[R] Function that create day of the year column.

[MacQueen, Don]

I would start with this example, which is available from base R, without
additional packages, to help understand the suggestions that follow.

> unclass(as.POSIXlt(Sys.Date()))
$sec
[1] 0

$min
[1] 0

$hour
[1] 0

$mday
[1] 5

$mon
[1] 10

$year
[1] 114

$wday
[1] 3

$yday
[1] 308

$isdst
[1] 0

attr(,"tzone")
[1] "UTC"

And then see the $yday element

For example:


>   as.POSIXlt( as.Date('2014-9-13') )$yday
[1] 255
>   as.POSIXlt( as.Date('2014-1-1') )$yday
[1] 0

Note that the year starts with day 0, which might not be what you expect.

If you have three columns try
  as.POSIXlt( as.Date( paste(year, month, day, sep='-') )$yday


Which can be illustrated by
>   as.POSIXlt( as.Date( paste(2014, 1, 31, sep='-') ) )$yday
[1] 30

If your ³date² column is already of class Date
   > class(Sys.Date())
   [1] "Date"

then
  as.POSIXlt( date )$yday
is sufficient. Otherwise you have to convert it to Date class.


-- 
Don MacQueen

Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062





On 11/4/14, 2:01 AM, "Frederic Ntirenganya" <ntfredo at gmail.com> wrote:

>Dear All,
>
>I would like to make a function that create Day of the year column in a
>dataset.
>State of the problem: write a function that create a day column either
>from
>a single date column (string/or factors) or from date in 3 columns (year,
>month, day).
>
>I made the following function for a single date. I would like to add a
>condition for date in 3 columns (year, month, day). My data is daily
>climate data.
>#write a function that create a day column either from a single date
>column
>(string/or factors)
>#or from date in 3 columns (year, month, day).
>
>DOY=function(data){
>
>#=================================================================
>#This function create day of teh year from a single date
>column(ex:2009-08-02) or/and
>#from the date in 3 columns (Year, month, Day).
>#================================================================
>  data$Rain=as.numeric(as.character(data$Rain))
>  dt=yday(data$Date) # single date of the data
>  datelp= dt>59 & !leap_year(data$Date)# tell us that the date occurs
>during a leap year
>  dt[datelp]=dt[datelp]+1 # add one for non leap_year
>  cbind(data, dt) # combining columns of data
>  conames(data)="DOY" # name of new column. ??I have a problem on how I
>can
>precise the column in gerenal.
>}
>
>ex: year  month day   Date         Rain Tmin Tmax
>      1971   1         1    1971-01-01   0     8.2  15
>       1971  1         2    1971-01-02   0     4.2  11
>        .        .          .       .               .      .      .
>        .        .          .       .               .      .      .
>        .        .          .       .               .      .      .
>
>Any ideal on how I can make this function is welcome. thanks!
>Frederic Ntirenganya
>Maseno University,
>African Maths Initiative,
>Kenya.
>Mobile:(+254)718492836
>Email: fredo at aims.ac.za
>https://sites.google.com/a/aims.ac.za/fredo/
>
>	[[alternative HTML version deleted]]
>
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