Hi guys i am trying to calculate transition matrix for every sequences which is presented by each row in a matrix. For example, i have a matrix:
dat<-matrix(c('a','b','c','a','a','a','b','b','a','c','a','a','c','c','a'),nrow = 3)
> `
[,1] [,2] [,3] [,4] [,5]
[1,] "a" "a" "b" "c" "c"
[2,] "b" "a" "b" "a" "c"
[3,] "c" "a" "a" "a" "a" `
I can easily calculate transition matrix for a single row using
> `mylistMc<-markovchainFit(data=dat[1,])`
I get a correct result
> `
a b c
a 0.5 0.5 0
b 0.0 0.0 1
c 0.0 0.0 1`
Then i tried to use
> `mc<-markovchainListFit(data=dat[1,])`
But the results are weird, it gives 4 matrixes instead of 3 (i suppose it takes columns instead of rows maybe), i believe MarkovChainListFit is used for a multiple sequences, and i just don't getting how to use it.
Any ideas guys? Thank you!
Try this:
mc <- apply(t(dat),2,function(x) markovchainFit(x))
trans_mat <- list(mc[[1]][[1]],mc[[2]][[1]],mc[[3]][[1]])
Related
Im trying to convert a data set in a long format panel structure to an adjacency matrix or edge list to make network graphs. The data set contains articles each identified by an ID-number. Each article can appear several times under a number of categories. Hence I have a long format structure at the moment:
ID <- c(1,1,1,2,2,2,3,3)
Category <- c("A","B","C","B","E","H","C","E")
dat <- data.frame(ID,Category)
I want to convert this into an adjacency matrix or edge list. Where the edge list such look something like this
A B
A C
B C
B E
B H
E H
C E
Edit: I have tried dat <- merge(ID, Category, by="Category") but it returns the error message Error in fix.by(by.x, x) : 'by' must specify a uniquely valid column
Thanks in advance
Update: I ended up using the crossprod(table(dat)) from the comments, but the solution suggested by Navy Cheng below works just as well
This code will work
do.call(rbind,lapply(split(dat, dat$ID), function(x){
t(combn(as.vector(x$Category), 2))
}))
Update
As #Parfait 's suggestion, you can have by instead of split+lapply.
1) Use by to group nodes ("A", "B", "C" ...) by Category;
2) Use combn to create edge between nodes in each group, and t to transform the matrix for further rbind
> edge.list <- by(dat, dat$ID, function(x) t(combn(as.vector(x$Category), 2)))
dat$ID: 1
[,1] [,2]
[1,] "A" "B"
[2,] "A" "C"
[3,] "B" "C"
------------------------------------------------------------
dat$ID: 2
[,1] [,2]
[1,] "B" "E"
[2,] "B" "H"
[3,] "E" "H"
------------------------------------------------------------
dat$ID: 3
[,1] [,2]
[1,] "C" "E"
3) Then merge the list
> do.call(rbind, edge.list)
[,1] [,2]
[1,] "A" "B"
[2,] "A" "C"
[3,] "B" "C"
[4,] "B" "E"
[5,] "B" "H"
[6,] "E" "H"
[7,] "C" "E"
So if you are willing to convert your data.frame to a data.table this problem can be solved pretty efficiently and cleanly and if you have many rows will be much faster.
library(data.table)
dat<-data.table(dat)
Basically you can apply functions to columns of the data.table in the j cell and group in the k cell. So you want all the combinations of categories taken two at a time for each ID which looks like this:
dat[,combn(Categories,2),by=ID]
However stopping at this point will keep the ID column and by default create a column called V1 that basically concatenates the array returned by combn into a vector of the categories and not the two-column adjacency matrix that you need. But by chaining another call to this you can create the matrix easily as you would with any single vector. In one line of code this will look like:
dat[,combn(Category,2),by=ID][,matrix(V1,ncol=2,byrow = T)]
Remember that the vector column we wish to convert to a matrix is called V1 by default and also we want the 2-column matrix to be created by row instead of the default which is by column. Hope that helps and let me know if I need to add anything to my explanation. Good luck!
This question already has an answer here:
Is there anything wrong with using T & F instead of TRUE & FALSE?
(1 answer)
Closed 4 years ago.
