I'm wanting to create a rolling function that conditionally counts the occurance of two columns in previous rows.
As an example, I have a dataset that looks like the following.
# Generate data
set.seed(123)
test <- data.frame(
Round = rep(1:5, times = 3),
Team = rep(c("Team 1", "Team 2", "Team 3"), each = 5),
Venue = sample(sample(c("Venue A", "Venue B"), 15, replace = T))
)
Round Team Venue
1 1 Team 1 Venue B
2 2 Team 1 Venue B
3 3 Team 1 Venue A
4 4 Team 1 Venue A
5 5 Team 1 Venue B
6 1 Team 2 Venue B
7 2 Team 2 Venue B
8 3 Team 2 Venue A
9 4 Team 2 Venue A
10 5 Team 2 Venue A
11 1 Team 3 Venue B
12 2 Team 3 Venue A
13 3 Team 3 Venue B
14 4 Team 3 Venue B
15 5 Team 3 Venue B
I want a new column that shows for each row, the number of times the team in that row has played at the venue in that row in the last 3 rounds.
I can do this quite easily with a for loop.
window <- 3
for (i in 1:nrow(dat)){
# Create index to search (if i is less than window, start at 1)
index <- max(i - window, 1):i
# Search when current row matches both team and venue
dat$VenueCount[i] <- sum(dat$Team[i] == dat$Team[index] & dat$Venue[i] == dat$Venue[index])
}
Round Team Venue VenueCount
1 1 Team 1 Venue B 1
2 2 Team 1 Venue B 2
3 3 Team 1 Venue A 1
4 4 Team 1 Venue A 2
5 5 Team 1 Venue B 2
6 1 Team 2 Venue B 1
7 2 Team 2 Venue B 2
8 3 Team 2 Venue A 1
9 4 Team 2 Venue A 2
10 5 Team 2 Venue A 3
11 1 Team 3 Venue B 1
12 2 Team 3 Venue A 1
13 3 Team 3 Venue B 2
14 4 Team 3 Venue B 3
15 5 Team 3 Venue B 3
However, I'm wanting to avoid a for loop (mostly as my actual dataset is relatively large at around ~30k rows). I'm thinking it should be doable with one of zoo
, dplyr
, purrr
or apply
but haven't been able to work it out.
Thanks
venturing a data.table
solution here. Will take it down if you are only looking for dplyr
solution
You can roll using a window of size 4, then count the number of occurrence matching the latest row.
library(data.table)
library(zoo)
setDT(test)
winsize <- 4
test[, .(Round,
Venue,
VenueCount=rollapplyr(c(rep("", winsize-1), Venue), winsize,
function(x) sum(x==last(x)))),
by=.(Team)]
result:
# Team Round Venue VenueCount
# 1: Team 1 1 Venue B 1
# 2: Team 1 2 Venue B 2
# 3: Team 1 3 Venue A 1
# 4: Team 1 4 Venue A 2
# 5: Team 1 5 Venue B 2
# 6: Team 2 1 Venue B 1
# 7: Team 2 2 Venue B 2
# 8: Team 2 3 Venue A 1
# 9: Team 2 4 Venue A 2
# 10: Team 2 5 Venue A 3
# 11: Team 3 1 Venue B 1
# 12: Team 3 2 Venue A 1
# 13: Team 3 3 Venue B 2
# 14: Team 3 4 Venue B 3
# 15: Team 3 5 Venue B 3
I actually worked out an answer using rollify
from the tibbletime
package with dplyr::mutate
. Will post here but still open to other responses!
library(dplyr)
library(tibbletime)
# Create data
set.seed(123)
test <- data.frame(
Round = rep(1:5, times = 3),
Team = rep(c("Team 1", "Team 2", "Team 3"), each = 5),
Venue = sample(sample(c("Venue A", "Venue B"), 15, replace = T))
)
Use rollify
to create custom function.
