Hi,
I would like to construct a transition matrix from a data frame with
annual transitions of marked plants.
plant<-c(1:6)
class<-c("seed","seed", "seed",
"veg", "rep", "rep")
fate<-c("dead", "veg","veg","rep",
"rep", "veg")
trans<-data.frame(plant, class, fate)
plant class fate
1 1 seed dead
2 2 seed veg
3 3 seed veg
4 4 veg rep
5 5 rep rep
6 6 rep veg
I have been using sql queries to do this, but I would like to construct
the matrix in R since I plan to resample transitions using
trans[sample(nrow(trans), 6, replace=T), ]
I know I can get the original size vector using table()
data.matrix(table(trans$class))
[,1]
rep 2
seed 3
veg 1
but I don't know how to get counts of each class-fate combination where
fate does NOT equal dead
seed veg = 2
veg rep = 1
rep rep = 1
rep veg = 1
or how to divide the class-fate count by the original class count in the
size vector to get survival probabilities
seed veg = 2 / 3 seed = 0.67
veg rep = 1 / 1 veg = 1
rep rep = 1 / 2 rep = 0.5
rep veg = 1 / 2 rep = 0.5
or construct the square matrix with rows and columns in the same
developmental sequence like dev<- c("seed","veg",
"rep").
seed veg rep
seed 0 0 0
veg 0.67 0 0.5
rep 0 1 0.5
Any help or suggestions would be appreciated.
Thanks,
Chris Stubben
--
Los Alamos National Lab
BioScience Division
MS M888
Los Alamos, NM 87545
Hi again,
I almost figured this out, but still need some help on the last part.
I can use prop.table to get survival probabilities...
A <- t(prop.table( table(trans$class, trans$fate),1) )
rep seed veg
dead 0.0000000 0.3333333 0.0000000
rep 0.5000000 0.0000000 1.0000000
veg 0.5000000 0.6666667 0.0000000
so now I just need to format the matrix. I thought I could create a
matrix of zeroes using size class names,
dev<- c("seed","veg", "rep").
A0<-matrix(numeric(9), nrow=3, dimnames=list(dev,dev) )
seed veg rep
seed 0 0 0
veg 0 0 0
rep 0 0 0
but how do I assign values in A to the corresponding rows and columns in
A0? I hope there is an easy solution that I'm overlooking.
seed veg rep
seed 0 0 0
veg 0.67 0 0.5
rep 0 1 0.5
Thanks,
Chris
Chris Stubben wrote:> Hi,
>
> I would like to construct a transition matrix from a data frame with
> annual transitions of marked plants.
>
> plant<-c(1:6)
> class<-c("seed","seed", "seed",
"veg", "rep", "rep")
> fate<-c("dead",
"veg","veg","rep", "rep",
"veg")
>
> trans<-data.frame(plant, class, fate)
>
> plant class fate
> 1 1 seed dead
> 2 2 seed veg
> 3 3 seed veg
> 4 4 veg rep
> 5 5 rep rep
> 6 6 rep veg
>
> I have been using sql queries to do this, but I would like to construct
> the matrix in R since I plan to resample transitions using
> trans[sample(nrow(trans), 6, replace=T), ]
>
> I know I can get the original size vector using table()
>
> data.matrix(table(trans$class))
> [,1]
> rep 2
> seed 3
> veg 1
>
>
> but I don't know how to get counts of each class-fate combination where
> fate does NOT equal dead
>
> seed veg = 2
> veg rep = 1
> rep rep = 1
> rep veg = 1
>
>
> or how to divide the class-fate count by the original class count in the
> size vector to get survival probabilities
>
> seed veg = 2 / 3 seed = 0.67
> veg rep = 1 / 1 veg = 1
> rep rep = 1 / 2 rep = 0.5
> rep veg = 1 / 2 rep = 0.5
>
>
> or construct the square matrix with rows and columns in the same
> developmental sequence like dev<- c("seed","veg",
"rep").
