R 基于时间范围创建列
我一直在尝试根据时间范围创建一个新列:准时(12:00:00之前)、迟到(12:00:00到15:00:00之间)和非常晚(15:00:00之后)。我可以根据固定的时间而不是范围创建列 资料 期望输出R 基于时间范围创建列,r,time,R,Time,我一直在尝试根据时间范围创建一个新列:准时(12:00:00之前)、迟到(12:00:00到15:00:00之间)和非常晚(15:00:00之后)。我可以根据固定的时间而不是范围创建列 资料 期望输出 time worker Day Arrival 1 2020-07-21 15:25:00 Ryan Tue very late 2 2020-07-21 11:20:00 Tim Tue punctual 3 2020-07-21 11:30:0
time worker Day Arrival
1 2020-07-21 15:25:00 Ryan Tue very late
2 2020-07-21 11:20:00 Tim Tue punctual
3 2020-07-21 11:30:00 John Tue punctual
4 2020-07-21 14:00:00 Adam Tue late
错误代码
df<-df %>% mutate(hour = lubridate::hour(time), minutes = lubridate::minutes(time),Arrival = case_when(hour <- 12 | (hour == 12 & minutes <= 30) ~ 'punctual',
hour <- 15 | (hour == 15 & minutes <= 30) ~ 'late',
TRUE ~ 'very late'))
df%变异(小时=lubridate::小时(时间),分钟=lubridate::分钟(时间),到达=case_when(小时您可以使用case_when
并单独指定每个条件:
library(dplyr)
library(lubridate)
df %>%
mutate(hour = hour(time),
Arrival = case_when(hour < 12 ~ 'punctual',
hour < 15 ~ 'late',
TRUE ~ 'very late'))
为了将此扩展到最高达分钟级别的精度,我们可以在以下情况下使用case\u:
df %>%
mutate(hour = hour(time),
minutes = minute(time),
Arrival = case_when(
hour < 12 | (hour == 12 & minutes <= 30) ~ 'punctual',
hour < 15 | (hour == 15 & minutes <= 30) ~ 'late',
TRUE ~ 'very late'))
谢谢Ronak Shah!如果是12:30:00和15:30:00呢?根据这里的逻辑,他们将分别迟到和非常晚。至于标准,如果我也想包括会议记录呢?在这种情况下,你可以在
方法时使用
case\u。类似于hour
还有另一个分钟用lubridate::minutes
设置s
列,并更改条件。例如,如果您希望12:30之前的时间是“准时的”。Dohour<12 |(hour==12&minutes我在lapply(件,如.numeric)中遇到此错误:强制引入的NAs
。我将插入当前代码以解决上述问题
df$Arrival <- cut(as.integer(format(df$time, '%H')), c(0, 11, 14, 23),
c('punctual', 'late', 'very late'))
df
# time worker Day Arrival
#1 2020-07-21 15:25:00 Ryan Tue very late
#2 2020-07-21 11:20:00 Tim Tue punctual
#3 2020-07-21 11:30:00 John Tue punctual
#4 2020-07-21 14:00:00 Adam Tue late
df %>%
mutate(hour = hour(time),
minutes = minute(time),
Arrival = case_when(
hour < 12 | (hour == 12 & minutes <= 30) ~ 'punctual',
hour < 15 | (hour == 15 & minutes <= 30) ~ 'late',
TRUE ~ 'very late'))
df <- structure(list(time = structure(c(1595345100, 1595330400, 1595331000,
1595340000), class = c("POSIXct", "POSIXt"), tzone = "UTC"),
worker = c("Ryan", "Tim", "John", "Adam"), Day = c("Tue",
"Tue", "Tue", "Tue")), row.names = c("1", "2", "3", "4"), class = "data.frame")