Hi Kristi,
This is a bit hard to follow, but I'll try. As you are replacing the
numeric values of the intermediate table with the character values of the
factor A$date, it looks to me as though the answer is as it should be. 2 ->
ENL, 3 -> ENM. I suspect that the solution is not difficult, but I can't
quite make out what you are trying to accomplish.
Jim
On Tue, Oct 13, 2015 at 11:44 AM, Kristi Glover <kristi.glo...@hotmail.com>
wrote:
> Hi Jim,
>
> Thank you very much for your suggestions. It seems very easy but it is
> frustrating as it did not work me. with creating factors and rearranging
> the columns, still z value (site) did change.
>
> for example
>
> Tag site DATE
> a1 2C7 t1
> a1 ENL t3
> a1 ENM t2
>
> ENL is supposed to be assigned for the period t3. ENM should be assigned
> in t2, but using the code, the table gave wrong information as the z value
> (site) did not move to the corresponding column. ENL is in t2, ENM is inn
> t3 coumns, which is wrong.
>
> > tmp1
> t1 t2 t3
> a1 "2C7" "ENL" "ENM"
>
> I have included the code if any one help me to solve the problem. This is
> a just example, I have a very big data set so that I could not check
> it manually therefore, I just checked few rows but it did not work. Your
> help is highly appreciated.
>
>
> Here is the example
>
> A<-structure(list(Tag = structure(c(1L, 1L, 1L), .Label = "a1", class =
> "factor"),
> site = structure(1:3, .Label = c("2C7", "ENL", "ENM"), class =
> "factor"),
> DATE = structure(c(1L, 3L, 2L), .Label = c("t1", "t2", "t3"
> ), class = "factor"), date = structure(c(1L, 3L, 2L), .Label = c("t1",
> "t2", "t3"), class = "factor")), .Names = c("Tag", "site",
> "DATE", "date"), row.names = c(NA, -3L), class = "data.frame")
>
> A$date<-factor(A$DATE, levels=c("t1","t2","t3"))
> tmp <- split(A, A$Tag)
> head(tmp)
> tail(tmp)
> tmp1 <- do.call(rbind, lapply(tmp, function(x){
> tb <- table(A$date)
> idx <- which(tb>0)
> tb1 <- replace(tb, idx, as.character(A$site))
> }))
>
> tmp1
>
>
>
>
>
> ------------------------------
> *From:* Jim Lemon <drjimle...@gmail.com>
> *Sent:* October 12, 2015 4:22 AM
> *To:* Kristi Glover
> *Cc:* R-help
> *Subject:* Re: [R] 3D matrix columns messed up
>
> Hi Kristi,
> The first part is relatively easy:
>
> # change first line to
> x$time<-factor(x$time,levels=c("t1","t2","t3","t4","t10","t21"))
>
> As you have specified "site" as the second element in "x", not the third,
> perhaps you just want:
>
> x<-structure(list(vs = structure(c(1L, 1L, 2L, 3L, 4L, 2L, 3L, 1L, 1L),
> .Label = c("vs1", "vs2", "vs3", "vs4"), class = "factor"),
> time = structure(c(1L, 3L, 5L, 1L, 5L, 1L, 6L, 2L, 4L),
> .Label = c("t1", "t10", "t2", "t21", "t3", "t4"), class = "factor")),
> site = structure(c(1L, 2L, 3L, 1L, 3L, 1L, 3L, 1L, 2L),
> .Label = c("A", "B", "D"), class = "factor"),
> .Names = c("vs", "time", "site"), class = "data.frame",
> row.names = c(NA, -9L))
>
> Jim
>
> On Mon, Oct 12, 2015 at 7:41 PM, Kristi Glover <kristi.glo...@hotmail.com>
> wrote:
>
>> Hi R Users,
>> I was trying to make a matrix with three variables (x,y, z), but y
>> variable (columns) names did not stay in its sequential order,
>> t1,t2,t3,---t21; rather the matrix columns automatically appeared as
>> a t1,t10, t2,t20 etc. Besides these, z value (sites) did not come in the
>> right place (meaning in right columns name). I am wondering how I can make
>> the matrix with the sequential order with right z value (site). I tried it
>> several ways but did not work. One of the examples I used is given here.
>> Would you mind to give me a mints?
>>
>> x<-structure(list(vs = structure(c(1L, 1L, 2L, 3L, 4L, 2L, 3L, 1L,
>> 1L), .Label = c("vs1", "vs2", "vs3", "vs4"), class = "factor"),
>> site = structure(c(1L, 2L, 3L, 1L, 3L, 1L, 3L, 1L, 2L), .Label =
>> c("A",
>> "B", "D"), class = "factor"), time = structure(c(1L, 3L,
>> 5L, 1L, 5L, 1L, 6L, 2L, 4L), .Label = c("t1", "t10", "t2",
>> "t21", "t3", "t4"), class = "factor")), .Names = c("vs",
>> "site", "time"), class = "data.frame", row.names = c(NA, -9L))
>>
>>
>> x$time<-factor(x$time)
>> tmp <- split(x, x$vs)
>> tmp1 <- do.call(rbind, lapply(tmp, function(x){
>> tb <- table(x$time)
>> idx <- which(tb>0)
>> tb1 <- replace(tb, idx, as.character(x$site))
>> }))
>>
>>
>> tmp1
>>
>>
>> ## I want the z (site) in respective columns.
>>
>> ______________________________________________
>> 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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