-10-2015    3
42 31-07-2015    1

>subset(releasedate$x,releasedate$freq==3)
>[1] 30-10-2015
42 Levels: 01-05-2015 02-10-2015 03-07-2015 04-09-2015 04-12-2015 ... 31-07-2015

Is there any other way I can search elements in vector by their no of occurence ?

What exactly do you need that's different from the solution you have? There are other ways to do anything. — effel, Jun 08 '16 at 21:24
I need to find out the date where 3 movies where released.I found out one solution by using plyr package then using count function to store x and freq in releasedate and then subsetting x where freq=3.Is there any other way of doing it ? — Deepak Bhattacharya, Jun 08 '16 at 21:33
You could do this with `aggregate` in base R, or with `data.table` via `.N`. — effel, Jun 08 '16 at 21:35
Can u please write the example for above code how to do using aggregate and data.table.Thanks — Deepak Bhattacharya, Jun 08 '16 at 21:42
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example — Bulat, Jun 08 '16 at 22:13
As effel said, what is wrong with your solution? Speed? Readability? — timcdlucas, Jun 08 '16 at 22:27

eipi10 · Accepted Answer · 2016-06-10T18:34:24.017

With dplyr:

library(dplyr)

date_three = bollywood %>% count(Rdate) %>% filter(n >= 3)

With data.table:

library(data.table)

date_three = setDT(bollywood)[ , list(freq=.N), by = Rdate ][freq >= 3]

or slightly more directly

date_three = setDT(bollywood)[, if (.N >= 3L) .(freq = .N), by = Rdate]

FWIW, here are some timings:

# Fake data
set.seed(2488)
bollywood=data.frame(Rdate=sample(seq(as.Date("2015-01-01"), as.Date("2015-12-31"), "1 day"), 
                                  1e6, replace=TRUE))

microbenchmark::microbenchmark(
  eipiDplyr = bollywood %>% count(Rdate) %>% filter(n >= 3),
  eipiDT = setDT(bollywood)[ , list(freq=.N), by = Rdate ][freq >= 3],
  ArunDT = setDT(bollywood)[, if (.N >= 3L) .(freq = .N), by = Rdate],
  times=20)

Unit: milliseconds
      expr      min       lq     mean   median       uq      max neval cld
 eipiDplyr 47.76676 51.21090 56.37334 53.48006 62.16901 71.94527    20   b
    eipiDT 43.41946 45.22264 47.57584 46.37179 47.97606 58.91733    20  a 
    ArunDT 42.97207 44.62598 47.76645 46.40803 51.46064 56.89516    20  a

Thanks @Arun. My `data.table` skills are somewhat, ahem, rudimentary. — eipi10, Jun 10 '16 at 18:21
no worries.. just wanted to show that we can avoid the intermediate data and get to the result directly. — Arun, Jun 10 '16 at 19:08

how to search with number of occurrences in factor

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