Hi Kristi,
I'm only guessing here, but if I change the order of the levels of "site",
it comes out the way you want. I don't think that this is a solution, but
let me know if it helps.
A<-structure(list(Tag = structure(c(1L, 1L, 1L), .Label = "a1", class =
"factor"),
site = structure(1:3, .Label = c("2C7", "ENM", "ENL"), 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$site))
}))
tmp1
Jim
On Wed, Oct 14, 2015 at 3:17 AM, Kristi Glover <kristi.glo...@hotmail.com>
wrote:
> Hi Jim,
>
> Thank you very much for the message. Sorry for the email that was not
> clear. Yes, you are right, in A$date, should be t2> ENM, t3> ENL. but using
> the code it gave t2>ENL, t3>ENM. I am struggling to fix it. If the data set
> was small, I could do it manually.
>
>
> would you mind to try this example?
>
>
> 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$site))
> }))
>
> tmp1
> =========
>
>
>
> ------------------------------
> *From:* Jim Lemon <drjimle...@gmail.com>
> *Sent:* October 13, 2015 4:24 AM
> *To:* Kristi Glover
> *Cc:* R-help
> *Subject:* Re: [R] 3D matrix columns messed up _ looking for your help
>
> Hi Kristi,
> This is a bit hard to follow, but I'll try. As you are replacing the
> numeric values of the intermediate table with the character values of the
> factor A$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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