[scheme][Gauche]append2, reverse, find

「プログラミングGauche」P.54 ～ 55

 

• リストとリストを連結するappend2

;; append list (define (append2 a b)   (if (pair? a)       (cons (car a)(append2 (cdr a) b))       b))   (append2 '(1 2 3) '(4 5 6)) ;; => (1 2 3 4 5 6) (append2 '(1 (2 3) (4 5 (6))) '(((7 8) 9) 10)) ;; => (1 (2 3) (4 5 (6)) ((7 8) 9) 10)

 

 

• リストを反転させるreverse

;; reverse list (define (reverse ls)   (fold cons '() ls)) (reverse '(1 2 3 4 5)) ;; =>  (5 4 3 2 1)   ;; reverse list 2 (define (reverse2 ls)   (if (null? (cdr ls))       ls       (append2 (reverse2 (cdr ls))(list (car ls)))))   (reverse2 '(1 2 3 4 5)) ;; => (5 4 3 2 1)       (define (reverse2 ls)   (if (null? (cdr ls))       ls       (append2 (reverse2 (cdr ls)) (cons (car ls) '()))))     (reverse2 '(1 2 3 4 5)) ;; => (5 4 3 2 1)

 

 

• リストから最初に条件に合致した要素を探し出すfind

;; find (define (find predicate ls)   (if (null? ls)       #f       (if (predicate (car ls))       (car ls)       (find predicate (cdr ls)))))     (find odd? '(1 2 3 4 5)) ;; => 1 (find odd? '(1)) ;; => 1 (find odd? '()) ;; => #f (find odd? '(2 4 6 8 10)) ;; => #f (find odd? (cons 1 2)) ;; => 1     ;; find - cond (define (find predicate ls)   (cond    ((null? ls) #f)    ((predicate (car ls)) (car ls))    (else (find predicate (cdr ls)))))     (find even? '(1 2 3 4 5)) ;; => 2 (find even? '(1)) ;; => #f (find even? '(1 3 5 7)) ;; => #f

 

 

• リストから条件に合致したすべての要素をリストにして返すfind

(define (find predicate ls)   (cond    ((null? ls) '())    ((predicate (car ls)) (cons (car ls) (find predicate (cdr ls))))    (else (find predicate (cdr ls)))))     (find odd? '(1 2 3 4 5 6 7 8 9 10)) ;; => (1 3 5 7 9) (find even? '(1 2 3 4 5 6 7 8 9 10)) ;; => (2 4 6 8 10)