determine the year of a date

Not exactly.

If you have been "Excel"ed and have both formats in one column then you
really have a data cleanup task to do.

fix the source of the data (e.g. go back into Excel and reformat the column
with consistent formatting).

If you need to do this on an ongoing basis and cannot fix the data source,
then you might need to choose the correct format on the fly... something
like:

yYdates <- function( ds ) {
  fmt <- ifelse( grepl( "^\\d{1,2}/\\d{1,2}/\\d{2}$", ds )
               , "%m/%d/%y"
               , "%m/%d/%Y"
               )
  as.Date( ds, format=fmt )
}

yYdates( c( "02/18/2015", "03/21/16" ) )
[1] "2015-02-18" "2016-03-21"

but this would have to be tailored to the formatting problems in your data
and is not really a general solution.

On Wed, 3 Feb 2016, Boris Steipe wrote:

As the documentation says, that's exactly what the expansion will do by
default.

B.


On Feb 3, 2016, at 2:31 PM, carol white <wht_crl at yahoo.com> wrote:

yes, some of them are like 09/01/15 and some others are 09/01/2015 and
all of them range between 2015 and 2016. The goal was to convert 09/01/15 to
09/01/2015. I don't know if the fact that they range between 2015 and 2016
make the task easier.

Best wishes
Carol


On Wednesday, February 3, 2016 7:34 PM, Boris Steipe
<boris.steipe at utoronto.ca> wrote:


Sorry - messed up example: corrected here...

d <- "7/27/77"
strptime(d, format="%m/%d/%y")  #  "1977-07-27 EDT"
x <- strptime(d, format="%m/%d/%y")

strftime(x, format="%m/%d/%Y")  # "07/27/1977"


On Feb 3, 2016, at 1:29 PM, Boris Steipe <boris.steipe at utoronto.ca>
wrote:

It seems your autocorrect is playing tricks on you but if I understand
you correctly you have a two digit year and want to convert that to a four
digit year? That's not uniquely possible of course; by convention, as the
documentation to strptime() says:

 On input, values 00 to 68 are prefixed by 20 and 69 to 99 by 19 ? that

 is the behaviour specified by the 2004 and 2008 POSIX standards...

If this is correct for you, you need to convert the string to a time
object and the time object back to string. The format specifier %Y prints
four-digit years:


d <- "7/27/59"
strptime(d, format="%m/%d/%y")  # "2059-07-27 EDT"
x <- strptime(d, format="%m/%d/%y")

strftime(x, format="%m/%d/%Y")  # "07/27/1977"



B.







On Feb 3, 2016, at 11:03 AM, carol white via R-help
<r-help at r-project.org> wrote:

Hi,might be trivial but how to determine the year of a date which is in
the %m/%d/%y format and those whose year is century should be modified to
ISO so that all date will have with year in ISO?
Regards,
Carol

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