repeat_prod_mat = function(mat1,mat2){
prod = 0
repeat{
if (ncol(mat1) == ncol(mat2) & nrow(mat1) == nrow(mat2)){
prod = mat1*mat2
}
else{
break
}
}
return(prod)
}
I have to do this with repeat loop but no idea why it won't work.
Related
For a project, I am trying to replicate/rewrite the following Visual Basic code from a case study so I can run it in R.
For j=1 to N
For I=1 to M
While I>j
If b(I,j) c=P then
a(I,j)=l else
if b(1, j)<= [P+ (l-P ) /21 then
a(I,j)=3 else
a(1, j)=O
end if
end if
wend
while Icj
if a(I,j)=l then
a(j,i)=l else
if a(I,j)=3 then
a(j,I)=O else
a( j ,I) =3
end if
end if
wend
next
next
The goal is to compare the values b(j,I) from matrix B(M,N) and fill matrix A(M,N) with either 0,1 or 3. It should represent the results of a football game.
This is what I got so far:
M <- 3
N <- 3
A <- matrix(0,M,N)
B <- matrix(sample(0:4,25, replace=TRUE),nrow=M, ncol=N)
for (i in 1:nrow(B)) {
for (j in 1:ncol(B)) {
while (i > j) {
if (B[i,j] > B[j,i]) {
A[i,j] = 3 & A[j,i] = 0
}
if (B[i,j] < B[j,i]) {
A[i,j] = 0 & A[j,i] = 3
}
else {
A[i,j] = 1 & A[j,i] = 1
}
}
}
}
I would really appreciate any help or tips!
I expect the given code to output the answer : 1. However the loop runs forever. I am trying to see if while loop works for such a case. The use of while loop is necessary for the solution.
a = list(1,2,3,4)
for(i in a){
while(i != 2){
print(i)
}
}
Here are two solutions that work with while. The first one with a 'flag' set as TRUE, and the index as 1, based on the condition, set the 'flag' to FALSE
flag <- TRUE
i <- 1
while(flag) {
print(a[[i]])
i <- i + 1
if(a[[i]] == 2) {
flag <- FALSE
}
}
#[1] 1
Or we add a break
i <- 1
while(TRUE) {
print(a[[i]])
i <- i + 1
if(a[[i]] == 2) {
break
}
}
#[1] 1
The value of i does not change inside the while loop, so you never advance to the next item in list. You need if instead of while:
a = list(1,2,3,4)
for(i in a){
if(i != 2){
print(i)
}
}
I am learning how to use while and for loops, and am having difficulty executing a for loop within a while loop. I am trying to recurse a simple procedure over a vector, and have the while loop set the conditionals for which parts of the vector are operated upon. This is really just meant as an exercise to understand how to use for and while loops in conjunction.
data <- c(1:200)
n=0
while(n < 100){
for(x in data){
n=x+1
if(x >= 10 & x < 100){
print("double digit")
} else {
print("single digit")
}}}
I want to stop the while loop when the procedure hits x=100 of the vector that runs from 1:200. But instead, the for loop runs through every element within the vector, from 1:200, failing to stop executing when n hits 100.
Would appreciate any advice to help out a new coder, thanks.
I have also tried
for(x in data){
n=x+1
while(n < 100){
if(x >= 10 & x < 100){
print("double digit")
} else {
print("single digit")
}}}
But the code does not stop executing.
First let's try a for loop. Each time through it n will be set to the loop counter plus 1. If this result is between 10 and 100, print a message, if not print something else. Note that no loop depends on n .
data <- c(1:200)
n = 0
for (x in data) {
n = x + 1
if (n < 100) {
if (x >= 10 && x < 100) {
print("double digit")
} else {
print("single digit")
}
}
}
x
#[1] 200
n
#[1] 201
Now for a while loop. I believe it is much simpler, it only envolves one variable, n.
n <- 0
while (n < 100) {
n = n + 1
if (n < 100) {
if (n >= 10) {
print("double digit")
} else {
print("single digit")
}
}
}
n
#[1] 100
I was asked to implement function that calculates n-dimensional matrix determinant using Laplace expansion. This involves recursion. I developed this:
minor<-function(A,i,j) {
return(A[c(1:(i-1),(i+1):dim(A)[1]),c(1:(j-1),(j+1):dim(A)[2])])
}
determinantRec<-function(X,k) {
if (dim(X)[1] == 1 && dim(X)[2] == 1) return(X[1][1])
else {
s = 0
for (i in 1:dim(X)[2]) {
s = s + X[k][i]*(-1)^(k+i)*determinantRec(minor(X,k,i),k)
}
return(s)
}
}
where k in determinantRec(X,k) function indicates which row I want to use Laplace expansion along of.
