结合日、月、年、小时到R中的日期时间

Question

day   month     year    hour
01     05      2017     00
05     12      2017     01
10     07      2017     23

i don't have column minute and seconds now i am trying to get a single variable as date_time in the format of

2017-05-01 00:00:00
2017-12-05 01:00:00
2017-07-10 23:00:00

getting the error while using the below code

df$date <- as.Date(with(df, paste(year, month, day,hour,sep="-")),
           "%Y-%m-%d %H:%M:%S") 

thanks in advance

Answer 1

We can try paste the columns together with sprintf , then convert to datetime with as.POSIXct

as.POSIXct(do.call(sprintf, c(df1, fmt = c("%02d-%02d-%4d %02d"))), format = "%d-%m-%Y %H")
#[1] "2017-05-01 00:00:00 IST" "2017-12-05 01:00:00 IST" "2017-07-10 23:00:00 IST"

---

Or use lubridate

library(lubridate)
with(df1, ymd_h(paste(year, month, day, hour, sep= ' ')))

Answer 2

since year, month, day, hour are stored in different columns, why not just use ISOdate function? for instance:

> ISOdate("2017","12", "05", "01")
[1] "2017-12-05 01:00:00 GMT"

if you don't want to see time zone info, just wrap ISOdate with as.Date

> as.Date(ISOdate("2017","12", "05", "01"))
[1] "2017-12-05"

But this changes the class of object from POSIXct to Date and (correct me if I'm wrong) does not really remove time zone info.

This was discussed in another question How to convert in both directions between year,month,day and dates in R?

Answer 3

You can do this directly without making a string using lubridate .

library(lubridate)

df1 %>%
  mutate(date = make_datetime(year, month, day, hour))

This is probably more reliable than using string parsing.