Given a matrix with one row, one column, or one cell, I need to reorder the rows while keeping the matrix structure. I tried adding drop=F but it doesn't work! What did I do?
test = matrix(letters[1:5]) # is a matrix
test[5:1,,drop=F] # not a matrix
test2 = matrix(letters[1:5],nrow=1) # is a matrix
test2[1:1,,drop=F] # not a matrix
test3 = matrix(1) # is a matrix
test3[1:1,,drop=F] # not a matrix
I'd guess it was an overwritten F; F can be set as a variable, in which case it's no longer false. Always write out FALSE fully, it can't be set as a variable.
See Is there anything wrong with using T & F instead of TRUE & FALSE?
Also the R Inferno, section 8.1.32, is a good reference.
> F <- 1
> test = matrix(letters[1:5]) # is a matrix
> test[5:1,,drop=F] # not a matrix
[1] "e" "d" "c" "b" "a"
> test[5:1,,drop=FALSE] # but this is a matrix
[,1]
[1,] "e"
[2,] "d"
[3,] "c"
[4,] "b"
[5,] "a"
> rm(F)
> test[5:1,,drop=F] # now a matrix again
[,1]
[1,] "e"
[2,] "d"
[3,] "c"
[4,] "b"
[5,] "a"
The code in your question works fine in a fresh R session:
test = matrix(letters[1:5]) # is a matrix
result = test[5:1,,drop=F]
result
# [,1]
# [1,] "e"
# [2,] "d"
# [3,] "c"
# [4,] "b"
# [5,] "a"
class(result) # still a matrix
# [1] "matrix"
dim(result)
# [1] 5 1
Even on the 1x1 matrix:
test3 = matrix(1) # is a matrix
result3 = test3[1:1,,drop=F]
class(result3)
# [1] "matrix"
dim(result3)
# [1] 1 1
Maybe you've loaded other packages that are overriding the default behavior? What makes you think you don't end up with a matrix?
The following works:
test <- matrix(test[5:1,, drop = F], nrow = 5, ncol = 1)
When you use is.matrix to test it, the output is a matrix. At the same time, you specify the number of rows (nrow) and number of columns (ncol) to coerce it to the number of rows and columns you require.
This question already has answers here:
How to generate all possible combinations of vectors without caring for order?
(2 answers)
Closed 5 years ago.
What R command generates all possible ordered combinations of length k?
For example from this vector:
a,b,c,d
It want to generate all combinations of length 3 but only those ones where the order is conserved:
a,b,c
a,b,d
a,c,d
b,c,d
Or If I have this vector
a,b,7,d,e
I want to do the same for length 2:
a,b
a,7
a,d
a,e
b,7
b,d
b,e
7,d
7,e
d,e
combn doesn't work here because it gives you all possible combinations including reversed ones such as
c,b
In simple cases I could try to do it with expand.grid but both methods would need further processing.
Maybe there is a base function (or package) able to do what I want or even accepting more complex conditions.
PD: When I say "ordered" I'm speaking about the order of appearance in the starting vector. I don't mean the typographic order, though in my example they are the same.
You can use combn in base R:
vec <- c("a", "b", "c", "d")
len <- 2
combn(length(vec), len, function(x) vec[x])
# [,1] [,2] [,3] [,4] [,5] [,6]
#[1,] "a" "a" "a" "b" "b" "c"
#[2,] "b" "c" "d" "c" "d" "d"
Of length 3:
combn(length(vec), 3, function(x) vec[x])
# [,1] [,2] [,3] [,4]
#[1,] "a" "a" "a" "b"
#[2,] "b" "b" "c" "c"
#[3,] "c" "d" "d" "d"
OR as #Sotos pointed out in the comments:
combn(vec, len)
UPDATE: FIXED
This is fixed in the upcoming release of R 3.1.0. From the CHANGELOG:
combn(x, simplify = TRUE) now gives a factor result for factor input
x (previously user error).
Related to PR#15442
I just noticed a curious thing. Why does combn appear to unclass factor variables to their underlying numeric values for all except the first combination?
x <- as.factor( letters[1:3] )
combn( x , 2 )
# [,1] [,2] [,3]
#[1,] "a" "1" "2"
#[2,] "b" "3" "3"
This doesn't occur when x is a character:
x <- as.character( letters[1:3] )
combn( x , 2 )
# [,1] [,2] [,3]
#[1,] "a" "a" "b"
#[2,] "b" "c" "c"
Reproducible on R64 on OS X 10.7.5 and Windows 7.
