On Wed, Feb 3, 2016 at 3:43 PM, Jeff Newmiller <jdnew...@dcn.davis.ca.us> wrote:
> 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.
No need to reinvent the wheel. Just do
library(lubridate)
mdy(c("09/01/15", "09/01/2015"))
Best,
ista
The best route to resolution is to
> 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_...@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.ste...@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.ste...@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@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
>>>>>
>>>>> [[alternative HTML version deleted]]
>>>>>
>>>>> ______________________________________________
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>>>>> and provide commented, minimal, self-contained, reproducible code.
>>>
>>>
>>>>
>>>> ______________________________________________
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>>>
>>>
>>>
>>
>> ______________________________________________
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>> and provide commented, minimal, self-contained, reproducible code.
>>
>
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