有没有办法在 R 中将单列从军事时间转换为标准时间?
Is there a way to convert a single column from military time to standard time within R?
我有数据集
ID Date
A 2019-12-13 20:19:24
A 2019-12-13 20:19:24
A 2019-12-13 20:19:24
我想要这个结果,其中小时数转换为标准时间
ID Date
A 2019-12-13 8:19:24
A 2019-12-13 8:19:24
A 2019-12-13 8:19:24
I have tried this:
library('lubridate)
format(strptime(data$Date, format ='%H:%M:%S'), '%I:%M:%S %p')
I keep getting error: 3: In doTryCatch(return(expr), name, parentenv, handler) :
invalid graphics
在基础 R 中,您可以使用 %r
transform(df,Date = format(strptime(Date,"%F %T"),"%F %r"))
ID Date
1 A 2019-12-13 08:19:24 PM
2 A 2019-12-13 08:19:24 PM
3 A 2019-12-13 08:19:24 PM
我有数据集
ID Date
A 2019-12-13 20:19:24
A 2019-12-13 20:19:24
A 2019-12-13 20:19:24
我想要这个结果,其中小时数转换为标准时间
ID Date
A 2019-12-13 8:19:24
A 2019-12-13 8:19:24
A 2019-12-13 8:19:24
I have tried this:
library('lubridate)
format(strptime(data$Date, format ='%H:%M:%S'), '%I:%M:%S %p')
I keep getting error: 3: In doTryCatch(return(expr), name, parentenv, handler) :
invalid graphics
在基础 R 中,您可以使用 %r
transform(df,Date = format(strptime(Date,"%F %T"),"%F %r"))
ID Date
1 A 2019-12-13 08:19:24 PM
2 A 2019-12-13 08:19:24 PM
3 A 2019-12-13 08:19:24 PM