last_n_games = 3
count_games <- rollify(function(x) sum(last(x) == x), window = last_n_games)
Now use mutate to run the function. This returns NA for the first 2 rows (ie last_n_games - 1
). I can then use group_by
and row_number
to count these first occurrences
test <- test %>%
group_by(Team) %>%
mutate(VenueCount = count_games(Venue)) %>%
group_by(Team, Venue) %>%
mutate(VenueCount = ifelse(is.na(VenueCount), row_number(Team), VenueCount))
This returns the following
# A tibble: 15 x 4
# Groups: Team, Venue [6]
Round Team Venue VenueCount
<int> <fct> <fct> <int>
1 1 Team 1 Venue B 1
2 2 Team 1 Venue B 2
3 3 Team 1 Venue A 1
4 4 Team 1 Venue A 2
5 5 Team 1 Venue B 1
6 1 Team 2 Venue B 1
7 2 Team 2 Venue B 2
8 3 Team 2 Venue A 1
9 4 Team 2 Venue A 2
10 5 Team 2 Venue A 3
11 1 Team 3 Venue B 1
12 2 Team 3 Venue A 1
13 3 Team 3 Venue B 2
14 4 Team 3 Venue B 2
15 5 Team 3 Venue B 3
So I like using data.table, it's fast, versatile.
The idea is to join itself on 2 times, with 2 lag (round+1)
and (round+2)
, so here is what I did.
> test1<-test
> test2<-test
> test<-as.data.table(test)
> test1<-as.data.table(test1)
> test2<-as.data.table(test2)
Get these data.frames into data.table after get replicas
> test1[,Round:=Round+1,]
> test2[,Round:=Round+2,]
Round with lags then join them together like this:
> test2[test1,on=c('Round','Team')][test,on=c('Round','Team')]
Round Team Venue i.Venue i.Venue.1
1: 1 Team 1 NA NA Venue B
2: 2 Team 1 NA Venue B Venue B
3: 3 Team 1 Venue B Venue B Venue A
4: 4 Team 1 Venue B Venue A Venue A
5: 5 Team 1 Venue A Venue A Venue B
6: 1 Team 2 NA NA Venue B
7: 2 Team 2 NA Venue B Venue B
8: 3 Team 2 Venue B Venue B Venue A
9: 4 Team 2 Venue B Venue A Venue A
10: 5 Team 2 Venue A Venue A Venue A
11: 1 Team 3 NA NA Venue B
12: 2 Team 3 NA Venue B Venue A
13: 3 Team 3 Venue B Venue A Venue B
14: 4 Team 3 Venue A Venue B Venue B
15: 5 Team 3 Venue B Venue B Venue B
Since this result a lot of NAs, here we use a function from R-Cookbook.com ben mentioned in his answer
compareNA <- function(v1,v2) {
# This function returns TRUE wherever elements are the same, including NA's,
# and false everywhere else.
same <- (v1 == v2) | (is.na(v1) & is.na(v2))
same[is.na(same)] <- FALSE
return(same)
}
we can get our end result:
> end <-
test2[test1, on = c('Round', 'Team')][test, on = c('Round',
'Team')][, VenueCount :=
(1 + compareNA(i.Venue.1, i.Venue) + compareNA(i.Venue.1, Venue)), ]
Explanation: test2
right join test1
, on Round
and Team
, and right join test
on Round
and Team
, so you get:
i.Venue.1
is current Venue of the Team
, i.Venue
is last Venue of the Team
, Venue
is last 2 Venue of the Team
,
with a logical
(1 + compareNA(i.Venue.1, i.Venue) + compareNA(i.Venue.1, Venue))
You can count how many times the team played on this venue in last 3 rounds.
> end
Round Team Venue i.Venue i.Venue.1 VenueCount
1: 1 Team 1 NA NA Venue B 1
2: 2 Team 1 NA Venue B Venue B 2
3: 3 Team 1 Venue B Venue B Venue A 1
4: 4 Team 1 Venue B Venue A Venue A 2
5: 5 Team 1 Venue A Venue A Venue B 1
6: 1 Team 2 NA NA Venue B 1
7: 2 Team 2 NA Venue B Venue B 2
8: 3 Team 2 Venue B Venue B Venue A 1
9: 4 Team 2 Venue B Venue A Venue A 2
10: 5 Team 2 Venue A Venue A Venue A 3
11: 1 Team 3 NA NA Venue B 1
12: 2 Team 3 NA Venue B Venue A 1
13: 3 Team 3 Venue B Venue A Venue B 2
14: 4 Team 3 Venue A Venue B Venue B 2
15: 5 Team 3 Venue B Venue B Venue B 3
hope this helps
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