Question

我在R中有这种格式的数据集：

#!/usr/bin/expect -f

#Usage sshsudologin.expect <host> <ssh user> <ssh password> <foldername>

set timeout 20

spawn scp  -r "/sourcefolder/[lindex $argv 3]" [lindex $argv 1]@[lindex $argv 0]:"/export/home/[lindex $argv 1]/"

expect "yes/no" {
send "yes\r"
expect "*?assword" { send "[lindex $argv 2]\r" }
} "*?assword" { send "[lindex $argv 2]\r" }


expect eof

我需要将其转换为以下格式：

+----------+-------+-----------+
|  Person  | Group | Timestamp |
+----------+-------+-----------+
| Person A | X     | 12:00 PM  |
| Person A | X     | 12:01 PM  |
| Person A | X     | 12:03 PM  |
| Person A | Y     | 12:10 PM  |
| Person A | Y     | 12:11 PM  |
| Person A | Y     | 12:12 PM  |
| Person A | X     | 12:20 PM  |
| Person A | X     | 12:21 PM  |
| Person A | X     | 12:22 PM  |
| …        |       |           |
+----------+-------+-----------+

（将所有相似的条目分组为1-同一组可以在另一组之后重复就像上面的例子一样-组是X> Y> X）

我有数百个人，大约有2000万条记录。我尝试运行for循环，但这只花了太多时间。

请告诉我是否有更简单的方法来实现这一目标。

感谢您的帮助。预先感谢。

Answer 1

这是一个data.table解决方案，应该相当快。

library(data.table)

dt[, .(Ranking = rleid(Group), Group), by = .(Person)][, .SD[1], by = .(Ranking, Person)]
#      Person Ranking Group
# 1: Person A       1     X
# 2: Person A       2     Y
# 3: Person A       3     X

（原始方法未单独计算每个人的Rleid，已对其进行编辑以进行修复。）

另一种方法。不知道这样做是否会更快，但是我们可以将问题概念化为保留“人员”或“组”与上一行不同的行，然后按组编号：

dt[is.na(shift(Person)) | shift(Person) != Person | shift(Group) != Group, .(Person, Group)][, Ranking := 1:.N, by = .(Person)][]
#      Person Group Ranking
# 1: Person A     X       1
# 2: Person A     Y       2
# 3: Person A     X       3

使用此数据：

dt = fread("  Person  | Group | Timestamp
 Person A | X     | 12:00 PM  
 Person A | X     | 12:01 PM  
 Person A | X     | 12:03 PM  
 Person A | Y     | 12:10 PM  
 Person A | Y     | 12:11 PM  
 Person A | Y     | 12:12 PM  
 Person A | X     | 12:20 PM  
 Person A | X     | 12:21 PM  
 Person A | X     | 12:22 PM", sep = "|")

Answer 2

library(dplyr)
library(tidyr)
d %>%
    group_by(Person) %>%
    mutate(Ranking = sequence(rle(Group)$lengths) == 1) %>%
    ungroup() %>%
    select(-Timestamp) %>%
    filter(Ranking) %>%
    mutate(Ranking = cumsum(Ranking))
## A tibble: 3 x 3
#  Person   Group Ranking
#  <chr>    <chr>   <int>
#1 Person A X           1
#2 Person A Y           2
#3 Person A X           3

在基本R中

do.call(rbind, lapply(split(d, d$Person), function(x){
    data.frame(Person = x$Person[1],
               with(rle(x$Group),
                    data.frame(Group = values,
                               Ranking = seq_along(values))))}))

数据

d = structure(list(Person = c("Person A", "Person A", "Person A", 
                              "Person A", "Person A", "Person A",
                              "Person A", "Person A", "Person A"),
                   Group = c("X", "X", "X", "Y", "Y", "Y", "X", "X", "X"),
                   Timestamp = c("12:00 PM", "12:01 PM", "12:03 PM", "12:10 PM",
                                 "12:11 PM", "12:12 PM", "12:20 PM", "12:21 PM",
                                 "12:22 PM")),
              class = "data.frame",
              row.names = c(NA, -9L))

Answer 3

这是一个整洁的解决方案，可确保在返回排名之前，在Person中将时间戳按升序排序。

library(tidyverse)

get_ranking <- function(data) {
  grps <- rle(data$Group)$values
  data.frame(Group = grps, Ranking = seq_along(grps))
}

dat %>%
  group_by(Person) %>%
  arrange(Timestamp) %>%
  group_modify(~ get_ranking(.x))

使用此数据：

dat <- data.frame(Person= 'Person A', 
                  Group=rep(c('X','Y','X'),each=3), 
                  Timestamp=as.POSIXct('2010-01-01 12:00 PM')+(1:9)*60,
                  stringsAsFactors = FALSE)

要产生此输出：

# A tibble: 3 x 3
# Groups:   Person [1]
  Person   Group Ranking
  <chr>    <fct>   <int>
1 Person A X           1
2 Person A Y           2
3 Person A X           3

R中群组的汇总和排名

3 个答案: