[R] 3D matrix columns messed up _ looking for your help

[Kristi Glover]

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 <drjimlemon at 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.glover at hotmail.com<mailto:kristi.glover at 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 <drjimlemon at gmail.com<mailto:drjimlemon at 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.glover at hotmail.com<mailto:kristi.glover at 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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