This is a homework problem. I am brand new to R just FYI
This is the problem:
R does have a built-in constant pi. Here we will use random numbers to estimate the value of
π. Create a function approx.pi() that takes a parameter N. Inside this function, code the
following steps:
(1) set j equal to 0.
(2) start a for loop with counter i that repeats N times.
(3) Inside the for loop, generate two random uniform numbers x and y between -1 and +1
using runif().
(4) If x^2 + y^2 < 1 then add one to j.
(5) End the for loop.
(6) Return the estimate of π which is 4×j/N
and this is the code I have:
approx.pi <- function(N) {
j <- 0
for (i in N) {
x <- runif(1,-1,1)
y <- runif(1,-1,1)
if (x^2+y^2< 1) {
j=j+1
}
}
return(4*j/N)
}
approx.pi(N=5)
I am getting some number values returned but they are no where near pi can anyone help? Thank you.
This code will only run for 1 value i = N.
for (i in N) {}
You need to change to
for (i in 1:N){}
I have tried edited code and got the output
approx.pi <- function(N) {
j <- 0
for (i in 1:N) {
x <- runif(1,-1,1)
y <- runif(1,-1,1)
if (x^2+y^2< 1) {
j=j+1
}
}
return(4*j/N)
}
Here is my output. (you use 5 is too small).
> approx.pi(5000)
[1] 3.188
> approx.pi(5000)
[1] 3.1488
> approx.pi(5000)
[1] 3.1344
> approx.pi(5000)
[1] 3.1672
> approx.pi(5000)
[1] 3.1632
> approx.pi(5000)
[1] 3.1152
Related
I need help with coding this into R. I'm not well versed with R's library functions.
Given - Let X be a Bernoulli random variable which is defined as X = 1 if we get a head with probability p ## and X = 0 if we get a tail with probability 1 − p.
##This is what I have so far. Not sure if I'm taking the right way in doing this
p = runif(1,0,1)
if (p<0.5){
X=1
} else {
X=0
}
print(p); print(X)
## Calculate E(2X)
E(2X) = E(X)*2 = 2p
## Calculate var(3X + 4)
## Calculate cov(3X, 5)
## Calculate cov(4X, 6X + 1000)
There is a built in function rbinom(n,size, prob) that can be utilized to return a vector of zeroes and ones of length n if size=1. Alternatively, you can mimic this following the logic that you started above, if you like, doing something like this
Your logic, using uniform distribution
flip_coin_n_times_with_head_prop_p <- function(n,p) {
uniforms = runif(n,0,1)
heads = sum(uniforms<=p)
return(heads)
}
Using R's built in rbinom()
built_in_flip <- function(n,p) {
heads=sum(rbinom(n,1,p))
return(heads)
}
Now, you can see that these are equivalent, by flipping a coin with probability 0.3, 10000 times:
> flip_coin_n_times_with_head_prop_p(10000,.3)
[1] 2967
> built_in_flip(10000,.3)
[1] 3058
Of course, you may want the sequence of flips themselves.. Either function can be adapated, but I've gone with the built_in_flip() for illustration purposes.
Add parameter return_trials, default is TRUE; will return the sequence of flips by default, but set to FALSE to get the number
of heads
built_in_flip <- function(n,p, return_trials=T) {
trials = rbinom(n,1,p)
if(return_trials) return(trials)
else return(sum(trials))
}
Now, when we call this function we get the sequence of flips
built_in_flips(n=10, p=0.5)
[1] 1 1 0 0 1 0 0 0 0 1
We can assign a long sequence of flips to X like this, and then
take the mean(X) or the mean(2*X)
X = built_in_flips(10000,0.3, return_trials=T)
# "Expectation of X"
mean(X)
[1] 0.3026
# "Expectation of 2X"
mean(2*X)
[1] 0.6052
I'm trying to convert a while loop to a recursion.
I know the while loop is more efficient, but I'm trying to understand how to convert a for/while loop to recursion, and recursion to a for/while/if loop.
my function as I'm using a while loop:
harmon_sum <- function(x){
n <- 1
sum <- 0
while (sum < x)
{
sum <- sum + (1/n)
n <- (n +1)
}
return(n)
}
This function takes some numeric value, suppose x=2, and returns the number of objects for the harmonic sum that you need to sum up in order to create a greater number then x. (for x=2, you'd need to sum up the first 5 objects of the harmonic sum)
[![harmonic sum][1]][1]
**example**: `harmon_sum <- function(x){
n <- 1
sum <- 0
while (sum < x)
{
sum <- sum + (1/n)
print(sum)
n <- (n +1)
print(n)
}
return(n)
}
> harmon_sum(x =2)
[1] 1
[1] 2
[1] 1.5
[1] 3
[1] 1.833333
[1] 4
[1] 2.083333
[1] 5
[1] 5`
my version for the recursive function:
harmon_sum2 <- function(x, n =1){
if( x<= 0){
return(n-1)
}
else {
x <- (x- (1/(n)))
harmon_sum2(x, n+1)
}
}
which returns me the wrong answer.
