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$date, it looks to me as though the answer is as it should be. 2 -> ENL, 3 ->
ENM. I suspect that the solution is not difficult, but I can't quite make out
what you are trying to accomplish.
Jim
On Tue, Oct 13, 2015 at 11:44 AM, Kristi Glover
<kristi.glo...@hotmail.com<mailto:kristi.glo...@hotmail.com>> wrote:
Hi Jim,
Thank you very much for your suggestions. It seems very easy but it is
frustrating as it did not work me. with creating factors and rearranging the
columns, still z value (site) did change.
for example
Tag site DATE
a1 2C7 t1
a1 ENL t3
a1 ENM t2
ENL is supposed to be assigned for the period t3. ENM should be assigned in t2,
but using the code, the table gave wrong information as the z value (site) did
not move to the corresponding column. ENL is in t2, ENM is inn t3 coumns, which
is wrong.
> tmp1
t1 t2 t3
a1 "2C7" "ENL" "ENM"
I have included the code if any one help me to solve the problem. This is a
just example, I have a very big data set so that I could not check it manually
therefore, I just checked few rows but it did not work. Your help is highly
appreciated.
Here is the example
A<-structure(list(Tag = structure(c(1L, 1L, 1L), .Label = "a1", class =
"factor"),
site = structure(1:3, .Label = c("2C7", "ENL", "ENM"), class = "factor"),
DATE = structure(c(1L, 3L, 2L), .Label = c("t1", "t2", "t3"
), class = "factor"), date = structure(c(1L, 3L, 2L), .Label = c("t1",
"t2", "t3"), class = "factor")), .Names = c("Tag", "site",
"DATE", "date"), row.names = c(NA, -3L), class = "data.frame")
A$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<mailto: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<mailto: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.
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