Statistical Programming with R

Transcription

1 Statistical Programming with R Lecture 6: Programming Examples Bisher M. Iqelan biqelan@iugaza.edu.ps Department of Mathematics, Faculty of Science, The Islamic University of Gaza , Semester 1

2 Hilbert Matrix We have explained the structure of Hilbert matrices of dierent orders in Lecture 4. Here we are going to write an R code to produce the Hilbert matrix of any order. H=function(n){ H=matrix(0,n,n) for(i in 1:n){ for(j in 1:n){ H[i,j]=1/i+j-1 H You can try it now: > H(7) Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

3 The Low of Big Number Think with the following R code about the probability of obtaining one of the six faces when tossing a fair die. It explain the fact that they should be equal for large and large number of tosses. w=list() freq=list() n=c(20,100,500,1000,5000,20000) for(i in 1:6){ w[[i]]=sample(1:6,n[i],replace=true) freq[[i]]=table(w[[i]])/length(w[[i]]) freq Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

4 The Low of Big Number Continue We can easily see the bar plot of the frequencies of these trials by applying the following R code (it depends on the previous code) par(mfrow=c(3,2)) for(i in 1:6){ barplot(freq[[i]]) title(main=substitute(paste('no. of trials=',a),list(a=n[i]))) Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

5 Root Finding Finding the solutions of equations of the form h(x) = c is a very important problem in applied mathematics (and statistics). By writing f (x) = h(x) c, it is clear that solving the above equation is equivalent to solving f (x) = 0, and we can restrict our attention to nding the solutions of this type of equation. The solutions are called roots or zeros of the function f (x). A function may have zero, one, or many roots. Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

6 The Roots of a General Polynomial R has a built-in function, called polyroot(), for nding all the roots of a polynomial. The coecients of the polygon a n x n + + a 1 x + a 0 are passed in a vector (a 0, a 1,..., a n ) to polyroot. The computed roots are returned in a vector. Since, in general, the roots of polynomials are complex-valued, this vector has mode "complex". Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

7 Examples Let us nd the roots or zeros of the polynomial > polyroot(c(1,-2,1)) [1] 1+0i 1-0i and for the polynomial p(x) = 1 2x + x 2 p(x) = 3 4x + 5x 2 7x 3 + x 4 > polyroot(c(3,-4,5,-7,1)) [1] i [2] i [3] i [4] i Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

8 where f denotes the derivative of f. Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38 Finding square roots Suppose we want to create a function mysqrt for nding the square root of its argument x. Our purpose of this example is to introduce a general numerical method for solving equations: The NewtonRapson method. We want to nd x, i.e. the number(s) z such that z 2 = x. (Recall that 4 = ±2.) Thus for a given value x (e.g. x = 10) we want to solve the equation f (z) = z 2 x = 0 Consider the function f (z) = z The function f is nonlinear but it can be approximated by a linear function (a straight line) in a given point. E.g. close to z = 2 the function is approximated by the line call it l(z). The linear approximation to f close to a specic point z 0 (e.g. z 0 = 2) is given by: f (z) f (z 0 ) + f (z 0 )(z z 0 ) = [f (z 0 ) f (z 0 )z 0 ] + f (z 0 )z

9 NewtonRapson method Hence the linear approximation is l(z) = [f (z 0 ) f (z 0 )z 0 ] + f (z 0 )z = a + bz, For the function f (z) = z 2 10 we have f (z) = 2z. With z 0 = 2 we have f (z 0 ) = 4 and f (z 0 ) = 6 so l(z) becomes l(z) = z The idea in NewtonRapson is the following: Suppose z 0 = 2 is our initial guess for a solution to the equation f (z) = 0 (which is dicult to solve). Now l(z) is an approximation to f and it is easy to solve l(z) = 0 : z = a b = 14 4 = 3.5 So we may take 3.5 to be a new approximation to the solution to f (z) = 0. We write z 1 = 3.5. Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

10 NewtonRapson method Now we nd the line which approximates f close to z 1 = 3.5. The line is (check this for yourself): Solving l(z) = 0 gives l(z) = z. z = 22.25/7 = So the next approximation is z 2 = Iterate this scheme until the values for z stop changing. This is in essence the NewtonRapson method. Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

11 NewtonRapson method Before going to programming let us again consider the solution to l(z) = a + bz = 0, i.e. z = a b. where, a = f (z 0 ) f (z 0 )z 0 and b = f (z 0 ). Inserting the latter in the former gives the form in which the NewtonRapson method is usually presented in the literature: z = a b = f (z 0) f (z 0 )z 0 f (z 0 ) = z 0 f (z 0) f (z 0 ) Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

12 Implementing mysqrt We shall content ourself with an implementation which only returns the positive root. First, we need to implement our function f = z 2 x which is a function of z and x: f <- function(z, x) { value <- z^2 - x return(value) > f(z = 2, x = 10) [1] -6 We also need the derivative f deriv.f <- function(z) { value <- 2 * z return(value) of f (which here doesn't depend on x): Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

