Convert a date vector into Julian day in R

问题

I have a column of dates in the format: 16Jun10 and I would like to extract the Julian day. I have various years.

I have tried the functions julian and mdy.date and it doesn\'t seem to work.

回答1:

Try the following to convert from class character(i.e. text) to class POSIXlt, and then extract Julian day (yday):

tmp <- as.POSIXlt("16Jun10", format = "%d%b%y")
tmp$yday
# [1] 166

For more details on function settings:

?POSIXlt
?DateTimeClasses

Another option is to use a Date class, and then use format to extract a julian day (notice that this class define julian days between 1:366, while POSIXlt is 0:365):

tmp <- as.Date("16Jun10", format = "%d%b%y")
format(tmp, "%j")
# [1] "167"

回答2:

Similarly:

require(lubridate)
x = as.Date('2010-06-10')
yday(x)

[1] 161

Also note, using lubridate:

> dmy('16Jun10')
[1] "2010-06-16 UTC"

回答3:

You can use R's insol package which has a JD(x, inverse=FALSE) function which converts POSIXct to Julian Day Number (JDN).

insol package also has JDymd(year,month,day,hour=12,minute=0,sec=0) for custom dates.

To display the whole Julian Date (JD) you possibly have to set options(digits=16).

回答4:

my.data = read.table(text = "
     OBS  MONTH1  DAY1  YEAR1
       1       3     1   2012
       2       3    31   2012
       3       4     1   2012
       4       4    30   2012
       5       5     1   2012
       6       5    31   2012
       7       6     1   2012
       8       6    30   2012
       9       7     1   2012
      10       7    31   2012    
", header = TRUE, stringsAsFactors = FALSE)

my.data$MY.DATE1 <- do.call(paste, list(my.data$MONTH1, my.data$DAY1, my.data$YEAR1))
my.data$MY.DATE1 <- as.Date(my.data$MY.DATE1, format=c("%m %d %Y"))

my.data$my.julian.date <- as.numeric(format(my.data$MY.DATE1, "%j"))
my.data

Returns, which technically is incorrect since Julian dates do not return to 1 on the first day of each January:

http://en.wikipedia.org/wiki/Julian_day

The dates below are Ordinal dates:

   OBS MONTH1 DAY1 YEAR1   MY.DATE1 my.julian.date
1    1      3    1  2012 2012-03-01             61
2    2      3   31  2012 2012-03-31             91
3    3      4    1  2012 2012-04-01             92
4    4      4   30  2012 2012-04-30            121
5    5      5    1  2012 2012-05-01            122
6    6      5   31  2012 2012-05-31            152
7    7      6    1  2012 2012-06-01            153
8    8      6   30  2012 2012-06-30            182
9    9      7    1  2012 2012-07-01            183
10  10      7   31  2012 2012-07-31            213

回答5:

Here are my R versions of code originally written in APL and converted to J. We call this pseudo-Julian because it is only intended for dates after October 15, 1582 which is when calendar reform, in some parts of the Western world, arbitrarily changed the date.

#* toJulian: convert 3-element c(Y,M,D) timestamp into pseudo-Julian day number.
toJulian<- function(TS3)
{   mm<- TS3[2]
    xx<- 0
    if( mm<=2) {xx<- 1}
    mm<- (12*xx)+mm
    yy<- TS3[1]-xx
    nc<- floor(0.01*yy)
    jd<- floor(365.25*yy)+floor(30.6001*(1+mm))+TS3[3]+1720995+(2-(nc-floor(0.25*nc)))
    return(jd)
#EG toJulian c(1959,5,24) -> 2436713
#EG toJulian c(1992,12,16) -> 2448973
}

Here's the inverse function:

#* toGregorian: convert pseudo-Julian day number to timestamp in form c(Y,M,D)
# (>15 Oct 1582).  Adapted from "Numerical Recipes in C" by Press,
# Teukolsky, et al.
toGregorian<- function(jdn)
{   igreg<- 2299161       # Gregorian calendar conversion day c(1582,10,15).
    ja<- floor(jdn)
    xx<- 0
    if(igreg<=ja){xx<- 1}
    jalpha<- floor((floor((xx*ja)-1867216)-0.25)/36524.25)
    ja<- ((1-xx)*ja) + ((xx*ja)+1+jalpha-floor(0.25*jalpha))
    jb<- ja+1524
    jc<- floor(6680+((jb-2439870)-122.1)/365.25)
    jd<- floor(365.25*jc)
    je<- floor((jb-jd)/30.6001)
    id<- floor((jb-jd)-floor(30.6001*je))
    mm<- floor(je-1)
    if(12<mm){mm<- mm-12}
    iyyy<- floor(jc-4715)
    if(mm>2){iyyy<- iyyy-1}
    if(0>iyyy){iyyy<- iyyy-1}
    gd<- c(iyyy, mm, id)
    return(gd)
#EG toGregorian 2436713 -> c(1959,5,24)
#EG toGregorian 2448973 -> c(1992,12,16)
}

来源：https://stackoverflow.com/questions/21414847/convert-a-date-vector-into-julian-day-in-r

标签

date

julian-date