My problem is when I run determinantRec(matrix(c(1,2,3,4),nrow = 2,ncol = 2),1) this error appears:
C stack usage 7970628 is too close to the limit
What is wrong with my code?
#julia, there is one simple type in your code. Just remove the '*' at the end of the definition of 's'. And don't indent the recursion.
determinantRek<-function(X,k) {
if (dim(X)[1] == 1 && dim(X)[2] == 1)
return(X[1,1])
if (dim(X)[1] == 2 && dim(X)[2] == 2)
return(X[1,1]*X[2,2]-X[1,2]*X[2,1])
else
s = 0
for (i in 1:dim(X)[2]) {
s = s + X[k,i]*(-1)^(k+i)
determinantRek(X[-k,-i],k)
}
return(s)
}
I did this way and works just fine, although it is super slow, compared to the det function in base R
laplace_expansion <- function(mat){
det1 <- function(mat){
mat[1]*mat[4]-mat[2]*mat[3]
}
determinant <- 0
for(j in 1:ncol(mat)){
mat1 <- mat[-1,-j]
if(nrow(mat1) == 2){
determinant <- determinant+mat[1,j]*(-1)^(1+j)*det1(mat1)
}else{
val <- mat[1,j]*(-1)^(1+j)
if(val != 0){
determinant <- determinant+val*laplace_expansion(mat1)
}
}
}
return(determinant)
}
This is my approach, I think it's cleaner.
deter <- function(X) {
stopifnot(is.matrix(X))
stopifnot(identical(ncol(X), nrow(X)))
if (all(dim(X) == c(1, 1))) return(as.numeric(X))
i <- 1:nrow(X)
out <- purrr::map_dbl(i, function(i){
X[i, 1] * (-1)^(i + 1) * deter(X[-i, -1, drop = FALSE])
})
return(sum(out))
}
Thank you #ArtemSokolov and #MrFlick for pointing the problem cause, it was it. I also discovered that this code does not calculate properly the determinant of 2x2 matrix. After all it looks like that:
determinantRek<-function(X,k) {
if (dim(X)[1] == 1 && dim(X)[2] == 1)
return(X[1,1])
if (dim(X)[1] == 2 && dim(X)[2] == 2)
return(X[1,1]*X[2,2]-X[1,2]*X[2,1])
else
s = 0
for (i in 1:dim(X)[2]) {
s = s + X[k,i]*(-1)^(k+i)*
determinantRek(X[-k,-i],k)
}
return(s)
}
Debuging with browser() was also helpful :)
I am trying to build a matrix model which ends if certain conditions are invoked - however for some reason the break() command isn't working, although stop() does. Unfortunately stop() is not what I need as I need to run the model a number of times.
The first break command in the model works, but I have left it in with dth>100 so that you can see for yourselves
n.steps <- 200
ns <- array(0,c(14,n.steps))
ns[13,1]<-rpois(1,3)
ns[14,1] <- 1
k<-0
for (i in 1:n.steps){
k<-k+1
ns[13,1]<-rpois(1,2)
death<-sample(c(replicate(1000,
sample(c(1,0), prob=c(surv.age.a, 1-surv.age.a), size = 1))),1)
ns[14,k] <- death
if (death == 0) {
dth <- sample(1:100, 1)
if (dth > 100) {
ns[14,k]<-0
print("stop.1")
break()
} else {
while (death == 0) {
if (ns[13, k] > 0) {
rep.vec[i]<-ns[13,k]
ns[13, k] <- ns[13, k] - 1
ns[14,k+1]<-1
print("replace")
} else {
if (ns[13, k] == 0) {
print("stop.2")
ns[14,k+1]<-0
break()
}
}
}
}
}
}
Try this (only showing the relevant portions):
for (i in 1:n.steps){
# ...
break.out.of.for <- FALSE
while (death == 0) {
if (ns[13, k-1] > 0) {
# ...
} else {
if (ns[13, k] == 0) {
# ...
break.out.of.for = TRUE
break
}
}
if (break.out.of.for) {
break
}
}