Hi Marna,
This is a condition that the function cannot handle. It would be
possible to reformat the result based on the time intervals, but the
stretch_df function doesn't try to interpret the values, just
stretches them out to a wide format.
Jim
On Wed, May 2, 2018 at 9:16 AM, Marna Wagley <marna.wag...@gmail.com> wrote:
> Hi Jim,
> The data set is correct. I took two readings from the "SITE A" within a
> short time interval, therefore I want to take the first value if there are
> repeated within a same group of "timeGroup".
> Therefore I wanted following
>
> FinalData1
>
> B1 B2
> id_X "A" "B"
> id_Y "A" "B"
>
> thanks,
>
>
>
> On Tue, May 1, 2018 at 4:05 PM, Jim Lemon <drjimle...@gmail.com> wrote:
>>
>> Hi Marna,
>> I think this is due to having three rows for id_X and only two for
>> id_Y. The function creates a data frame with enough columns to hold
>> the greatest number of values for each ID variable. Notice that the
>> SITE_n columns contain three values for id_X (A, A, B) and two for
>> id_Y (A, B, NA) as there was no third occasion of measurement for the
>> latter. Even though there are only two _values_ for SITE, there must
>> be enough space for three. In your desired output, SITE for the second
>> occasion of measurement is wrong (it should be "A"), and for the third
>> occasion it is unknown. Even if there was only one value for SITE in
>> the original data frame, it should be repeated for the correct number
>> of observations. I think you may be mixing up case ID with location of
>> observation.
>>
>> Jim
>>
>>
>> On Wed, May 2, 2018 at 8:48 AM, Marna Wagley <marna.wag...@gmail.com>
>> wrote:
>> > Hi Jim,
>> > Thank you very much for your suggestions. I used it but it gave me three
>> > sites. But actually I do have only two sites "Id_X" and "Id_y" . In fact
>> > "A" is repeated two times for "Id_X". If it is repeated, I would like to
>> > take the first one among many repeated values.
>> >
>> > dat<-structure(list(ID = structure(c(1L, 1L, 1L, 2L, 2L), .Label =
>> > c("id_X",
>> >
>> > "id_Y"), class = "factor"), EventDate = structure(c(4L, 5L, 2L,
>> >
>> > 3L, 1L), .Label = c("9/15/16", "9/15/17", "9/7/16", "9/8/16",
>> >
>> > "9/9/16"), class = "factor"), timeGroup = structure(c(1L, 1L,
>> >
>> > 2L, 1L, 2L), .Label = c("B1", "B2"), class = "factor"), SITE =
>> > structure(c(1L,
>> >
>> > 1L, 2L, 1L, 2L), .Label = c("A", "B"), class = "factor")), .Names =
>> > c("ID",
>> >
>> > "EventDate", "timeGroup", "SITE"), class = "data.frame", row.names =
>> > c(NA,
>> >
>> > -5L))
>> >
>> > library(prettyR)
>> >
>> > stretch_df(dat,idvar="ID",to.stretch=c("EventDate","SITE"))
>> >
>> >
>> > ID timeGroup EventDate_1 EventDate_2 EventDate_3 SITE_1 SITE_2 SITE_3
>> > 1 id_X B1 9/8/16 9/9/16 9/15/17 A A
>> > B
>> > 2 id_Y B1 9/7/16 9/15/16 <NA> A B
>> > <NA>
>> >>
>> >
>> > Basically I am looking for like following table
>> >
>> > ID timeGroup EventDate_1 EventDate_2 EventDate_3 SITE_1 SITE_2
>> > 1 id_X B1 9/8/16 9/9/16 9/15/17 A B
>> > 2 id_Y B1 9/7/16 9/15/16 <NA> A B
>> >
>> > Thanks
>> >
>> >
>> > On Tue, May 1, 2018 at 3:32 PM, Jim Lemon <drjimle...@gmail.com> wrote:
>> >>
>> >> Hi Marna,
>> >> Try this:
>> >>
>> >> library(prettyR)
>> >> stretch_df(dat,idvar="ID",to.stretch=c("EventDate","SITE"))
>> >>
>> >> Jim
>> >>
>> >>
>> >> On Wed, May 2, 2018 at 8:24 AM, Marna Wagley <marna.wag...@gmail.com>
>> >> wrote:
>> >> > Hi R user,
>> >> > I was trying to convert a long matrix to wide? I have an example and
>> >> > would
>> >> > like to get a table (FinalData1):
>> >> >
>> >> >
>> >> > FinalData1
>> >> > B1 B2
>> >> > id_X "A" "B"
>> >> > id_Y "A" "B"
>> >> >
>> >> > but I got the following table using the following code.
>> >> >
>> >> > FinalData1
>> >> >
>> >> > B1 B2
>> >> >
>> >> > id_X "A" "A"
>> >> >
>> >> > id_Y "A" "B"
>> >> >
>> >> >
>> >> > the code and the example data I used are given below. Is there any
>> >> > suggestions to fix the problem?
>> >> >
>> >> >
>> >> > dat<-structure(list(ID = structure(c(1L, 1L, 1L, 2L, 2L), .Label =
>> >> > c("id_X",
>> >> >
>> >> >
>> >> > "id_Y"), class = "factor"), EventDate = structure(c(4L, 5L, 2L,
>> >> >
>> >> > 3L, 1L), .Label = c("9/15/16", "9/15/17", "9/7/16", "9/8/16",
>> >> >
>> >> > "9/9/16"), class = "factor"), timeGroup = structure(c(1L, 1L,
>> >> >
>> >> > 2L, 1L, 2L), .Label = c("B1", "B2"), class = "factor"), SITE =
>> >> > structure(c(
>> >> > 1L,
>> >> >
>> >> > 1L, 2L, 1L, 2L), .Label = c("A", "B"), class = "factor")), .Names =
>> >> > c("ID",
>> >> >
>> >> > "EventDate", "timeGroup", "SITE"), class = "data.frame", row.names =
>> >> > c(NA,
>> >> >
>> >> > -5L))
>> >> >
>> >> >
>> >> > tmp <- split(dat, dat$ID)
>> >> >
>> >> > tmp1 <- do.call(rbind, lapply(tmp, function(dat){
>> >> >
>> >> > tb <- table(dat$timeGroup)
>> >> >
>> >> > idx <- which(tb>0)
>> >> >
>> >> > tb1 <- replace(tb, idx, as.character(dat$SITE))
>> >> >
>> >> > }))
>> >> >
>> >> >
>> >> > tmp1
>> >> >
>> >> > FinalData<-print(tmp1, quote=FALSE)
>> >> >
>> >> > [[alternative HTML version deleted]]
>> >> >
>> >> > ______________________________________________
>> >> > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> >> > https://stat.ethz.ch/mailman/listinfo/r-help
>> >> > PLEASE do read the posting guide
>> >> > http://www.R-project.org/posting-guide.html
>> >> > and provide commented, minimal, self-contained, reproducible code.
>> >
>> >
>
>
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