如何将数字转换为正常日期？

Question

I have got a netcdf file to extract some variables from it. the provider told me that the time variable (A) is hours since 1900-01-01 00:00:0.0 . I need to convert this variable to a normal date and write to another txt file.

f1 <- open.ncdf("C:\\Users\\data.nc")
# [1] "file C:\\Users\\data.nc has 3 dimensions:"
# [1] "longitude   Size: 2"
# [1] "latitude   Size: 2"
# [1] "time   Size: 2920"
# [1] "file C:\\Users\\data.nc has 5 variables:"
# [1] "short stm[longitude,latitude,time]  Longname:soil textre Missval:-32767"

time:

A <- get.var.ncdf(nc=f1,varid="time")
head(A)
## [1] 990552 990558 990564 990570 990576 990582

another variable:

B1 <- get.var.ncdf(nc=f,varid="stm")
write.table(t(rbind(A,B1)),file="output.txt")

I need your help to convert the time variable to a normal date before writing to the text output file.

Answer 1

Date

If A are hours since "1900-01-01" , just divide them by 24, and pass this date into the origin parameter within the as.Date function, as in

A <- c(990552, 990558, 990564, 990570, 990576, 990582)
as.Date(A/24, origin = "1900-01-01")
## [1] "2013-01-01" "2013-01-01" "2013-01-01" "2013-01-01" "2013-01-02" "2013-01-02"

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Date + time

If you want the hours too, use as.POSIXct instead and multiply by 3600 (You should probably specify your time zone too within tz parameter)

as.POSIXct(A*3600, origin = "1900-01-01")
## [1] "2013-01-01 02:00:00 IST" "2013-01-01 08:00:00 IST" "2013-01-01 14:00:00 IST" "2013-01-01 20:00:00 IST" "2013-01-02 02:00:00 IST" "2013-01-02 08:00:00 IST"

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Time zone

In order to know your time zone, type

Sys.timezone()

For a full list of possible time zones, type

OlsonNames()

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Adding/Removing hours

Either way, you can remove/add hours by multiplying the number of hours by 3600 and adding/removing from the results, for example

as.POSIXct(A*3600, origin = "1900-01-01") - 2*3600