13 Implementing mysqrt So now we are almost there. The NewtonRapson steps (2) become: > z0 <- 2 > x <- 10 > z1 <- z0 - f(z0, x)/deriv.f(z0) > z1 [1] 3.5 > z2 <- z1 - f(z1, x)/deriv.f(z1) > z2 [1] > z3 <- z2 - f(z2, x)/deriv.f(z2) > z3 [1] Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

14 Implementing mysqrt It would be nice to write a single function for doing the iterations above, and such a function can be made in dierent ways: We create a simple function which performs 5 NewtonRapson steps: NRsteps <- function(x, z0) { z <- z0 for (i in 1:5) { z <- z - f(z, x)/deriv.f(z) print(z) return(z) Having done so, it is easy to implement our mysqrt function: mysqrt <- function(x, z0) { z <- NRsteps(x, z0) return(z) Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

15 Implementing mysqrt Calling the function gives > mysqrt(x = 10, z0 = 2) [1] 3.5 [1] [1] [1] [1] [1] Clearly, no changes after 4 iterations! We want to rene our NRsteps function such that it does not always make 5 iterations. Instead it should iterate as many times as needed to obtain that the changes in the estimates is smaller than e.g We would also like the function to print the number of iterations: Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

16 Rening mysqrt This can be done as follows: NRsteps <- function(x, z0) { z <- z0 itcount <- 1 repeat { z.new <- z - f(z, x)/deriv.f(z) if (abs(z.new - z) < 0.001) { break itcount <- itcount + 1 z <- z.new cat("nr iterations:", itcount, "\n") return(z.new) Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

17 Rening mysqrt Having to specify a starting value is annoying. An option could be to take x itself as a default starting value. This can be implemented as: mysqrt <- function(x, z0=x) { z <- NRsteps(x, z0) return(z) > mysqrt(10) NR iterations: 5 [1] Notice that, our method is not working for x < 0. Moreover, for x = 0 the function should return an informative error message. Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

18 Rening mysqrt This can be done as follows: mysqrt<-function(x, z0=x) { if (x < 0) { cat("can not take square root of a -ve number...\n") else { if (x == 0) { return(0) else { z <- NRsteps(x, z0) return(z) Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

19 Testing mysqrt function Now, calling our function mysqrt() to obtain > mysqrt(-7) Can not take square root of a negative number... > mysqrt(17) NR iterations: 6 [1] > mysqrt(17) [1] 0 Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

20 Try this one, it is as the R output This can be done as follows: mysqrt<-function(x, z0=x) { if (x < 0) { cat("[1] NaN\n","Warning message:\n","in sqrt(",x,"): NaNs produced\n") else { if (x == 0) { return(0) else { z <- NRsteps(x, z0) return(z) Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

21 Loops in R For Loop The most commonly used loop structures in R are for, while and apply loops. Less common are repeat loops. OR, similarly; for(i in 1:n){ statements for(variable in sequence) { statements Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

22 For loops: Examples Example: Print a statement for(i in 1:3){ print("hello R Students!") cat("i=",i,"\n") [1] "Hello R Students!" i= 1 [1] "Hello R Students!" i= 2 [1] "Hello R Students!" i= 3 Everything inside the curly brackets {... is done 3 times. Looped commands can depend on i (or anyother counter). R creates a vector i with 1:3 in it. For Loop is exible, but slow when looping over large number of elds (e.g. thousands of rows or columns) Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

23 For loops with break statement: Example As we can see from the output, the loop terminates when it encounters the break statement. for (i in 1:5) { for (j in 1:5) { for (k in 1:5) { cat(i," ",j," ",k,"\n") if (k ==3) break Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

24 For loops with next statement: Example A next statement is useful when we want to skip the current iteration of a loop without terminating it. On encountering next, R skips further evaluation and starts next iteration of the loop. x <- 1:5 for (i in x) { if (i == 3){ next print(i) [1] 1 [1] 2 [1] 4 [1] 5 Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

25 For loops: Examples Calculating factorials for an integer x: x = x(x 1)(x 2) fact=function(x){ fact=1 if (x<2) {fact=1 else { for (i in 2:x) fact=fact*i return (fact) > fact(5) [1] 120 Another function without loop My.factorial <- function(x) { if (x == 0) return (1) else return (x * My.factorial(x-1)) > My.factorial(5) [1] 120 Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

26 unique() function: Examples unique() function removes duplicated elements/rows from a vector, data frame or array. > x <- c(2:8,4:10) > unique(x) [1] > A=rbind(1:3,11:13,1:3,15:17,1:3) > unique(a) [,1] [,2] [,3] [1,] [2,] [3,] Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

