I'm still learning Scheme.
If have these two list of lists:
'((1 2 (3 4) 5) (12 13 4))
'((3 4 9) (7 6 5 4))
I want to get this list:
'((1 2 (3 4) 5) (12 13 4) (3 4 9) (7 6 5 4))
But with cons:
(cons '((1 2 (3 4) 5) (12 13 4)) '((3 4 9) (7 6 5 4)))
I get this list:
'(((1 2 (3 4) 5) (12 13 4)) (3 4 9) (7 6 5 4))
NOTE:
In this example the both lists has two sublists. But they can have n sub-lists.
I have tried with append but it doesn't work when one of the list of lists is only a list:
(append '(1 2 3 4) '((23 24 25 26) (a b c)))
> '(1 2 3 4 (23 24 25 26) (a b c))
Is there a function that do it or do I have to implement it?
The behavior that you want is very specific, you won't find it in the standard library. Good news is, it's simple to implement in a portable and efficient way that covers all the possible cases, assuming that the input lists are non-empty:
(define (my-append lst1 lst2)
(cond ((and (pair? (car lst1)) (pair? (car lst2)))
(append lst1 lst2))
((pair? (car lst1))
(append lst1 (list lst2)))
((pair? (car lst2))
(append (list lst1) lst2))
(else
(append (list lst1) (list lst2)))))
For example:
(my-append '((1 2 (3 4) 5) (12 13 4)) '((3 4 9) (7 6 5 4)))
=> '((1 2 (3 4) 5) (12 13 4) (3 4 9) (7 6 5 4))
(my-append '(1 2 3 4) '((23 24 25 26) (a b c)))
=> '((1 2 3 4) (23 24 25 26) (a b c))
Related
I am trying to count how many times the first number of the list of list called one, two or three is repeated, the idea of the code that I made is for example to choose the list of list called "one" and then take the first letter that has "a" inside and then compare in this case the first value that would be "10" with the others in this list , in this case I should get it to say that the "10" was repeated once, then it would continue with the other two "b" and "c", but when doing the following code
(define a (list 10 2 3 54 6 9 7 10))
(define b (list 5 1 8 6 5 5 4 77 8 6))
(define c (list 80 80 80))
(define e (list 99 156 54 48 99))
(define d (list 16 94 75 30 56 16 8 16))
(define one (list a b c))
(define two (list c e b a))
(define three (list b c d e))
; receives 'one', 'two' or 'three' and finds how many times the first number of the list is repeated in the list of lists
(define (find-repeated-number list)
(define (find-repeated-number-helper list)
(if (null? list)
0
(if (equal? (car list) (car (cdr list)))
(+ 1 (find-repeated-number-helper (cdr list)))
; displays the number of times the first number of the list is repeated in the list of lists
(find-repeated-number-helper (cdr list)))))
(find-repeated-number-helper list))
(find-repeated-number one)
I get the following error, why is it? how can i get what i want to do? I would appreciate any help
*** ERROR: pair required, but got ()
While loading "./jdoodle.sc" at line 19
Stack Trace:
_______________________________________
0 (car (cdr list))
at "./jdoodle.sc":14
1 (equal? (car list) (car (cdr list)))
at "./jdoodle.sc":14
Command exited with non-zero status 70
I'm still not sure if the main point of this exercise is to write low-level recursive code or use some functions provided by the standard library (in this case, count or a combination of filter and length), but here's my wild guess:
(define a (list 10 2 3 54 6 9 7 10))
(define b (list 5 1 8 6 5 5 4 77 8 6))
(define c (list 80 80 80))
(define e (list 99 156 54 48 99))
(define d (list 16 94 75 30 56 16 8 16))
(define one (list a b c))
(define two (list c e b a))
(define three (list b c d e))
(define (find-count lst element)
(count (lambda (n) (= n element))
lst))
(define (print-count lst)
(let ((number-count (find-count (cdr lst) (car lst))))
(write (string-append
"number of times the first number is repeated: "
(number->string (car lst))
" is "
(number->string number-count)
(if (= number-count 1)
" time"
" times")))
(newline)))
(for-each print-count one)
Output:
"number of times the first number is repeated: 10 is 1 time"
"number of times the first number is repeated: 5 is 2 times"
"number of times the first number is repeated: 80 is 2 times"
I am trying to display a Fibonacci sequence less than an upper bound, but I'm having trouble printing out the series that is constrained by that upper bound.
Without using imperative programming principles with declaring variables such as setf, setq, and set, etc. How can I solve this problem?
So far I have
(defun fibonacci (n &optional (a 0) (b 1))
(if (or (zerop n) (< a n)
nil)
(cons a (fibonacci (1- n) b (+ a b)))))
Expected output of (fibonacci 100): (0 1 1 2 3 5 8 13 21 34 55 89). However, what I am getting is (0 1 1 2 3 5 8 13 21 34 55).