I'd rather find a solution with just one variable (x), instead of using two variables (x, n), but I couldn't figure a way to do that.
It seems to me that if you change return(n-1) to return(n) you do get the right results.
harmon_sum2 <- function(x, n=1){
if( x <= 0){
return(n)
}
else {
x <- (x- (1/(n)))
harmon_sum2(x, n+1)
}
}
harmon_sum(2)
[1] 5
harmon_sum2(2)
[1] 5
harmon_sum(4)
[1] 32
harmon_sum2(4)
[1] 32
Your function needs to know n. If you don't want to pass it, you need to store it somewhere where all functions on the call stack can access it. For your specific case you can use sys.nframe instead:
harmon_sum2 <- function(x){
if( x<= 0){
return(sys.nframe())
}
else {
x <- (x- (1/(sys.nframe())))
harmon_sum2(x)
}
}
harmon_sum(8)
#[1] 1675
harmon_sum2(8)
#[1] 1675
However, this doesn't work if you call your function from within another function:
print(harmon_sum2(8))
#[1] 4551
Another alternative is the approach I demonstrate in this answer.
I am writing a function to perform bit inversion for each row of a binary matrix which depends on a predefined n value. The n value will determine the number of 1 bits for each row of the matrix.
set.seed(123)
## generate a random 5 by 10 binary matrix
init <- t(replicate(5, {i <- sample(3:6, 1); sample(c(rep(1, i), rep(0, 10 - i)))}))
n <- 3
## init_1 is a used to explain my problem (single row matrix)
init_1 <- t(replicate(1, {i <- sample(3:6, 1); sample(c(rep(1, i), rep(0, 10 - i)))}))
The bit_inversion function does this few things:
If the selected row has number of 1's lesser than n, then it randomly select a few indices (difference) and invert them. (0 to 1)
Else if the selected row has number of 1's greater than n, then it randomly select a few indices (difference) and invert them. (1 to 0)
Else do nothing (when the row has number of 1's equals to n.)
Below is the function I implemented:
bit_inversion<- function(pop){
for(i in 1:nrow(pop)){
difference <- abs(sum(pop[i,]) - n)
## checking condition where there are more bits being turned on than n
if(sum(pop[i,]) > n){
## determine position of 1's
bit_position_1 <- sample(which(pop[i,]==1), difference)
## bit inversion
for(j in 1:length(bit_position_1)){
pop[bit_position_1[j]] <- abs(pop[i,][bit_position_1[j]] - 1)
}
}
else if (sum(pop[i,]) < n){
## determine position of 0's
bit_position_0 <- sample(which(pop[i,]==0), difference)
## bit inversion
for(j in 1:length(bit_position_0)){
pop[bit_position_0[j]] <- abs(pop[bit_position_0[j]] - 1)
}
}
}
return(pop)
}
Outcome:
call <- bit_inversion(init)
> rowSums(call) ## suppose to be all 3
[1] 3 4 5 4 3
But when using init_1 (a single row matrix), the function seems to work fine.
Outcome:
call_1 <- bit_inversion(init_1)
> rowSums(call)
[1] 3
Is there a mistake in my for and if...else loop?
Change the line in 'j' for loop
pop[bit_position_1[j]] <- abs(pop[i,][bit_position_1[j]] - 1)
into
pop[i,bit_position_1[j]] <- abs(pop[i,][bit_position_1[j]] - 1)
You forgot the row index.
And, here is a more compact version of your for loop:
for(i in 1:nrow(pop)){
difference <- abs(sum(pop[i,]) - n)
logi <- sum(pop[i,]) > n
pop[i,sample(which(pop[i,]==logi), difference)] <- !logi
}
a) Create a vector X of length 20, with the kth element in X = 2k, for k=1…20. Print out the values of X.
b) Create a vector Y of length 20, with all elements in Y equal to 0. Print out the values of Y.
c) Using a for loop, reassigns the value of the k-th element in Y, for k = 1…20. When k < 12, the kth element of Y is reassigned as the cosine of k. When the k ≥ 12, the kth element of Y is reassigned as the value of integral sqrt(t)dt from 0 to K.
for the first two questions, it is simple.
> x1 <- seq(1,20,by=2)
> x <- 2 * x1
> x
[1] 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40
> y <- rep(0,20)
> y
[1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
i got stuck on the last one,
t <- function(i) sqrt(i)
for (i in 1:20) {
if (i < 12) {
y[i] <- cos(i)
}
else if (i >= 12) {
y[i] <- integral(t, lower= 0, Upper = 20)
}
}
y // print new y
Any suggestions? thanks.
