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基于R(dplyr,lubridate)中的多个条件创建开始和结束时间列

lubridate dplyr loops r

1

Lynn · 技术社区 · 5 年前

 Read      Box       ID      Time
 T         out               10/1/2019 9:00:01 AM
 T         out               10/1/2019 9:00:02 AM
 T         out               10/1/2019 9:00:03 AM
 T         out               10/1/2019 9:02:59 AM
 T         out               10/1/2019 9:03:00 AM
 F                           10/1/2019 9:05:00 AM
 T         out               10/1/2019 9:06:00 AM
 T         out               10/1/2019 9:06:02 AM
 T         in                10/1/2019 9:07:00 AM
 T         in                10/1/2019 9:07:02 AM
 T         out               10/1/2019 9:07:04 AM
 T         out               10/1/2019 9:07:05 AM
 T         out               10/1/2019 9:07:06 AM
           hello             10/1/2019 9:07:08 AM

基于此数据集中的某些条件,我想创建startime列和endtime列。当发生以下情况时,我想创建一个“starttime”:Read==“T”,Box==“out”,ID=>” 当此条件的第一个实例发生时,将生成starttime。例如,对于这个数据集,starttime将是10/1/2019 9:00:01 AM,因为这是我们首先看到所需条件的地方(Read=T,Box=out,ID=) 然而,当这些条件中的任何一个都不真实时,将创建结束时间。因此,第一个结束时间正好在第6行之前,时间是2019年10月1日上午9:03:00。我的最终目标是为此创建一个持续时间列。

这是我想要的输出:

  starttime                    endtime                     duration

  10/01/2019 9:00:01 AM        10/01/2019 9:03:00 AM       179 secs
  10/1/2019 9:06:00 AM         10/1/2019 9:06:02 AM        2 secs
  10/1/2019 9:07:04 AM         10/1/2019 9:07:06 AM        2 secs

  structure(list(Read = structure(c(3L, 3L, 3L, 3L, 3L, 2L, 3L, 
  3L, 3L, 3L, 4L, 4L, 3L, 1L), .Label = c("", "F", "T", "T "), class = "factor"), 
  Box = structure(c(3L, 3L, 3L, 3L, 3L, 1L, 3L, 3L, 2L, 2L, 
  3L, 3L, 3L, 1L), .Label = c("", "in", "out"), class = "factor"), 
  ID = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
  1L, 1L, 1L, 2L), .Label = c("", "hello"), class = "factor"), 
  Time = structure(1:14, .Label = c("10/1/2019 9:00:01 AM", 
 "10/1/2019 9:00:02 AM", "10/1/2019 9:00:03 AM", "10/1/2019 9:02:59 AM", 
 "10/1/2019 9:03:00 AM", "10/1/2019 9:05:00 AM", "10/1/2019 9:06:00 AM", 
 "10/1/2019 9:06:02 AM", "10/1/2019 9:07:00 AM", "10/1/2019 9:07:02 AM", 
 "10/1/2019 9:07:04 AM", "10/1/2019 9:07:05 AM", "10/1/2019 9:07:06 AM", 
 "10/1/2019 9:07:08 AM"), class = "factor")), class = "data.frame", row.names = c(NA, 
 -14L))

我认为总的来说,我必须创建一个循环。我相信我的思维过程是正确的,只是不确定如何制定代码。这就是我要尝试的:

 df2 <- mutate(df,
      Date = lubridate::mdy_hms(Date))




   for ( i in 2:nrow(df2))
    {
  if(df2$Read[[i]] == 'T')

     }

我认为这可能是一个开始(只是把我的条件放在循环中,我不知道如何完成这个)

1 回复 | 直到 5 年前

1

Ronak Shah 5 年前

你可以不用循环来做这个。使用 dplyr 因为使用管道很容易做多种事情。

Time 列到 POSIXct cond 根据要检查的条件提供逻辑值的列,使用 康德 first 和 last 时间 以及每组的差异。

library(dplyr)

df %>%
  mutate(Time = lubridate::mdy_hms(Time), 
         cond = Read == "T" & Box == "out" & ID == "", 
         grp = cumsum(!cond)) %>%
  filter(cond) %>%
  group_by(grp) %>%
  summarise(starttime = first(Time), 
            endtime = last(Time), 
            duration = difftime(endtime, starttime, units = "secs")) %>%
  select(-grp)

# A tibble: 3 x 3
#  starttime           endtime             duration
#  <dttm>              <dttm>              <drtn>  
#1 2019-10-01 09:00:01 2019-10-01 09:03:00 179 secs
#2 2019-10-01 09:06:00 2019-10-01 09:06:02   2 secs
#3 2019-10-01 09:07:04 2019-10-01 09:07:06   2 secs

数据

我整理了一下你的资料 df

df <- structure(list(Read = c("T", "T", "T", "T", "T", "F", "T", "T", 
"T", "T", "T", "T", "T", ""), Box = c("out", "out", "out", "out", 
"out", "", "out", "out", "in", "in", "out", "out", "out", "hello"
), ID = c("", "", "", "", "", "", "", "", "", "", "", "", "", 
""), Time = c("10/1/2019 9:00:01 AM", "10/1/2019 9:00:02 AM", 
"10/1/2019 9:00:03 AM", "10/1/2019 9:02:59 AM", "10/1/2019 9:03:00 AM", 
"10/1/2019 9:05:00 AM", "10/1/2019 9:06:00 AM", "10/1/2019 9:06:02 AM", 
"10/1/2019 9:07:00 AM", "10/1/2019 9:07:02 AM", "10/1/2019 9:07:04 AM", 
"10/1/2019 9:07:05 AM", "10/1/2019 9:07:06 AM", "10/1/2019 9:07:08 AM"
)), row.names = c(NA, -14L), class = "data.frame")