将时间对象转换为R中的分类（早上，下午，晚上，晚上）变量？答案

【问题标题】：Convert time object to categorical (morning, afternoon, evening, night) variable in R?将时间对象转换为R中的分类（早上，下午，晚上，晚上）变量？
【发布时间】：2018-03-19 18:52:51
【问题描述】：

我在时间数据框中有这个向量，格式为小时：分钟，我想转换为一天中的分类时间：

    time <- c("15:03", "08:01", "11:59", "23:47", "14:20")
    df$time <- format(strptime(df$time, tz = "" , format = "%H: %M"), format = "%H: %M")
    df <- data.frame(time)

我想我会考虑早上 5 点到 11 点，下午 11 点到 16 点，晚上 16 点到 19 点，以及其他任何时间，直到晚上 5 点。原始数据的时间格式为带 strptime() 的小时：分钟。

我在论坛上发现了一些类似的问题，但我似乎无法调整代码来处理我的数据。

【问题讨论】：

标签： r time

【解决方案1】：

time <- as.POSIXct(strptime(c("15:03", "08:01", "11:59", "23:47", "14:20"),"%H:%M"),"UTC")

x=as.POSIXct(strptime(c("050000","105959","110000","155959","160000",
                        "185959"),"%H%M%S"),"UTC")
library(tidyverse)
case_when(
between(time,x[1],x[2]) ~"morning",
between(time,x[3],x[4]) ~"afternoon",
between(time,x[5],x[6]) ~"evening",
TRUE ~"night")
[1] "afternoon" "morning"   "afternoon" "night"     "afternoon"

使用基础 R：

time <- as.POSIXct(strptime(c("15:03", "08:01", "11:59", "23:47", "14:20"),"%H:%M"),"UTC")

x=as.POSIXct(strptime(c("000000","050000","110000","160000","190000","235959"),
                      "%H%M%S"),"UTC")
labs=c("night","morning","afternoon","evening","night")
labs[findInterval(time,x)]
[1] "afternoon" "morning"   "afternoon" "night"     "afternoon"

【讨论】：

【解决方案2】：

我认为这可以完成，我不确定如何接受重复标签，但也许其他人会。关键是使用chron::times() 创建一个按时间顺序排列的对象，而不是一个日期时间对象。

time <- c("15:03", "08:01", "11:59", "23:47", "14:20")
timep <- as.POSIXct(time, format = "%H:%M") %>% format("%H:%M:%S")
cut(chron::times(timep) , breaks = (1/24) * c(0,5,11,16,19,24), 
    labels = c("night", "morning", "afternoon", "evening", "night1"))

# [1] afternoon morning   afternoon night1    afternoon
# Levels: night morning afternoon evening night1

更新：

tod <- cut(chron::times(timep) , breaks = (1/24) * c(0,5,11,16,19,24))
c("night","morning","afternoon","evening","night")[as.numeric(tod)]
# "afternoon" "morning"   "afternoon" "night"     "afternoon"

【讨论】：

【解决方案3】：

使用一些regex 和ifelse

df$hour <- as.numeric(gsub("\\:.*$", "", df$time))
df$cat <- with(df,  ifelse(hour >= 5 & hour<=11, "morning",
                            ifelse(hour>11 & hour<=16, "evening", "night")))
df
   time hour     cat
1 15:03   15 evening
2 08:01    8 morning
3 11:59   11 morning
4 23:47   23   night
5 14:20   14 evening

【讨论】：

【解决方案4】：

这个和@Onyambu类似，只是用了plyr的mapvalues()和lubridate的hour()：

library(lubridate)
library(plyr)
df$timeofdat<-   mapvalues(hour(df$time),from=c(0:23),
  to=c(rep("night",times=5), rep("morning",times=6),rep("afternoon",times=5),rep("night", times=8)))

【讨论】：

【解决方案5】：

我能够使用ifelse 语句来创建类别。我将strptime 更改为as.POSIXct，并且只保留了制作组的时间。在 df 中有 3 列代表原始时间，只有小时，然后是组。如果类别需要成为一个因素，您可以使用as.factor 将其更改为一个因素。

time <- c("15:03", "08:01", "11:59", "23:47", "14:20")

time2 <- format(as.POSIXct(time, tz = "" , format = "%H: %M"), "%H")

df <- data.frame(time, time2 = as.numeric(time2))

df$time_category <- ifelse(df$time2 >= 05 & df$time2 <= 11, "Morning",
            ifelse(df$time2 > 11 & df$time2 <= 16, "Afternoon",
            ifelse(df$time2 > 16 & df$time2 <= 19, "Evening", "Night")))

【讨论】：