R中的滚动条件计数答案

【问题标题】：Rolling conditional count in RR中的滚动条件计数
【发布时间】：2018-04-04 14:44:08
【问题描述】：

我想创建一个滚动函数，有条件地计算前行中两列的出现次数。

例如，我有一个如下所示的数据集。

# 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

我想要一个新列，显示每一行中该行的球队在过去 3 轮比赛中在该行的场地进行比赛的次数。

我可以很容易地用 for 循环做到这一点。

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

但是，我想避免 for 循环（主要是因为我的实际数据集相对较大，约为 30k 行）。我认为使用zoo、dplyr、purrr 或apply 之一应该是可行的，但无法解决。

谢谢

【问题讨论】：

标签： r dplyr

【解决方案1】：

在这里尝试data.table 解决方案。如果您只是在寻找dplyr 解决方案，将其删除

您可以使用大小为 4 的窗口滚动，然后计算匹配最新行的出现次数。

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)]

结果：

#       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

【讨论】：

或者没有动物园test[, n := .SD[.(Team = Team, Venue = Venue, r_dn = Round - 3L, r_up = Round), on=.(Team, Venue, Round >= r_dn, Round <= r_up), .N, by=.EACHI]$N]
谢谢！我对data.table 不是很熟悉，但很高兴将其留给其他人

【解决方案2】：

我实际上使用tibbletime 包中的rollify 和dplyr::mutate 得出了一个答案。将在此处发布，但仍对其他回复开放！

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))
)

使用rollify 创建自定义函数。

last_n_games = 3
count_games <- rollify(function(x) sum(last(x) == x), window = last_n_games)

现在使用 mutate 来运行函数。这将返回前 2 行的 NA（即last_n_games - 1）。然后我可以使用group_by 和row_number 来计算这些第一次出现的次数

test <- test %>%
  group_by(Team) %>%
  mutate(VenueCount = count_games(Venue)) %>%
  group_by(Team, Venue) %>%
  mutate(VenueCount = ifelse(is.na(VenueCount), row_number(Team), VenueCount))

这将返回以下内容

# 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

【讨论】：

【解决方案3】：

所以我喜欢使用 data.table，它速度快、用途广泛。

这个想法是加入 2 次，有 2 个滞后 (round+1) 和 (round+2)，所以这就是我所做的。

> test1<-test
> test2<-test
> test<-as.data.table(test)
> test1<-as.data.table(test1)
> test2<-as.data.table(test2)

获取副本后将这些data.frames放入data.table中

> test1[,Round:=Round+1,]
> test2[,Round:=Round+2,]

有滞后的回合，然后像这样将它们连接在一起：

> 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

由于这个结果很多NA，这里我们使用来自R-Cookbook.com ben mentioned in his answer的函数

  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)
   }

我们可以得到最终结果：

 > 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)), ]

说明： test2 右加入test1，在Round 和Team，右加入test 在Round 和Team，所以你得到：

i.Venue.1 是Team 的当前地点， i.Venue 是Team 的最后一个地点， Venue是Team的最后2个场地，

有逻辑的

(1 + compareNA(i.Venue.1, i.Venue) + compareNA(i.Venue.1, Venue))

您可以计算球队在过去 3 轮比赛中在这个场地打了多少次。

> 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

希望对你有帮助

【讨论】：