>
> seed veg rep
> seed 0 0 0
> veg 0.67 0 0.5
> rep 0 1 0.5
>
> Any help or suggestions would be appreciated.
> Thanks,
>
>
> Chris Stubben
>
>
> --
> Los Alamos National Lab
> BioScience Division
> MS M888
> Los Alamos, NM 87545
>
>
>
If you order your factor levels in your vectors in the order you want in the output, then the prop.table(prop()) command will give you what you want. But you have to reorder the factor levels so that the levels commands give the following output:>levels(trans$class)[1] "seed" "veg" "repr">levels(trans$fate)[1] "seed" "veg" "repr" "dead" That means that you need to include "seed" as a factor level in fate (even if that level is unused). The prop.table(table()) command will then produce a 3 by 4 table. Remove the last row (that contains the proportion dead), and your set. Hans Gardfjell Dept of Ecology and Environmental science Ume?? University, Sweden Chris Stubben wrote: Hi again, I almost figured this out, but still need some help on the last part. I can use prop.table to get survival probabilities... A <- t(prop.table( table(trans$class, trans$fate),1) ) rep seed veg dead 0.0000000 0.3333333 0.0000000 rep 0.5000000 0.0000000 1.0000000 veg 0.5000000 0.6666667 0.0000000 so now I just need to format the matrix. I thought I could create a matrix of zeroes using size class names, dev<- c("seed","veg", "rep"). A0<-matrix(numeric(9), nrow=3, dimnames=list(dev,dev) ) seed veg rep seed 0 0 0 veg 0 0 0 rep 0 0 0 but how do I assign values in A to the corresponding rows and columns in A0? I hope there is an easy solution that I'm overlooking. seed veg rep seed 0 0 0 veg 0.67 0 0.5 rep 0 1 0.5 Thanks, Chris Chris Stubben wrote: >/ Hi, />/ />/ I would like to construct a transition matrix from a data frame with />/ annual transitions of marked plants. />/ />/ plant<-c(1:6) />/ class<-c("seed","seed", "seed", "veg", "rep", "rep") />/ fate<-c("dead", "veg","veg","rep", "rep", "veg") />/ />/ trans<-data.frame(plant, class, fate) />/ />/ plant class fate />/ 1 1 seed dead />/ 2 2 seed veg />/ 3 3 seed veg />/ 4 4 veg rep />/ 5 5 rep rep />/ 6 6 rep veg />/ />/ I have been using sql queries to do this, but I would like to construct />/ the matrix in R since I plan to resample transitions using />/ trans[sample(nrow(trans), 6, replace=T), ] />/ />/ I know I can get the original size vector using table() />/ />/ data.matrix(table(trans$class)) />/ [,1] />/ rep 2 />/ seed 3 />/ veg 1 />/ />/ />/ but I don't know how to get counts of each class-fate combination where />/ fate does NOT equal dead />/ />/ seed veg = 2 />/ veg rep = 1 />/ rep rep = 1 />/ rep veg = 1 />/ />/ />/ or how to divide the class-fate count by the original class count in the />/ size vector to get survival probabilities />/ />/ seed veg = 2 / 3 seed = 0.67 />/ veg rep = 1 / 1 veg = 1 />/ rep rep = 1 / 2 rep = 0.5 />/ rep veg = 1 / 2 rep = 0.5 />/ />/ />/ or construct the square matrix with rows and columns in the same />/ developmental sequence like dev<- c("seed","veg", "rep"). />/ />/ seed veg rep />/ seed 0 0 0 />/ veg 0.67 0 0.5 />/ rep 0 1 0.5 />/ />/ Any help or suggestions would be appreciated. />/ Thanks, />/ />/ />/ Chris Stubben />/ />/ />/ -- />/ Los Alamos National Lab />/ BioScience Division />/ MS M888 />/ Los Alamos, NM 87545 />/ />/ />