I have the following code in R:
z <- scale(x) / sqrt(n-1) # standardized matrix x such that z'z=correlation matrix
R <- t(z) %*% z # correlation matrix
I <- diag(py - 1) # identity matrix(py defined before)
df <- rep(0, length(k)) # k=seq(0,5,0.001)
for (i in seq(0,5,0.001)) {
H <- z %*% solve(R+(i*I)) %*% t(z)
tr <- sum(diag(H))
df <- c(df,tr) ## problem here
}
The last line in the code is not good, as what I want is a vector (df) that reads each number from tr for each i, so that df returns a vector containing all tr.
Any help is appreciated.
Thanks
Separate the points that you want to solve at from the loop index.
solve_points <- seq(0,5,0.001)
for(i in seq_along(solve_points))
{
H=z%*%solve(R+(solve_points*I))%*%t(z)
tr=sum(diag(H))
df[i] <- tr
You want to fill in the vector df, not concatenate it all the time. That will slow R down a lot as it has to copy the object each iteration of the loop.
I think you probably want something like this:
for (i in seq_along(k)) { ## loop over 1:length(k)
H <- z %*% solve(R+(k[i]*I)) %*% t(z) ## use i to index into k
tr <- sum(diag(H))
df[i] <- tr ## add `tr` to the ith element of df
}
but a reproducible example would have helped. For example, you might not need to index k, depends on what your code is really doing and you don;t provide all the objects to check.
Related
all
I'm new to R. I try many ways and still cannot solve it. Can anyone help to check??
I am trying to produce 3 times 100 random values that follow a chisquare distribution. Console says ''number of items to replace is not a multiple of replacement length''. Any hint to fix it??
for(i in 1:3) {
x1[i] <- rchisq(100, df=2)
n1[i] <- length(x1[i])
}
As an explanation for your problem: You are trying to store a vector of 100 elements into a single element, the ith element, of a vector, x1. To illustrate, you could put a vector of values into a vector of the same length:
x <- rnorm(6, 0, 1)
x[1:3] <- c(1,2,3)
x
## [1] 1.0000000 2.0000000 3.0000000 -0.8652300 1.3776699 -0.8817483
You could to store them into a list, each element of a list is a vector that can be of any length. You will need double square brackets.
x1 <- list()
for(i in 1:3) {
x1[[i]] <- rchisq(100, df=2)
n1[i] <- length(x1[[i]])
}
Lists and vectors are different types of data structures in R, you can read a lot about them in advanced R.
It depends on what containers you want to use. There are two containers that come to mind, either a list or matrix.
# list format
x1 = list();
n1 = vector();
for(i in 1:3) {
x1[[i]] <- rchisq(100, df=2)
n1[i] <- length(x1[[i]])
}
note the double brackets [[i]] as mentioned in the comments
# matrix format
x1 = matrix(NA, nrow = 100, ncol = 3)
n1 = vector();
for(i in 1:3) {
x1[,i] <- rchisq(100, df=2)
n1[i] <- length(x1[,i])
}
Title's a little rough, open to suggestions to improve.
I'm trying to calculate time-average covariances for a 500 length vector.
This is the equation we're using
The result I'm hoping for is a vector with an entry for k from 0 to 500 (0 would just be the variance of the whole set).
I've started with something like this, but I know I'll need to reference the gap (i) in the first mean comparison as well:
x <- rnorm(500)
xMean <-mean(x)
i <- seq(1, 500)
dfGam <- data.frame(i)
dfGam$gamma <- (1/(500-dfGam$i))*(sum((x-xMean)*(x[-dfGam$i]-xMean)))
Is it possible to do this using vector math or will I need to use some sort of for loop?
Here's the for loop that I've come up with for the solution:
gamma_func <- function(input_vec) {
output_vec <- c()
input_mean <- mean(input_vec)
iter <- seq(1, length(input_vec)-1)
for(val in iter){
iter2 <- seq((val+1), length(input_vec))
gamma_sum <- 0
for(val2 in iter2){
gamma_sum <- gamma_sum + (input_vec[val2]-input_mean)*(input_vec[val2-val]-input_mean)
}
output_vec[val] <- (1/length(iter2))*gamma_sum
}
return(output_vec)
}
Thanks
Using data.table, mostly for the shift function to make x_{t - k}, you can do this:
library(data.table)
gammabar <- function(k, x){
xbar <- mean(x)
n <- length(x)
df <- data.table(xt = x, xtk = shift(x, k))[!is.na(xtk)]
df[, sum((xt - xbar)*(xtk - xbar))/n]
}
gammabar(k = 10, x)
# [1] -0.1553118
The filter [!is.na(xtk)] starts the sum at t = k + 1, because xtk will be NA for the first k indices due to being shifted by k.
