I am doing 'find majority element' using R. And when the function me is running, it would not stop and got "Error: C stack usage 7971200 is too close to the limit"
I have checked single majority element comparison and it works.
So which step i went wrong?
library(Dict)
maj.com <- function(a,b) {
if (is.null(a)&& is.null(b)) {
return(NULL)
}
if (!(is.null(a)) && is.null(b)) {
return(a)
}
if (is.null(a) && !(is.null(b)) ){
return(b)
}
if (unlist(a[1]) == unlist(b[1])) {
a[2] = as.integer(a[2])+ as.integer(b[2])
return(a)
}
if (unlist(a[1]) != unlist(b[1]) && as.integer(a[2]) > as.integer(b[2])) {
a[2] = as.integer(a[2]) - as.integer(a[2])
return(a)
}
if (unlist(a[1]) != unlist(b[1]) && as.integer(a[2]) < as.integer(b[2])) {
b[2] = as.integer( b[2] )- as.integer(a[2])
return(b)
}
if (unlist(a[1]) != unlist(b[1]) && as.integer(a[2]) == as.integer(b[2])) {
return(NULL)
}
}
mydict <- function(key,value) {
return(list(key,value))
}
me <- function(arr) {
n = length(arr)
m = round(n/2)
if (n>1) {
return(maj.com(me(arr[1:m]),me(arr[m+1:n])))
}
else {
print(arr)
return(mydict(arr,1))
}
}
i <- 2
j <- 0
for (i in 2:1000) {
if(return.prime(i)){j = j + 1}
i = i + 1
}
I want to check how many prime numbers there are in 1 to 1000 using my own function return.prime which returns TRUE when the number input is a prime and FALSE when the number input is not prime. The return.prime function is the function below and it is correct.
return.prime <- function(d){
if(d ==1 ){print(FALSE)}
if (d == 2){
print(TRUE)
}
if(d !=2 && d!=1){
if(any(d %% (2:(d-1)) == rep(0,d-2))==TRUE){
print(FALSE)}
else
print(TRUE)
}
}
The problem is when I run my program it says:
[1] TRUE
Error in if (return.prime(i)) { : argument is of length zero
I do not know what causes the length zero.
R doesn't work that way. You're just having the function print the word "TRUE" or "FALSE". Instead, you need to ?return TRUE or FALSE. Consider:
return.prime <- function(d){
if(d==1){ return(FALSE) }
if(d==2){ return(TRUE) }
if(d !=2 && d!=1){
if(any(d %% (2:(d-1)) == rep(0,d-2))==TRUE){
return(FALSE)
} else{
return(TRUE)
}
}
}
i <- 2
j <- 0
for (i in 2:1000) {
if(return.prime(i)){j = j + 1}
i = i + 1
}
j # [1] 168
This recursion code doesn't give an output for n > 2. It works if I include the last if condition as else. Why does this happen?
fib <- function(n){
if(n == 0){
0
}
if(n ==1 ){
1
}
if(n == 2){
1
}
if (n > 2) {
print(n)
fib(n-1) + fib(n-2)
}
}
A function by default returns the result of the last value evaluated inside the function. Here you have a bunch of separate if statements that will all be evaluated (your function doesn't return early at any point. Observe
f <- function(a) {
-99
if (a==1) {111}
if (a==2) {22}
}
f(1)
# -- doesn't return anything --
f(2)
# [1] 22
f(3)
# -- doesn't return anything --
When 'a==1', That middle if statement would evaluate to 111 but that value is then discarded when you run the next if statement just like the -99 value was before it. The last expression that was evaluated in this function is the if (a==2) {22} and that will either return 22 or it will return NULL. It won't return the previous values because it has no idea how those would be related without the else statement.
fib <- function(n){
if(n == 0){
return(0)
}
if(n ==1 ){
return(1)
}
if(n == 2){
return(1)
}
if (n > 2) {
print(n)
return(fib(n-1) + fib(n-2))
}
}
Or you could turn those into if/else staments so only one "thing" will be returned
fib <- function(n){
if(n == 0){
0
} else if(n ==1 ){
1
} else if(n == 2){
1
} else if (n > 2) {
print(n)
fib(n-1) + fib(n-2)
}
}
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 generate a vector of random numbers based on a finite random variable X
With probGen function I generate a variable X, l1 is the first line and l2 is the second one.
And at this point if(sum1 >= U) I recive this error Error in if (sum1 >= U) { : argument is of length zero
This is my code:
probGen=function(n)
{
v=vector()
k=sample(1:n,1)
v=rep(0,k)
for(i in 1:n)
{
aux=sample(1:k,1)
v[aux]=v[aux]+1
}
vfinal=vector()
klen=0
for(i in 1:k)
{
if(v[i]!=0) klen=klen+1
}
for(i in 1:k)
{
if(v[i]!=0)
vfinal=c(vfinal,rep(1/(klen*v[i]),v[i]))
}
vfinal=sample(vfinal)
return (vfinal)
}
n=22
l1=c(1:n)
l2=probGen(n)
l1
l2
simVar=function(l1,l2)
{
variante=vector()
U=runif(1,0,1)
for(i in 1:length(l1))
{
sum1=1-1
for(j in 1:i-1)
{
if(i-1>=1)
{
sum1=sum1+l2[j]
}
}
sum2=0.0
for(j in 1:i)
{
sum2=sum2+l2[j]
}
if(sum1 >= U)
{
if(U<sum2)
{
variante=c(variante,l1[i])
}
}
}
return (variante)
}
varR=simVar(l1,l2)
varR
Any idea?
Thanks!
The for(j in 1:i-1) near the top of the code for simVar is evaluating as (1:i)-1, resulting in a zero j which produces a NA value of sum1. Use for(j in 1:(i-1)) instead.