I have a time series of n columns, where n >=2. For each row I am trying to generate a matrix by applying a function to each permutation of the n elements. for example:
perm <- function(x, y) x-y
row = c(1, 2, 3)
sapply(row,function(x) sapply(row, function(y) perm(x,y)))
> sapply(row,function(x) sapply(row, function(y) perm(x,y)))
[,1] [,2] [,3]
[1,] 0 1 2
[2,] -1 0 1
[3,] -2 -1 0
This approach works. But I know that it is not right. I just have this feeling that there is a better way to do this using the matrix math functions and I am not quite figuring it out. The apply approach is ~fine, but I know I can do better. In this case, reproducing the output for a simple subtraction like the one illustrated would be a fine answer but ideally it would come with some pointers to where to look in the docs for other types of operations. For what its worth, if anyone cares, the actual data is a list of interest rates of various tenors and the final output is going to be a matrix of forward-rates, so only the upper triangular matters, the diagonal and lower triangular won't have any information.
There is no way to eliminate the loops entirely in this case (that I know of). In all cases you will have to calculate the function for each element pair. R does have a function specifically built for this however outer
outer(row, row, perm)
[,1] [,2] [,3]
[1,] 0 -1 -2
[2,] 1 0 -1
[3,] 2 1 0
It is not necessarily faster to use outer compared to a normal loop, but it is a builtin method designed for the purpose.
Edit: Vectorized functions
There is a method one could use however, if the function is vectorized (like perm). Generating all permutations and then adding dimensions afterwards
n <- length(row)
perm1 <- row[rep(1:n, n)]
perm2 <- row[rep(1:n, each = n)]
res <- perm(perm1, perm2)
dim(res) <- c(n, n) #change to a 2 dimensional array (matrix)
res
[,1] [,2] [,3]
[1,] 0 -1 -2
[2,] 1 0 -1
[3,] 2 1 0
Which will be much faster (but more memory intensive) if the function is optimized for vectorization
Related
I have a big list (10,000 items) of string vectors (50 elements each).
Here is a small sample from a vector:
c("0101000000000000000000000000000000000000000000001000000000000000000000000000000100000000000100000000000000100000000000000000110000000000000000000000",
"0000000000000000000000000000000010000000000000000000000000110000000000000000000000000001000000000000001000000010000000000000000000000000000000000001",
"0000000000000000000100010000000010000000000000010000000000100000000000000100000001000001100000000000001000001000000100000000000000000000000000000100")
In every vector, I would like to estimate all pairwise differences (ie number of different characters in each pair) and then estimate the mean.
Given the big size of the list, what would be a cost effective way to do so?
thank you!!
EDIT- UPDATE
I am adding an example with shorter vectors and expected output (same as my comment below):
a=c("0010100101",
"1001011101",
"1111111010")
expected result (mean of pairwise differences):
6.666667
Many thanks to #Roland for providing the solution, based on Levenshtein distance!
Your question is not terribly clear but you appear to want a Levenshtein distance:
x = c("0010100101",
"1001011101",
"1111111010")
#switch off deletions and insertions:
d <- adist(x, costs = list(ins=Inf, del=Inf, sub=1))
# [,1] [,2] [,3]
#[1,] 0 6 8
#[2,] 6 0 6
#[3,] 8 6 0
mean(d[upper.tri(d)])
#[1] 6.666667
Well, I realised the small data sample was not that short, so I repeat the example with shorter strings:
a=c("0010100101",
"1001011101",
"1111111010")
I came up with this solution:
library(stringdist)
# matrix with no. of pairwise different characters
xx=(1-outer(a,a,"stringsim",method="hamming"))*nchar(a[1])
> xx
[,1] [,2] [,3]
[1,] 0 6 8
[2,] 6 0 6
[3,] 8 6 0
# mean no. of pairwise differences
xx1=mean(xx[lower.tri(xx,diag = FALSE)])
> xx1
[1] 6.666667
The result is fine, but I am not sure if this is an ideal solution, though..
(and if the "hamming" method is the correct one for any case).
