计算R中的序列

Question

id  random  count
a   0        -1
a   1         1
a   1         2
a   0        -1
a   0        -2
a   1         1
a   0        -1
a   1         1
a   0        -1
b   0        -1
b   0        -2
b   1         1
b   0        -1
b   1         1
b   0        -1
b   0        -2
b   0        -3

id是一个玩家，random是二进制0或1，我想创建一个计数列，按玩家计算1和0的序列，最好没有循环，因为数据库非常大。

Answer 1

这是一个dplyr解决方案

dat %>%
  transform(idx = c(0,cumsum(random[-1L] != random[-length(random)]))) %>%
  group_by(id, idx) %>%
  mutate(count = -1*cumsum(random == 0) + cumsum(random == 1)) %>%
  ungroup() %>%
  select(-idx)

Source: local data frame [17 x 3]

   id random count
1   a      0    -1
2   a      1     1
3   a      1     2
4   a      0    -1
5   a      0    -2
6   a      1     1
7   a      0    -1
8   a      1     1
9   a      0    -1
10  b      0    -1
11  b      0    -2
12  b      1     1
13  b      0    -1
14  b      1     1
15  b      0    -1
16  b      0    -2
17  b      0    -3

Answer 2

我想这就是你要找的东西：

library(data.table)
setDT(DF)[, count := seq_len(.N), by=.(id,rleid(random))]

这使

    id random count
 1:  a      0     1
 2:  a      1     1
 3:  a      1     2
 4:  a      0     1
 5:  a      0     2
 6:  a      1     1
 7:  a      0     1
 8:  a      1     1
 9:  a      0     1
10:  b      0     2
11:  b      0     3
12:  b      1     1
13:  b      0     1
14:  b      1     1
15:  b      0     1
16:  b      0     2
17:  b      0     3

（在data.table包的下一个版本1.9.8中，会有一个小的快捷方式setDT(DF)[, count := rowid(rleid(random)), by=id]。我正在制作这个注释，所以我可以稍后更新答案。）

---

您可能还需要运行组的标识符：

DF[, rid := rleid(random), by=id]

这使

    id random count rid
 1:  a      0     1   1
 2:  a      1     1   2
 3:  a      1     2   2
 4:  a      0     1   3
 5:  a      0     2   3
 6:  a      1     1   4
 7:  a      0     1   5
 8:  a      1     1   6
 9:  a      0     1   7
10:  b      0     1   1
11:  b      0     2   1
12:  b      1     1   2
13:  b      0     1   3
14:  b      1     1   4
15:  b      0     1   5
16:  b      0     2   5
17:  b      0     3   5

如果你读完introductory materials on the package，你会发现这些变量也可以在一个步骤中创建。

Answer 3

我认为最简单的方法是使用streak_run包中的runner函数。 streak_run也是最快的，如下节所示

解

library(runner)
df <- data.frame( id = 1:10, random = sample(c(0,1), 10, replace=T))

df$count <- streak_run(df$random) 
df$count[df$random==0] <- -df$count[df$random==0]

df

#       id random count
#1   1      0    -1
#2   2      0    -2
#3   3      1     1
#4   4      1     2
#5   5      1     3
#6   6      1     4
#7   7      0    -1
#8   8      0    -2
#9   9      0    -3
#10 10      0    -4

---

基准

runner_example <- function(df){
  df$count <- streak_run(df$random) 
  df$count[df$random==0] <- -df$count[df$random==0]
  return(df)}

dplyr_example <- function(df){
  df %>%
    transform(idx = c(0,cumsum(random[-1L] != random[-length(random)]))) %>%
    group_by(id, idx) %>%
    mutate(count = -1*cumsum(random == 0) + cumsum(random == 1)) %>%
    ungroup() %>%
    select(-idx)
  return(df)}

dt_example <- function(df){
  setDT(df)[, count := seq_len(.N), by=.(id,rleid(random))]
  return(df)}


library(dplyr);library(data.table)
library(microbenchmark); library(magrittr)
df <- data.frame( id = 1:2000L, random = sample(letters[1:2], 2000L, replace=T))

microbenchmark(
  dplyr  = dplyr_example(df),
  dt     = dt_example(df),
  runner = runner_example(df),
  times=100
)

#Unit: microseconds
#   expr        min         lq       mean      median         uq        max neval
#  dplyr 134388.839 164274.611 204478.048 188548.4975 222777.298 526019.563   100
#     dt   1306.139   1710.665   2181.989   1941.3420   2380.953   5581.682   100
# runner    284.522    741.145   1022.456    853.5715   1004.553   7398.019   100