You cannot obtain a predictable result by sending invalid time
representation data to strptime... you have to work with valid time
representations.
See sample approach below:
############################
weekEnds <- function( DF ) {
d1_1 <- as.Date( sprintf( "%04d 1 1"
, DF[[ "Y" ]]
)
, format = "%Y %U %u"
)
d52_7 <- as.Date( sprintf( "%04d 52 7"
, DF[[ "Y" ]]
)
, format = "%Y %U %u"
)
week <- as.difftime( 7, units = "days" )
day <- as.difftime( 1, units = "days" )
d <- as.Date( sprintf( "%04d %d 1"
, DF[[ "Y" ]]
, DF[[ "wn" ]]
)
, format = "%Y %U %u"
)
before <- 0 == DF[[ "wn" ]]
after <- 53 == DF[[ "wn" ]]
d[ before ] <- d1_1[ before ] - week
d[ after ] <- d52_7[ after ] + day
DF[[ "weekBegin" ]] <- d
DF[[ "weekEnd" ]] <- d + week
DF
}
tst <- expand.grid( Y = 2000:2028
, wn = c( 0, 1, 53 )
)
result <- weekEnds( tst )
set.seed( 42 )
result[ sample( nrow( result ), 5 ), ]
#> Y wn weekBegin weekEnd
#> 80 2021 53 2021-12-27 2022-01-03
#> 81 2022 53 2022-12-26 2023-01-02
#> 25 2024 0 2024-01-01 2024-01-08
#> 70 2011 53 2011-12-26 2012-01-02
#> 54 2024 1 2024-01-08 2024-01-15
sum( is.na( result$weekBegin ) )
#> [1] 0
############################
On Tue, 16 Oct 2018, peter salzman wrote:
hi,
thanks for replying,
it has taken some time to understand
i have year+week and i need to find the 1st day and the last day of that week
i can decide when week starts
for example these 3 examples:
df <- data.frame(id = 1:3, year = c(2018, 2018, 2018), week=c(0,1,52))
## now run for all 3 rows:
for (kk in 1:3) {
print(df[kk,])
print('## version 1')
print(as.Date(paste(df$year[kk],df$week[kk],'Sun',sep=' '), format = "%Y %U
%a") )
print(as.Date(paste(df$year[kk],df$week[kk],'Mon',sep=' '), format = "%Y %U
%a") )
print(as.Date(paste(df$year[kk],df$week[kk],'Tue',sep=' '), format = "%Y %U
%a") )
print(as.Date(paste(df$year[kk],df$week[kk],'Wed',sep=' '), format = "%Y %U
%a") )
print(as.Date(paste(df$year[kk],df$week[kk],'Thu',sep=' '), format = "%Y %U
%a") )
print(as.Date(paste(df$year[kk],df$week[kk],'Fri',sep=' '), format = "%Y %U
%a") )
print(as.Date(paste(df$year[kk],df$week[kk],'Sat',sep=' '), format = "%Y %U
%a") )
print('## version 2')
print(as.Date(paste(df$year[kk],df$week[kk],'7',sep=' '), format = "%Y %U
%u") )
print(as.Date(paste(df$year[kk],df$week[kk],'1',sep=' '), format = "%Y %U
%u") )
print(as.Date(paste(df$year[kk],df$week[kk],'2',sep=' '), format = "%Y %U
%u") )
print(as.Date(paste(df$year[kk],df$week[kk],'3',sep=' '), format = "%Y %U
%u") )
print(as.Date(paste(df$year[kk],df$week[kk],'4',sep=' '), format = "%Y %U
%u") )
print(as.Date(paste(df$year[kk],df$week[kk],'5',sep=' '), format = "%Y %U
%u") )
print(as.Date(paste(df$year[kk],df$week[kk],'6',sep=' '), format = "%Y %U
%u") )
}
for week 0 we get NA for Sunday because it was Dec 31, 2017
similarly for week 52 we get NA for Tue,Wed, ... because these are in January
of 2019
my hope was to write
as.Date(paste(year,week,1), format "%Y %month %weekday")
as.Date(paste(year,week,7), format "%Y %month %weekday")
and get the first and last day of the given week even at the beginning and end
of year
for example
as.Date("2018 0 Sun","%Y %U %a") = '2017-12-31'
i hope this makes sense.
thanks for replying
peter
On Tue, Oct 16, 2018 at 2:11 PM Jeff Newmiller <jdnew...@dcn.davis.ca.us> wrote:
Er, my mistake, you are using %U not %W... but now I am really confused,
because the first Sunday is trivial with %U/%u.
Can you clarify what your actual upstream input is? Is it an invalid date
string as you say below, or is it year number?
On October 16, 2018 10:22:10 AM PDT, Jeff Newmiller
<jdnew...@dcn.davis.ca.us> wrote:
>If the date in your character representation does not exist then there
>is no requirement for a POSIX function to give any reliable answer...
>including NA. Using 00 as the week number won't always work.
>
>The first week/weekday combination that is guaranteed to exist by POSIX
>is 1/1 (first Monday). If the corresponding mon/mday is 1/1 then no
>days exist in week zero for that year and the first Sunday is 6 days
>more than the mday of the first Monday, else the mday of the first
>Sunday is one day less than the mday of the first Monday.
>
>You should if at all possible repair the computations that are creating
>the invalid string dates you mention.
>
>On October 16, 2018 8:11:12 AM PDT, peter salzman
><peter.salzmanu...@gmail.com> wrote:
>>it is simpler than i thought
>>
>>first day of given week is the last day minus 6days
>>
>>in other words:
>>d1 = as.Date('2018 00 Sat',format="%Y %U %a") - 6
>>d2 = as.Date('2018 00 Sun',format="%Y %U %a")
>>are the same as long both are not NA
>>
>>therefore to get the one that is not NA one can do
>>
>>max( c(d1,d2), na.rm=TRUE )
>>
>>maybe there is some other trick
>>
>>best,
>>peter
>>
>>
>>
>>
>>
>>
>>On Tue, Oct 16, 2018 at 10:22 AM peter salzman
>><peter.salzmanu...@gmail.com>
>>wrote:
>>
>>> hi,
>>>
>>> to turn year and week into the date one can do the following:
>>>
>>> as.Date('2018 05 Sun', "%Y %W %a")
>>>
>>> however, when we want the Sunday (1st day of week) of the 1st week
>of
>>2018
>>> we get NA because 1/1/2018 was on Monday
>>>
>>> as.Date('2018 00 Mon',format="%Y %U %a")
>>> ## 2018-01-01
>>> as.Date('2018 00 Sun',format="%Y %U %a")
>>> ## NA
>>>
>>> btw the same goes for last week
>>> as.Date('2017 53 Sun',format="%Y %U %a")
>>> ## 2017-12-31
>>> as.Date('2017 53 Mon',format="%Y %U %a")
>>> ## NA
>>>
>>> So my question is :
>>> how do i get
>>> from "2018 00 Sun" to 2018-12-31
>>> and
>>> from "2017 53 Mon" to 2018-01-01
>>>
>>> i realize i can loop over days of week and do some if/then
>>statements,
>>> but is there a built in function?
>>>
>>> thank you
>>> peter
>>>
>>>
>>>
>>>
>>>
>>> --
>>> Peter Salzman, PhD
>>> Department of Biostatistics and Computational Biology
>>> University of Rochester
>>>
--
Sent from my phone. Please excuse my brevity.
--
Peter Salzman, PhD
BMS
Greater Boston Area, MA
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