27 For loops: Examples Example: Apply this example from R documents: Exercises f = sample(letters[1:5], 10, replace=true) for( i in unique(f) ) print(i) Try these for loops: > for(objs in names(cars)) print(objs) > for(trig in c(sin,cos)) print(trig(pi)) > for (i in seq(2, 10, 3)) print(i) Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

28 while loops This kind of looping structure is suitable when the number of times the computations contained within the loop is repeated is not known in advance, and the termination of the loop is dependent on some other criteria. The general form of the while loop is: while(cond) expr The simple or compound R expression, expr, is repeatedly executed until the logical expression cond evaluates to a FALSE value. So, when using while, we need to set up an indicator variable and change its value within each iteration. Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

29 while loops: Examples Consider the following simplest example explaining while loop > y = 1 > while(y < 7){ print(y) y = y+1 [1] 1 [1] 2 [1] 3 [1] 4 [1] 5 [1] 6 Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

30 while loops: Examples Let us use break statement to end or stop the loop when x=3: x <- 1 > while(x < 5) {x <- x+1; if (x == 3) break; print(x) [1] 2 Now, let us use next statement to skip one step when x=3: > x <- 1 > while(x < 5) {x <- x+1; if (x == 3) next; print(x) [1] 2 [1] 4 [1] 5 Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

31 while loops: Examples Example:Calculation of the factorial int=4 fact=1 ind=int while (ind > 1){ fact=fact*ind ind=ind-1 > fact [1] 24 In the above example, ind is the indicator variable which change its value within each iteration. Exercise: Generalize the above code to a function call it Myfact to compute the factorial of any number. Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

32 while loops Examples In practice, a while loop is preferred when the cond expression is used for other purposes than just counting. It can be used, for example, to determine if a terminating condition such as that the error in a computed answer has decreased to be smaller than a pre-specied level and therefore is acceptable. For example, the following loop accumulates a sum to compute exp(5) using the power series 1+x+ x 2 /2! +... expansion: > i=0; term=1; sum=1; x=5 > while(term>.0001){ i=i+1 term=x^i/factorial(i) sum=sum+term > sum [1] > exp(5) [1] Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

33 while loops Examples In the previous example, the loop is terminated when the value of the next term to be added to the series is less than or equal to :0001. Note that this loop will not work for negative vales of x. Of course, if we knew that we need 21 terms in the series to achieve this accuracy of the result, we could have used the expression: > 1+sum(5^(1:20)/factorial(1:20)) [1] Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

34 repeat loops The repeat loop is similar to the while loop except that the condition for termination is tested inside the loop. This allows for more than a single condition to be checked and for these conditions to occur at dierent places in the loop. The general form of the repeat loop is: repeat { expression if(condition) { break where the expression is usually an R compound expression. The expression is evaluated repeatedly so that at least one break statement must be in the loop. The loop will be exited only when a given condition is satised. and any number of these statements may appear at dierent places in the loop with dierent cond logical expressions. Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

35 repeat loops Examples > x <- 1 > repeat { print(x) x = x+1 if (x == 7){ break [1] 1 [1] 2 [1] 3 [1] 4 [1] 5 [1] 6 Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

36 repeat loops Examples For example the while loop in the previous example may be rewritten as follows: i=0 term=1 sum=1 x=5 repeat{ i=i+1 term=x^i/factorial(i) if(term<=.0001) break sum=sum+term > sum [1] Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

37 repeat loops Examples Try this R code: i=0 term=1 sum=1 x=5 repeat{ i=i+1 term=x^i/factorial(i) sum=sum+term if(term<=.0001)break cat("i = ",i, "Term = ", term, "Sum = ",sum,fill=t) cat (" Final Value=", sum, fill=t) Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

38 Evaluating Polynomials A polynomial f (x) of degree n is a function of the form f (x) = a 0 + a 1 x + a 2 x a n x n. This form is the standard method of representing polynomial functions for mathematical purposes. Actually, this form is called the power form. f (x) above can be re-written in the following representation, called Horner's rule. f (x) = a 0 + (a 1 + (a 2 + (a n 1 + a n x)... )x). Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38

39 where a is a vector containing the coecients a 1, a 2,..., a n Bisher M. Iqelan (IUG) Lecture 6: Programming Examples 1 st Semester / 38 Evaluating Polynomials Examples Consider the evaluation of the polynomial with power form: f (x) = 7 + 3x + 4x 2 9x 3 + 5x 4 + 2x 5. An expression for evaluating f(x) using Horner's rule is (3.0 + (4.0 + ( ( x) x) x) x) x which requires only 5 multiplications. (How many multiplications are there in the power form?) This representation can have the following R code: n=6; x=2; a=c(7,3,4,-9,5,2); sum= a[n] for (i in (n-1):1){ sum=sum*x+a[i] cat(" Series sum = ",sum,fill=t)

40 End of lecture 6. Thank you.!!!