LISP 105
November 9th, 2021 EL Notes Guest on 23rd November 2021 06:45:35 AM
  1. ;;add all items in list
  2. (defun sumlist (l)
  3.   ( cond
  4.     ( (null l) 0)
  5.     ( t (+ (car l) (sumlist (cdr l))))
  6.     )
  7.   )
  8. ;;demo sumlist
  9. (sumlist '(2 4 6))
  10. ;12
  11. (sumlist '(2 4 6 10 12 14 16))
  12. ;64
  13.  
  14. ;;multiply all items in list
  15. (defun multlist (l)
  16.   ( cond
  17.     ( (null l) 1)
  18.     ( t (* (car l) (multlist (cdr l))))
  19.     )
  20.   )
  21. ;;demo multlist
  22. (multlist '(2 4 6))
  23. ;48
  24. (multlist '(2 4 6 10 12 14 16))
  25. ;1290240
  26.  
  27. ;;filer argument and stop after first argument and return the tail
  28. (defun rember (x l)
  29.   ( cond
  30.     ( (null l) l)
  31.     ( (eq x (car l))(cdr l))
  32.     ( t ( cons (car l) (rember x (cdr l))))
  33.     )
  34.   )
  35. ;;demo rember 1
  36. (rember 2 '(2 3 4))
  37. ;(3 4)
  38. (rember 2 '(3 2 4 5))
  39. ;(3 4 5)
  40. (rember 2 '(3 2 4 2 5))
  41. ;(3 4 2 5)
  42.  
  43. ;;recursively filter arguement out of  all input
  44. (defun rember (x l)
  45.   ( cond
  46.     ( (null l) l)
  47.     ( (eq x (car l))(rember x (cdr l)))
  48.     ( t ( cons (car l) (rember x (cdr l))))
  49.     )
  50.   )
  51. ;;demo rember 2
  52. (rember 2 '(2 3 4))
  53. ;(3 4)
  54. (rember 2 '(3 2 4 5))
  55. ;(3 4 5)
  56. (rember 2 '(3 2 4 2 5))
  57. ;(3 4 5)
  58.  
  59.  
  60. ;;recursively take head of inner lists
  61. (defun firsts (l)
  62.   (cond
  63.    ( (null l)())
  64.    ( t (cons (car (car l))(firsts (cdr l))))
  65.    )
  66.   )
  67. ;;demo firsts
  68. (firsts '( (4 5) (6 3)(8 3 1)))
  69. ;(4 6 8)
  70.  
  71. ;;replace second param with first param
  72. (defun replace (new old l)
  73.   (cond
  74.    ( (null l) l)
  75.    ( (eq old (car l)) (cons new( cdr l)))
  76.    ( t ( (cons (car l) ( replace new old (cdr l)))))
  77.    )
  78.   )
  79. ;;demo replace
  80. (replace 3 6 '(6 5 7))
  81. ;(3 5 7)
  82.  
  83. ;;recursion replace all second param with first param
  84. (defun replaceall (new old l)
  85.   (cond
  86.    ( (null l) l)
  87.    ( (eq old (car l)) (cons new (replaceall new old (cdr l))))
  88.    ( t (cons (car l) (replaceall new old (cdr l))))
  89.   )
  90. )
  91. ;;demo replaceall
  92. (replaceall 3 6 '(6 5 7 6))
  93. ;(3 5 7 3)
  94.  
  95. ;; insert new old etc
  96. (defun insertR (new old l)
  97.   (cond
  98.    ( (null l) l)
  99.    ( (eq old (car l)) (cons (car l) (cons new (cdr l))))
  100.    ( t (cons (car l) (insertR new old (cdr l))))
  101.    )
  102.   )
  103. ;;demo insertR
  104. (insertR 3 6 '(6 7 8))
  105. ;(6 3 7 8)
  106. (insertR 3 6 '(1 4 6 7 6))
  107. ;(1 4 6 3 7 6)
  108.  
  109. ;;insert all occurences
  110. ;; insert new old etc
  111. (defun insertallR (new old l)
  112.   (cond
  113.    ( (null l) l)
  114.    ( (eq old (car l)) (cons (car l) (cons new (insertallR new old (cdr l)))))
  115.    ( t (cons (car l) (insertallR new old (cdr l))))
  116.    )
  117.   )
  118. ;;demo insertallR
  119. (insertallR 3 6 '(6 7 8))
  120. ;(6 3 7 8)
  121. (insertallR 3 6 '(1 4 6 7 6))
  122. ;(1 4 6 3 7 6 3)

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