根据两个标准为每组获取多个值答案

【问题标题】：Getting multiple values per group based on two criteria根据两个标准为每组获取多个值
【发布时间】：2021-11-01 22:03:29
【问题描述】：

我正在尝试对数据集进行分组，并根据时间和速度的两个独立度量来获取第一个值和最高值。所以我需要每组最早记录的时间和速度，然后是每组最快记录的时间和速度。我已经走了这么远，但需要一些帮助......

library(tidyverse)
group <- c(1,1,1,1,1,2,2,3,3,4,4,4,4,4,4)
time <- c(1,6,4,5,7,12,10,2,3,8,9,11,13,14,15)
speed <- c(17,6, 99, 34, 12, 5, 67, 43, 23, 12, 15, 78, 61, 78, 20)
data = data.frame(group, time, speed)
summary = data %>% 
  group_by(group) %>%
  summarise(
    firstTime =  # lowest time
    HighestSpeedTime = , # time for highest speed
    firstSpeed = , #speed for lowest time
    highestSpeed = max(speed), # highest speed
  )

【问题讨论】：

标签： r dplyr

【解决方案1】：

更新： 这应该可行：在第 4 组中，我们有 2 行平局：（我们在两个时间点有最高速度）！

library(dplyr)
data %>% 
  group_by(group) %>%
  summarise(
    firstTime = min(time), # lowest time
    HighestSpeedTime = time[which(speed==max(speed))], # time for highest speed
    firstSpeed = speed[which(time==min(time))],#speed for lowest time
    highestSpeed = max(speed) # highest speed
  )

输出：

  group firstTime HighestSpeedTime firstSpeed highestSpeed
  <dbl>     <dbl>            <dbl>      <dbl>        <dbl>
1     1         1                4         17           99
2     2        10               10         67           67
3     3         2                2         43           43
4     4         8               11         12           78
5     4         8               14         12           78

【讨论】：

感谢 TarJae，这是完美的。还要感谢 Zaw、Wimpel 和 Paul Smith 的意见。

【解决方案2】：

这行得通吗？

library(tidyverse)
group <- c(1,1,1,1,1,2,2,3,3,4,4,4,4,4,4)
time <- c(1,6,4,5,7,12,10,2,3,8,9,11,13,14,15)
speed <- c(17,6, 99, 34, 12, 5, 67, 43, 23, 12, 15, 78, 61, 78, 20)
data = data.frame(group, time, speed)

summary <-  data |> 
  arrange(group, time) |> 
  group_by(group) |> 
  summarise(
    firsttime = min(time),
    highest_speed = max(speed)
  ) |> 
  left_join(data, by = c("group", "highest_speed" = "speed")) |> 
  group_by(group) |> 
  slice(1) |> 
  rename(highest_speed_time = time) |> 
  left_join(data, by = c("group", "firsttime" = "time")) |> 
  rename(first_speed = speed)

summary

  # group firsttime highest_speed highest_speed_time first_speed
  # <dbl>     <dbl>         <dbl>              <dbl>       <dbl>
  #     1         1            99                  4          17
  #     2        10            67                 10          67
  #     3         2            43                  2          43
  #     4         8            78                 11          12

【讨论】：

【解决方案3】：

这是一个 data.table 方法

library(data.table)
setDT(data)
temp <- data[data[, .I[speed == max(speed)], by = .(group)]$V1]
setnames(temp, new = c("group", "maxSpeedTime", "maxSpeed"))
# join together
data[, .(firstTime  = time[1],
         firstSpeed = speed[1]), 
     by = .(group)][temp, on = .(group)]
#    group firstTime firstSpeed maxSpeedTime maxSpeed
# 1:     1         1         17            4       99
# 2:     2        12          5           10       67
# 3:     3         2         43            2       43
# 4:     4         8         12           11       78
# 5:     4         8         12           14       78

【讨论】：

【解决方案4】：

另一种解决方案，使用链式inner_join：

library(tidyverse)

data %>% 
  group_by(group) %>% 
  summarise(firstTime = min(time)) %>% 
  inner_join(data,by=c("group", "firstTime"="time")) %>% 
  rename(firstSpeed=speed) %>% 
  inner_join(
    data %>% 
      group_by(group) %>% 
      summarise(highestSpeed = max(speed)) %>% 
      inner_join(data,by=c("group", "highestSpeed"="speed")) 
  ) %>% 
  relocate(highestTime=time, .before="highestSpeed")

#> Joining, by = "group"
#> # A tibble: 5 × 5
#>   group firstTime firstSpeed highestTime highestSpeed
#>   <dbl>     <dbl>      <dbl>       <dbl>        <dbl>
#> 1     1         1         17           4           99
#> 2     2        10         67          10           67
#> 3     3         2         43           2           43
#> 4     4         8         12          11           78
#> 5     4         8         12          14           78

另一种解决方案，基于purrr::map_dfr：

library(tidyverse)

data %>% 
  group_split(group) %>% 
  map_dfr(
    ~ data.frame(
      group = .x$group[1],
      firstTime = .x$time[min(.x$time) == .x$time],
      firstSpeed = .x$speed[min(.x$time) == .x$time],
      highestTime = .x$time[max(.x$speed) == .x$speed],
      highestSpeed = .x$speed[max(.x$speed) == .x$speed]))

#>   group firstTime firstSpeed highestTime highestSpeed
#> 1     1         1         17           4           99
#> 2     2        10         67          10           67
#> 3     3         2         43           2           43
#> 4     4         8         12          11           78
#> 5     4         8         12          14           78

更简洁：

library(tidyverse)

data %>% 
  group_split(group) %>% 
  map_dfr(~ data.frame(
    group = integer(), firstTime = integer(), firstSpeed = integer(),
    highestTime = integer(), highestSpeed = integer()) %>% 
      add_row(!!!setNames(c(.x$group[1],.x[min(.x$time) == .x$time, -1],
            .x[max(.x$speed) == .x$speed, -1]), names(.))))

#>   group firstTime firstSpeed highestTime highestSpeed
#> 1     1         1         17           4           99
#> 2     2        10         67          10           67
#> 3     3         2         43           2           43
#> 4     4         8         12          11           78
#> 5     4         8         12          14           78

【讨论】：