Suppose I have an arbitrary list representing Boolean expressions:
'(a and (b or c)) and another list of pairs representing the values of the literals
'((a . #t) (b . #f) (c . #t)))
What I want to do is replace the values in the first list using the values from the second list, and return a new list:
'(#t and (#f or #t))
So for example this code takes a list and an association list, and successfully replaces the values inside of it, though I'm trying to do the same thing given a list of conses instead of an association list. It works for '(a b) '((a 1) (b 2)) -> '(1 2), but I want to instead use '((a . 1) (b . 2)). Here is that code:
(define replace
(lambda (ls a-list)
(map (lambda (x)
(let ((lookup (assq x a-list)))
(if lookup
(cadr lookup)
x)))
ls)))
Would anyone know how to do this? Thanks so much!
Something like this:
(define bool '(a and (b or c)))
(define pairs '((a . #t) (b . #f) (c . #t)))
(define (value-for-key lst key)
(cond ((empty? lst) 'none)
((eq? key (caar lst))
(cdar lst))
(#t (value-for-key (cdr lst) key))))
(define (replace-bool lst vals)
(cond ((empty? lst) lst)
((pair? (car lst)) (cons (replace-bool (car lst) vals)
(replace-bool (cdr lst) vals)))
(#t (let ((val-for-key (value-for-key vals (car lst))))
(if (not (eq? val-for-key 'none))
(cons val-for-key (replace-bool (cdr lst) vals))
(cons (car lst) (replace-bool (cdr lst) vals)))))))
(replace-bool bool pairs)
-> (#t and (#f or #t))
First function is for searching in list of pairs key-value and returns value for given key or 'none, if that key isn't in that list.
Second function goes through Boolean expression list. Each atomic element is searched in key-value list and if it's found, is replaced. If Boolean expression contains nested expressions, they are traversed recursively.