I think it is due to the conversion to matrix done by the simplify parameter. If you don't use it you get:
combn( x , 2 , simplify=FALSE)
[[1]]
[1] a b
Levels: a b c
[[2]]
[1] a c
Levels: a b c
[[3]]
[1] b c
Levels: a b c
The fact that the first column is OK is due to the way combn works: the first column is specified separately and the other columns are then changed from the existing matrix using [<-. Consider:
m <- matrix(x,3,3)
m[,2] <- sample(x)
m
[,1] [,2] [,3]
[1,] "a" "1" "a"
[2,] "b" "3" "b"
[3,] "c" "2" "c"
I think the offending function is therefore [<-.
As Konrad said, the treatment of factors is often odd, or at least inconsistent. In this case I think the behaviour is weird enough to constitute a bug. Try submitting it, and see what the response is.
Since the result is a matrix, and there is no factor matrix type, I think that the correct behaviour would be to convert factor inputs to character somewhere near the start of the function.
I had the same problem. Coercing back to a character vector inside the combn command seems to work:
> combn(as.character(x),2)
[,1] [,2] [,3]
[1,] "a" "a" "b"
[2,] "b" "c" "c"
I am trying to find a way to get a list in R of all the possible unique permutations of A,A,A,A,B,B,B,B,B.
Combinations was what was originally thought to be the method for obtaining a solution, hence the combinations answers.
I think this is what you're after. #bill was on the ball with the recommendation of combining unique and combn. We'll also use the apply family to generate ALL of the combinations. Since unique removes duplicate rows, we need to transpose the results from combn before uniqueing them. We then transpose them back before returning to the screen so that each column represents a unique answer.
#Daters
x <- c(rep("A", 4), rep("B",5))
#Generates a list with ALL of the combinations
zz <- sapply(seq_along(x), function(y) combn(x,y))
#Filter out all the duplicates
sapply(zz, function(z) t(unique(t(z))))
Which returns:
[[1]]
[,1] [,2]
[1,] "A" "B"
[[2]]
[,1] [,2] [,3]
[1,] "A" "A" "B"
[2,] "A" "B" "B"
[[3]]
[,1] [,2] [,3] [,4]
[1,] "A" "A" "A" "B"
[2,] "A" "A" "B" "B"
[3,] "A" "B" "B" "B"
...
EDIT Since the question is about permuations and not combinations, the answer above is not that useful. This post outlines a function to generate the unique permutations given a set of parameters. I have no idea if it could be improved upon, but here's one approach using that function:
fn_perm_list <-
function (n, r, v = 1:n)
{
if (r == 1)
matrix(v, n, 1)
else if (n == 1)
matrix(v, 1, r)
else {
X <- NULL
for (i in 1:n) X <- rbind(X, cbind(v[i], fn_perm_list(n -
1, r - 1, v[-i])))
X
}
}
zz <- fn_perm_list(9, 9)
#Turn into character matrix. This currently does not generalize well, but gets the job done
zz <- ifelse(zz <= 4, "A", "B")
#Returns 126 rows as indicated in comments
unique(zz)
There's no need to generate permutations and then pick out the unique ones.
Here's a much simpler way (and much, much faster as well): To generate all permutations of 4 A's and 5 B's, we just need to enumerate all possible ways of placing 4 A's among 9 possible locations. This is simply a combinations problem. Here's how we can do this:
x <- rep('B',9) # vector of 9 B's
a_pos <- combn(9,4) # all possible ways to place 4 A's among 9 positions
perms <- apply(a_pos, 2, function(p) replace(x,p,'A')) # all desired permutations
Each column of the 9x126 matrix perms is a unique permutation 4 A's and 5 B's:
> dim(perms)
[1] 9 126
> perms[,1:4] ## look at first few columns
[,1] [,2] [,3] [,4]
[1,] "A" "A" "A" "A"
[2,] "A" "A" "A" "A"
[3,] "A" "A" "A" "A"
[4,] "A" "B" "B" "B"
[5,] "B" "A" "B" "B"
[6,] "B" "B" "A" "B"
[7,] "B" "B" "B" "A"
[8,] "B" "B" "B" "B"
[9,] "B" "B" "B" "B"