how can I convert a long to wide matrix?

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 <drjimlemon at 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.wagley at 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 <drjimlemon at 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.wagley at 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)

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