Reproducible x
x <- c(0.376972124936433, 0.301548373935665, -1.0980231706536, -1.13040590360378,
-2.79653431987176, 0.720573498411587, 0.93912102300901, -0.229377746707471,
1.75913134696347, 0.117366786802848, -0.853122822287008, 0.909259181618213,
1.19637295955276, -0.371583903741348, -0.123260233287436, 1.80004311672545,
1.70399587729432, -3.03876460529759, -2.28897494991878, 0.0583034949929225,
2.17436525195634, 1.09818265352131, 0.318220322390854, -0.0731475581637693,
0.834268741278827, 0.198750636733429, 1.29784138432631, 0.936718306241348,
-0.147433193833294, 0.110431994640128, -0.812504663900505, -0.743702167768748,
1.09534507180741, 2.43537370755095, 0.38811846676708, 0.290627670295127,
-0.285598287083935, 0.0760147178373681, -0.560298603759627, 0.447188372143361,
0.908501134499943, -0.505059597708343, -0.301004012157305, -0.726035976548133,
-1.18007702699501, 0.253074712637114, -0.370711296884049, 0.0221795637601637,
0.660044122429767, 0.48879363533552)
Given a data matrix with n rows and m columns, I would like to calculate the total sum of squares in R.
For this I've tried a loop that iterates through the rows of each column and saves the results in a vector. These are then added to the "TSS" vector where each value is the SS of one column. The sum of this vector should be the TSS.
set.seed(2020)
m <- matrix(c(sample(1:100, 80)), nrow = 40, ncol = 2)
tss <- c()
for(j in 1:ncol(m)){
tssVec <- c()
for(i in 1:nrow(m)){
b <- sum(((m[i,]) - mean(m[,j]))^2)
tssVec <- c(tssVec, b)
}
tss <- c(tss, sum(tssVec))
}
sum(tss)
The output is equal to 136705.6. This is not feasible at all. As a novice coder, I am unfortunately stuck.
Any help is appreciated!
There are many methods to evaluate the TSS, of course they will give you the same result. I would do something like:
Method 1 that implies the use of ANOVA:
n <- as.data.frame(m)
mylm <- lm(n$V1 ~ n$V2)
SSTotal <-sum(anova(mylm)[,2])
Method 2:
SSTotal <- var( m[,1] ) * (nrow(m)-1)
I'd like to convert a size n factor into a n×n binary matrix whose (i, j) element is 1 if i-th and j-th element of factor are same and 0 otherwise.
The following is a naive way to implement what I want to do but this code is quite slow. Is there any more efficient way to do the same thing?
size <- 100
id <- factor(sample(3, size, replace=TRUE))
mat <- matrix(0, nrow=size, ncol=size)
for(i in 1:size){
for(j in 1:size){
if(id[i] == id[j]){
mat[i, j] <- 1
}
}
}
Another alternative, which should be relatively fast
tcrossprod(model.matrix( ~ id + 0))
Similarly to Hong Ooi's answer you can use also sparse matrices
library(Matrix)
tcrossprod(sparse.model.matrix( ~ id + 0))
outer can be used for this.
mat <- outer(id, id, "==")
Since the output is a binary matrix, and O(N^2) objects are kind of large, this is a good use case for sparse matrices:
library(Matrix)
mat <- Matrix(nrow=100, ncol=100)
mat[] <- outer(id, id, "==") # [] means to assign into the existing 'mat' matrix
I'm having trouble efficiently loading data into a sparse matrix format in R.
Here is an (incomplete) example of my current strategy:
library(Matrix)
a1=Matrix(0,5000,100000,sparse=T)
for(i in 1:5000)
a1[i,idxOfCols]=x
Where x is usually around length 20. This is not efficient and eventually slows to a crawl. I know there is a better way but wasn't sure how. Suggestions?
You can populate the matrix all at once:
library(Matrix)
n <- 5000
m <- 1e5
k <- 20
idxOfCols <- sample(1:m, k)
x <- rnorm(k)
a2 <- sparseMatrix(
i=rep(1:n, each=k),
j=rep(idxOfCols, n),
x=rep(x, k),
dims=c(n,m)
)
# Compare
a1 <- Matrix(0,5000,100000,sparse=T)
for(i in 1:n) {
a1[i,idxOfCols] <- x
}
sum(a1 - a2) # 0
You don't need to use a for-loop. Yu can just use standard matrix indexing with a two column matrix:
a1[ cbind(i,idxOfCols) ] <- x