Will try not to complicate things too much with my explanations, but I'm confused how to best go about filling a triangulated correlation matrix with no repeat values with existing correlation values derived from another package. This involves extracting specific values from a list of text files. This is what I have done so far:
# read in list of file names (they are named '1_1', '1_2' .. so on until '47_48' with no repeat values generated)
filenames <- read_table('/home/filenames.txt', col_names = 'file_id')
# create symmetrical matrix
M <- diag(48)
ct <- 1
for (sub in (filenames$file_id)) {
subj <- read.table(paste0(dat_dir, '/ht_', sub, '.HEreg'), sep="", fill=TRUE)
ht <- as.character(subj$V2[grep("rG",sub$V1)]) # wanting to extract the specific value in that column for each text file
M[ct,] <- as.numeric(ht) #input this value into the appropriate location
ct <- ct + 1
}
This obviously does not give me the triangulated output I would envision - I know there is an error with inputting the variable 'ht' into the matrix, but am not sure how to solve this moving forward. Ideally, the correlation value of file 1_1 should be inserted in row 1, col 1, file 1_2 should be inserted in row 2, col 1, so on and so forth, and avoiding repeats (should be 0's)
Should I turn to nested loops?
Much help would be appreciated from this R newbie here, I hope I didn't complicate things unnecessarily!
I think the easiest way would be to read in all your values into a vector. You can do this using a variation of your existing loop.
Let us assume that your desired size correlation matrix is 5x5 (I know you have 48x48 judging by your code, but to keep the example simple I will work with a smaller matrix).
Let us assume that you have read all of your correlation values into the vector x in column major order (same as R uses), i.e. the first element of x is row 2 column 1, second element is row 3 column 1 etc. I am further assuming that you are creating a symmetric correlation matrix, i.e. you have ones on the diagonal, which is why the indexing starts the way it does, because of your use of the diag() function. Let's assume your vector x contains the following values:
x <- 1:10
I know that these are not correlations, but they will make it easy to see how we fill the matrix, i.e. which vector element goes into which position in the resulting matrix.
Now, let us create the identity matrix and zero matrices for the upper and lower triangular correlations (off diagonal).
# Assuming 5x5 matrix
n_elements <- 5
m <- diag(n_elements)
m_upper <- m_lower <- matrix(0, n_elements, n_elements)
To quickly fill the lower triangular matrix, we can use the lower.tri().
m_lower[lower.tri(m_lower, diag = FALSE)] <- x
This will yield the following output:
[,1] [,2] [,3] [,4] [,5]
[1,] 0 0 0 0 0
[2,] 1 0 0 0 0
[3,] 2 5 0 0 0
[4,] 3 6 8 0 0
[5,] 4 7 9 10 0
As you can see, we have successfully filled the lower triangular. Also note the order in which the elements of the vector is filled into the matrix. This is crucial for your results to be correct. The upper triangular is simply the transpose of the lower triangular, and then we can add our three matrices together to form your symmetric correlation matrix.
m_upper <- t(m_lower)
M <- m_lower + m + m_upper
Which yields the desired output:
[,1] [,2] [,3] [,4] [,5]
[1,] 1 1 2 3 4
[2,] 1 1 5 6 7
[3,] 2 5 1 8 9
[4,] 3 6 8 1 10
[5,] 4 7 9 10 1
As you see, there is no need to work with nested loops to fill these matrices. The only loop you need is to read in the results from files (which it appears you have a handle on). If you only want the triangulated output, you can simply stop at the lower triangular matrix above. If your vector of estimated correlations (in my example x) include the diagonal elements, simply set diag = TRUE in the lower.tri() function and you are good to go.
How can produce a matrix where the entries are, say, the product of the index of the row and column. For example:
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 2 4 6
[3,] 3 6 9
NB: this is not specific to multiplication. I actually need it to raise each entry to a power (row index - column index), and was looking to not have to induce loops (as I suspect there is a more R-friendly way).
Thanks!
