Question

如果包含类似

的数据的大型数据集

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1）我想将时间小时数据转移到时间

  df<-    Energy  Power    StartTime                 Timehour
          7.50    10     2017-10-18 09:47:10.073     0.75

2）然后我想将Time Hour值添加到starttime

 df<-    Energy  Power    StartTime                  Timehour
          7.50    10     2017-10-18 09:47:10.073     00:45:00

1）我尝试用Convert time from numeric to time format in R

更改时间表示法

 df<-    Energy  Power    StartTime                  Timehour     NewTime
          7.50    10     2017-10-18 09:45:10.000     00:45:00      2017-10-18 10:30:10.000

但我收到以下消息

       df$Timehour <- as.Date(df$Timehour)
       df$Timehour <- format(as.POSIXct(df$Timehour)*86400, origin = "1970-01-01",tz = "UTC", "%H:%M:%OS")

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有人可以帮助我解决第一和第二个问题吗？

Answer 1

基本解决方案：您需要在几秒钟内完成时间，将其简单地添加到Posixct

df$Timehour <- df$Timehour*3600
df$StartTime <- as.POSIXct(as.character(df$StartTime),format = "%Y-%m-%d %H:%M:%OS")
df$newtime <- df$StartTime + df$Timehour

  Energy Power           StartTime Timehour             newtime
1    7.5    10 2017-10-18 09:47:10     2700 2017-10-18 10:32:10

您可以通过这种方式将时间转换为所需的格式：

df$Timehour <- strftime(as.POSIXct(df$Timehour,origin='1900-01-01 00:00'),format="%H:%M:%S",tz = "GMT")

  Energy Power           StartTime Timehour             newtime
1    7.5    10 2017-10-18 09:47:10 00:45:00 2017-10-18 10:32:10

数据：

df<- read.table(text = "Energy,  Power,    StartTime,                 Timehour
7.50,    10 ,    2017-10-18 09:47:10,    0.75",header = TRUE ,sep = ",")

Answer 2

如果您使用lubridate将时间小时转换为分钟，则可以将其添加到starttime列

library(lubridate)
library(tidyverse)
library(readr)

df <- read_table("Energy  Power    StartTime                 Timehour
7.50    10     2017-10-18 09:47:10.073     0.75")

df %>% mutate(
  minutes = minutes(round(Timehour * 60)),
  newTime = StartTime + minutes)

# A tibble: 1 x 6
  Energy Power StartTime           Timehour minutes      newTime            
   <dbl> <int> <dttm>                 <dbl> <S4: Period> <dttm>             
1   7.50    10 2017-10-18 09:47:10    0.750 45M 0S       2017-10-18 10:32:10

如果需要，首先使用

将StartTime列转换为datatime类

df <- df %>% mutate(StartTime = ymd_hms(StartTime))

R如何将数字小时值转换为％H：％M：％S并为日期添加时间

2 个答案: