Racket, writing function that find nth element in list - functional-programming

Question

In order to understand functional programing, please help me to write a function that output nth element of a list,
Allowed command:
define lambda cond else empty empty? first rest cons list
list? = equal? and or not + - * / < <= > >=
Sample output:
(fourth-element '(a b c d e)) => d
(fourth-element '(x (y z) w h j)) => h
(fourth-element '((a b) (c d) (e f) (g h) (i j))) => (list 'g 'h)
or ‘(g h)
(fourth-element '(a b c)) => empty
I could write this in python, but I am not family with racket syntax,
def element(lst, x=0):
counter = x;
if (counter >= 3):
return lst[0]
else:
return element(lst[1:],x+1)
a = [1,2,3,4,5,6]
print(element(a))
The Output is 4
Comparing with code above in python. What is equivalent behavior in function that create local variable counter. What is "keyword" for return

Answer 1

It looks like you came up with an answer of your own. Nice work! I would recommend a more generic nth procedure that takes a counter as an argument. This allows you to get any element in the input list
(define (nth lst counter)
(cond ((null? lst) (error 'nth "index out of bounds"))
((= counter 0) (first lst))
(else (nth (rest lst) (- counter 1)))))
Now if you want a procedure that only returns the 4th element, we create a new procedure which specializes the generic nth
(define (fourth-element lst)
(nth lst 3))
That's it. Now we test them out with your inputs
(define a `(1 2 3 (4 5) 7))
(define b `(1 2 3))
(define c `((a b)(c d)(e f)(g h)(i j)))
(define d `(a b c))
(fourth-element a) ; '(4 5)
(fourth-element b) ; nth: index out of bounds
(fourth-element c) ; '(g h)
(fourth-element d) ; nth: index out of bounds
Note, when the counter goes out of bounds, I chose to raise an error instead of returning a value ("empty") like your program does. Returning a value makes it impossible to know whether you actually found a value in the list, or if the default was returned. In the example below, notice how your procedure cannot differentiate the two inputs
(define d `(a b c))
(define e `(a b c ,"empty"))
; your implementation
(fourth-element e) ; "empty"
(fourth-element d) ; "empty"
; my implementation
(fourth-element e) ; "empty"
(fourth-element d) ; error: nth: index out of bounds
If you don't want to throw an error, there's another way we can encode nth. Instead of returning nth element, we can return the nth pair whose head contains the element in question.
Below, nth always returns a list. If the list is empty, no element was found. Otherwise, the nth element is the first element in the result.
(define (nth lst counter)
(cond ((null? lst) '())
((= counter 0) lst)
(else (nth (rest lst) (- counter 1)))))
(define (fourth-element lst)
(nth lst 3))
(define a `(1 2 3 (4 5) 7))
(define b `(1 2 3))
(define c `((a b)(c d)(e f)(g h)(i j)))
(define d `(a b c))
(define e `(a b c ,"empty"))
(fourth-element a) ; '((4 5) 7)
(fourth-element b) ; '()
(fourth-element c) ; '((g h) (i j))
(fourth-element d) ; '()
(fourth-element e) ; '("empty")
Hopefully this gets you to start thinking about domain (procedure input type) and codomain (procedure output type).
In general, you want to design procedures that have natural descriptions like:
" nth takes a list and a number and always returns a list" (best)
" nth takes a list and a number and returns an element of the list or raises an exception if the element is not found" (good, but now you must handle errors)
Avoid procedures like
" nth takes a list and a number and returns an element of the list or a string literal "empty" if the element is not found" (unclear codomain)
By thinking about your procedure's domain and codomain, you have awareness of how your function will work as it's inserted in various parts of your program. Using many procedures with poorly-defined domains lead to disastrous spaghetti code. Conversely, well-defined procedures can be assembled like building blocks with little (or no) glue code necessary.

Answer 2

Here is how to write it in Python:
def nth(lst, idx=0):
if (len(lst) == 0):
return "empty"
elif (idx == 0):
return lst[0]
else:
return nth(lst[1:], idx - 1)
nth([1,2,3], 1)
# ==> 2
def fourth-element(lst):
return nth(lst, 4)
Same in Scheme/Racket:
(define (nth lst idx)
(cond ((empty? lst) empty) ; more effiecent than (= (length lst) 0)
((= idx 0) (first lst))
(else (nth (rest lst) (- idx 1))))
(nth '(1 2 3) 1)
; ==> 2
(define (fourth-element lst)
(nth lst 4))
There is no keyword for return. Every form returns the last evaluated code:
(if (< 4 x)
(bar x)
(begin
(display "print this")
(foo x)))
This if returns either the result of (bar x) or it prints "print this" then returns the result of (foo x). The reason is that for the two outcomes of the if they are the tail expressions.
(define (test x)
(+ x 5)
(- x 3))
This function has two expressions. The first is dead code since it has no side effect and since it's not a tail expression, but the (- x 3) is what this function returns.
(define (test x y)
(define xs (square x))
(define ys (square y))
(sqrt (+ xs ys)))
This has 3 expressions. The first two has side effects that it binds two local variables while the third uses this to compute the returned value.

Answer 3

(define a `(1 2 3 (4 5) 7))
(define b `(1 2 3))
(define c `((a b)(c d)(e f)(g h)(i j)))
(define d `(a b c))
(define (my-lst-ref lst counter)
(cond[(>= counter 3) (first lst)]
[else (my-lst-ref (rest lst)(+ counter 1))]
)
)
(define (fourth-element lst)
(cond[(>= (list-length lst) 4) (my-lst-ref lst 0)]
[else "empty"]))
(fourth-element a)
(fourth-element c)
(fourth-element d)
Output:
(list 4 5)
(list 'g 'h)
"empty"