M <- matrix(NA, 3,3)
Mrcprod <- row(M)*col(M)
Use the outer product of 1:3 and 1:3
outer(1:3,1:3)
# or
1:3 %o% 1:3
If you need the different of the row indices and column indices, use outer again
outer(1:3,1:3,"-")
I am trying to create an index vector for a programming problem. The idea is to be able to index the elements of a matrix so I can replace just these elements with another matrix.
nstks<- 2
stk<-1:nstks
nareas<-3
area<-1:nareas
eff<-c(10,10,10)
x<-matrix(1:6,nrow=nstks,ncol=nareas)
h<-matrix(0,nrow=length(eff)+nstks,ncol=nareas)
for(i in 1:nareas) h[i,i]<-1
This returns a 5 by 3 matrix with 1s on the diagonal of the first 3 rows. Now I want to replace the 4th and 5th rows with a 2 by 3 matrix returned by another function. One way I figured is to index the h matrix by:
hlen<-c(nareas + stk,(nareas+ stk +(nareas +nstks)),(nareas+stk +(nareas+nstks)+(nareas+nstks)))
h[hlen] <- x
This replaces the 4,5,9,10,14,15th elements of h with the elements of x in order.
However, I need to make this flexible for differing numbers of nstks and nareas. As an example, for nareas=4 and nstks=3, I need to spit out a vector: c(5,6,7,12,13,14,19,20,21,26,27,28)
To clarify: I need to create the jacobian matrix for a constrained optimization problem. The dimensions of the jacobian vary depending on the number of constraints, and number of variables. I want to write a function that will give the jacobian matrix for any specified number of dimensions.
The variable is eff, which has the same length as nareas. There are non-negativity constraints on eff, which are reflected in the first nareas*nareas sub matrix being a diagonal identity matrix. The last rows of the matrix reflect the constraint on the number of fish that can be caught, by stock. So, for one stock, there will only be 1 additional row, 2 stocks, 2 additional rows etc. etc.
I need to replace the elements in these last rows by the elements given by another matrix. In the example, x is just for illustration. The actual x is given by a function but will have these same dimensions. Does this clarify things?
Any ideas?
Thanks!
I believe I can use:
h[(length(eff)+1): (length(eff)+nstks),1:nareas]<-x
I was making it too complicated as usual. Thanks for the help.
Instead of trying to find the indices for the values you need to replace with a sub-matrix returned from another function, can you not just place in the sub-matrix directly?
E.g. if you have:
x <- matrix(c(0, 1, 1, 0, 0, 1), ncol=3)
x
[,1] [,2] [,3]
[1,] 0 1 0
[2,] 1 0 1
Identify the part of h you want to drop the sub-matrix in:
h[4:5, 1:3] <- x
h
[,1] [,2] [,3]
[1,] 1 0 0
[2,] 0 1 0
[3,] 0 0 1
[4,] 0 1 0
[5,] 1 0 1
Or, if x is a vector,
x <- c(0, 1, 1, 0, 0, 1)
x <- matrix(x, ncol=3, byrow=TRUE)
h[4:5, 1:3] <- x
I have a triangular matrix and would like to loop through all the elements efficiently. Is there a smart way that I simply don't see?
So here's a small example of my matrix
[,1] [,2] [,3] [,4]
[1,] 1 0 0 0
[2,] 2 1 0 0
[3,] 3 7 1 0
[4,] 4 4 10 1
What I want to do is perform some function f() on the elements in this matrix that are under (over) the main diagonal. (background: I know that I have a symmetric matrix and would like to perform some time consuming data manipulations and I'd like to 'mirror' my matrix)
As shown below, lower.tri() and upper.tri() provide an expressive (and fast) means of extracting and replacing elements in the relevant sectors a matrix. Since the function you're applying to the elements is presumably slow compared to indexing operations, there's probably no point in searching for faster indexing options (or in trying to avoid the single call to t()).
## Example data
m <- matrix(c(1,2,3,4,0,1,7,4,0,0,1,10,0,0,0,1), ncol=4)
## Example of a slow function
slowFun <- function(x) sapply(x, function(x) {Sys.sleep(0.1); x^2})
## Proposed strategy
m[lower.tri(m)] <- slowFun(m[lower.tri(m)])
m[upper.tri(m)] <- t(m)[upper.tri(m)]