What may help is that the command to calculate a one-dimensional integral is integrate not integral.
You have successfully completed the first two, so I'll demonstrate a different way of getting those vectors:
x <- 2 * seq_len(20)
y <- double(length = 20)
As for your function, you have the right idea, but you need to clean up your syntax a bit. For example, you may need to double-check your braces (using a set style like Hadley Wickham's will help you prevent syntax errors and make the code more readable), you don't need the "if" in the else, you need to read up on integrate and see what its inputs, and importantly its outputs are (and which of them you need and how to extract it), and lastly, you need to return a value from your function. Hopefully, that's enough to help you work it out on your own. Good Luck!
Update
Slightly different function to demonstrate coding style and some best practices with loops
Given a working answer has been posted, this is what I did when looking at your question. I think it is worth posting, as as I think that it is a good habit to 1) pre-allocate answers 2) prevent confusion about scope by not re-using the input variable name as an output and 3) use the seq_len and seq_along constructions for for loops, per R Inferno(pdf) which is required reading, in my opinion:
tf <- function(y){
z <- double(length = length(y))
for (k in seq_along(y)) {
if (k < 12) {
z[k] <- cos(k)
} else {
z[k] <- integrate(f = sqrt, lower = 0, upper = k)$value
}
}
return(z)
}
Which returns:
> tf(y)
[1] 0.540302306 -0.416146837 -0.989992497 -0.653643621 0.283662185 0.960170287 0.753902254
[8] -0.145500034 -0.911130262 -0.839071529 0.004425698 27.712816032 31.248114562 34.922139530
[15] 38.729837810 42.666671456 46.728535669 50.911693960 55.212726149 59.628486093
To be honest you almost have it ready and it is good that you have showed some code here:
y <- rep(0,20) #y vector from question 2
for ( k in 1:20) { #start the loop
if (k < 12) { #if k less than 12
y[k] <- cos(k) #calculate cosine
} else if( k >= 12) { #else if k greater or equal to 12
y[k] <- integrate( sqrt, lower=0, upper=k)$value #see below for explanation
}
}
print(y) #prints y
> print(y)
[1] 0.540302306 -0.416146837 -0.989992497 -0.653643621 0.283662185 0.960170287 0.753902254 -0.145500034 -0.911130262 -0.839071529 0.004425698
[12] 27.712816032 31.248114562 34.922139530 38.729837810 42.666671456 46.728535669 50.911693960 55.212726149 59.628486093
First of all stats::integrate is the function you need to calculate the integral
integrate( sqrt, lower=0, upper=2)$value
The first argument is a function which in your case is sqrt. sqrt is defined already in R so there is no need to define it yourself explicitly as t <- function(i) sqrt(i)
The other two arguments as you correctly set in your code are lower and upper.
The function integrate( sqrt, lower=0, upper=2) will return:
1.885618 with absolute error < 0.00022
and that is why you need integrate( sqrt, lower=0, upper=2)$value to only extract the value.
Type ?integrate in your console to see the documentation which will help you a lot I think.
I want to iterate a loop only for some values so I am using this:
present <- c(3,5,7,8)
for(i in present)
{
print(i)
}
which gives me
[1] 3
[1] 5
[1] 7
[1] 8
however I need to jump to the next value within the loop, say I dont want 5 to be printed in above example.
I cannot use next since I want it in nested for like this
present <- c(3,5,7,8)
for(i in present)
{
k <- i
"Jump to next value of present"
while(k < "The next value for i should come here")
{
k <- k + 1
print(k)
}
}
The output would be 3 4 5 6 7 8 but the condition must check value of k if it exceeds next value of i.
Is there anyway to accomplish this?
I'll take help of C to explain further,
for(i=0; i < 10; i++)
{
for(k=i;k <= i+1;k++)
{
printf("%d", k);
}
}
The link contains output of above code
http://codepad.org/relkenY3
It is easy in C since next value is in sequence, but here next value is not known, hence the problem.
What you should do is loop through two vectors:
x <- head(present, -1)
# [1] 3 5 7
y <- tail(present, -1)
# [1] 5 7 8
and the function to do that is mapply (have a look at ?mapply). A close translation of your pseudo-code would be:
invisible(mapply(function(x, y) while(x < y) {x <- x + 1; print(x)}, x, y))
but maybe you'll find this more interesting:
mapply(seq, x + 1, y)
I suspect the answer is to use seq_along and use it as an index into "present", but as others have pointed out your code does not promise to deliver what you expect, even with that simple modification. The K <- K=1 assignment jumps ahead too far to deliver a value of 3 at any point and the termination condition is likewise not clear. It turns into an infinite loop in the form you construct. Work with this;
present <- c(3,5,7,8)
for(i in seq_along(present))
{
k <- i
while(k < length(present) )
{
k <- k + 1
print(present[k])
}
}