You are decreasing upper bound for no reason. When you remove 1- and move one ) to right place, it works as expected:
(defun fibonacci (n &optional (a 0) (b 1))
(if (or (zerop n) (> a n))
nil
(cons a (fibonacci n b (+ a b)))))
Tests:
CL-USER 4 > (fibonacci 10)
(0 1 1 2 3 5 8)
CL-USER 5 > (fibonacci 100)
(0 1 1 2 3 5 8 13 21 34 55 89)
What you meant is
(defun fibonacci (n &optional (a 0) (b 1))
(if (or (zerop n) (> a n))
nil
(cons a (fibonacci n b (+ a b)))))
You had a ) placement typo / error, and the flipped test. Also, n is the upper limit, not a count. So there's no reason to decrease it.
The task I'm given is, in Racket, to "write a function, countIncreases, which takes a list of numbers and returns how many times the consecutive numbers increase in value. For example, countIncreases '(1 3 2 4 5 1) should return 3 because there are three increases: 1 3, 2 4, 4 5."
I've written a recursive function with the base case being if the list is empty, return 0. I believe my problem is properly comparing one value to the next. Should I be using a standard library list iteration function like foldr or map to accomplish this?
My code and tests are below, and a screenshot of error messages is attached here
(define (countIncreases aList)
(if (empty? aList)
0
(if (< (first aList) (rest aList))
(+ 1 (countIncreases (rest aList)))
(countIncreases (rest aList)))))
(check-expect (countIncreases '(1 3 2 4 5 1)) 3)
(check-expect (countIncreases '()) 0)
(check-expect (countIncreases '(1 2 3 4 5)) 4)
(check-expect (countIncreases '(5 4 3 2 1 2)) 1)
For my homework problem, I have to remove the duplicates in a list if part of the list is inside another. The expected outcome is supposed to be this:
(remove-redundant '((R (1 2 3 8 e 4 7 6 5))
(U (e 2 3 1 8 4 7 6 5))
(D (1 2 3 7 8 4 e 6 5)))
'((D (1 2 3 e 8 4 7 6 5))
(L (e 2 3 1 8 4 7 6 5))
(U (2 e 3 1 8 4 7 6 5))
(U (2 8 3 1 e 4 7 6 5))))
Returns:
((R (1 2 3 8 e 4 7 6 5)) (D (1 2 3 7 8 4 e 6 5)))
It is supposed to check to see if the list inside each list, for the 1st parameter appears anywhere in the 2nd. If it does appear, then remove that from the 1st list. Basically, if (1 2 3 e 8 4 7 6 5) matches something in the 1st list, remove it from the 1st list. I need to do this recursively, it is supposed to be a functional program.
I have already tried recursing through the list, but it doesn't reset (i.e. it will check the second list for the beginning of the first list but then will return.
(defun same-state (l1 l2)
(if (equal (cadr l1) (cadr l2)) t nil))
(defun remove-redundant (l1 l2)
(cond
((null l2) l1)
((null l1) nil)
((same-state (car l1) (car l2)) (remove-redundant (cdr l1) (cdr l2))
(T (remove-redundant l1 (cdr l2)))))
You're close, but there should be some accumulation of the cells you want to keep in the t case of your cond form,
eg.
;; ...
(t
(cons (car l1) (remove-redundant (cdr l1) (cdr l2))))
and when you check if the first element of l1 exists in l2 there should be another loop or recursion to check against all the elements in l2, eg.
(defun same-state (l1 l2)
(find l1 l2 :key #'cadr :test #'equal))
(defun remove-redundant (l1 l2)
(cond
((null l2) l1)
((null l1) nil)
((same-state (cadar l1) l2)
(remove-redundant (cdr l1) (cdr l2)))
(t (cons (car l1)
(remove-redundant (cdr l1) (cdr l2))))))
;; => ((R (1 2 3 8 e 4 7 6 5)) (D (1 2 3 7 8 4 e 6 5)))
You will need two loops: one to go through the first list, and then one per element of that to find matches in the second list.
I'm stuck 5 days with this task in Racket, does anybody know how can I approach it?
Given a function of arity 2 and a list of n elements, return the evaluation of the string function of all the elements, for example:
>(reduce + '(1 2 3 4 5 6 7 8 9 10))
55
> (reduce zip '((1 2 3) (4 5 6) (7 8 9)))
'((1 (4 7)) (2 (5 8)) (3 (6 9)))
Here you go.
(define (reduce func list)
(assert (not (null? list)))
(if (null? (cdr list))
(car list)
(func (car list) (reduce func (cdr list)))))
Tests:
> (reduce + '(1 2 3 4 5 6 7 8 9 10))
55
> (reduce zip '((1 2 3) (4 5 6) (7 8 9)))
((1 (4 7)) (2 (5 8)) (3 (6 9)))
For completeness, an implementation for zip (one that assumes two lists and that your lists are all the same length) is:
(define (zip l1 l2) (map list l1 l2))
You can express it in terms of foldl:
(define (reduce f xs)
(and (not (empty? xs)) (foldl f (first xs) (rest xs))))
(define reduce
(λ (f init ls)
(if (empty? ls)
init
(reduce f (f init (first ls)) (rest ls)))))