R 中平方根的 Newton-Raphson 方法

Question

I am a beginner in R and was asked to write the code to calculate square roots by the Newton–Raphson method in R. I approached it as follows:

square.root<-function(x,tol=1e-6,r=x/2)  #function to calculate the square roots
{
    n.iter=0  #number of iterations
    while(abs(r^2-x)>=tol)  #condition to check for a defined level of tolerance
    {
        r=(r+x/r)/2  #
        n.iter=n.iter+1  #number of iterations is incremented
    }
    output<-list(r,n.iter)
    names(output)<-c("x.sqrt","n.iter")
    return(output)
}

The above code is a prototype for the complete function in which, I plan to insert checks for the sanity of the arguments, eg, is the passed argument of the type a character or not, etc. Note that I haven't written those checks here. This code works fine as shown below:

square.root(10)
$x.sqrt
[1] 3.162278

$n.iter
[1] 4

square.root(99)
$x.sqrt
[1] 9.949874

$n.iter
[1] 6

square.root(100)
$x.sqrt
[1] 10

$n.iter
[1] 6

The only shortcoming of this code is: It doesn't work when the input is a numeric vector, as described by the output below:

square.root(c(10,99,100))
$x.sqrt
[1]  3.162278  9.979213 10.030495

$n.iter
[1] 4

Warning messages:
1: In while (abs(r^2 - x) >= tol) { :
  the condition has length > 1 and only the first element will be used
2: In while (abs(r^2 - x) >= tol) { :
  the condition has length > 1 and only the first element will be used
3: In while (abs(r^2 - x) >= tol) { :
  the condition has length > 1 and only the first element will be used
4: In while (abs(r^2 - x) >= tol) { :
  the condition has length > 1 and only the first element will be used
5: In while (abs(r^2 - x) >= tol) { :
  the condition has length > 1 and only the first element will be used

Now, I want to add the functionality that even a vector can be passed as an argument. My thoughts were that I create a blank data frame named as output . Then iterate over the vector x elements and go on appending new rows to the data frame output .

---

What I tried:

I wrote the following code concerning the above 'thought'.

square.root1<-function(x,tol=1e-6,r=x/2)#function to calculate the square roots
{
    output<-data.frame(x.sqrt=double(),n.iter=double())  #create a blank data frame
    n.iter=rep(0,times=length(x))  #a vector of zeroes
    i=1  #iteration counter
    while(i<=length(x))  #iterate through all the values of the vector
    {
        while(abs(r[i]^2-x[i])>=tol)  #the checking condition for the tolerance
        {
            r[i]=(r[i]+x[i]/r[i])/2
            n.iter[i]=n.iter[i]+1  #the number of iterations is incremented
        }
        result<-data.frame(r[i],n.iter)  #store the results in another data frame
        names(result)<-c("x.sqrt","n.iter")  #name the columns
        rbind(output,result)  #append the result data frame to the output data frame
        i=i+1  #iterate the looping counter
    }
    return(output)  #return the output data frame
}

---

Results and where I am stuck:

I seem to have messed up somewhere. The code creates an output that I was unprepared for, even for a numeric value:

square.root1(5)
[1] x.sqrt n.iter
<0 rows> (or 0-length row.names)

Looking at the above output it seems that the control is not going inside the first while loop. I am stuck as to how I can proceed to create this function that would allow a vector as an argument. Any help/hints are highly appreciated.

Note: I am bound to use the while loop in this function.

Another note: This question might be similar, but it didn't solve the problem. Below is the output I obtained from running this code:

MySqrt(c(99,10,100))
Iteration:  1           10  5.954545455  10.04545455
Iteration:  2         9.95  3.816967384  10.00010284
Iteration:  3  9.949874372  3.218424149           10
[1] 9.949874

This is pretty much the same result that I would have obtained from the first code I wrote for the square root with a vector as the argument.

Answer 1

Using Vectorize is perhaps the easiest way to do it as per the comment, but it is also possible to make a couple of small changes to your original code so that everything works on a vector...

square.root<-function(x, tol=1e-6, r=x/2)  #function to calculate the square roots
{
  n.iter=0  #number of iterations
  while(any(abs(r^2-x)>=tol))  #"any" added here - loops while any values greater than tol
  {
     n.iter=n.iter+(abs(r^2-x)>=tol)  #only increases n.iter for errors greater than tol
     r=(r+x/r)/2        #swapped with line above to get right value in n.iter calculation       
  }
  output<-list(r,n.iter)
  names(output)<-c("x.sqrt","n.iter")
  return(output)
}

square.root(c(10,99,100))
$x.sqrt
[1]  3.162278  9.949874 10.000000

$n.iter
[1] 4 6 6