LISP 69
November 9th, 2021 EL Notes By admin on 10th November 2021 02:07:08 AM
  1. ;;November 9th, 2021 EL Notes
  2.  
  3. ;;add all items in list
  4. (defun sumlist (l)
  5.   ( cond
  6.     ( (null l) 0)
  7.     ( t (+ (car l) (sumlist (cdr l))))
  8.     )
  9.   )
  10. ;;demo sumlist
  11. (sumlist '(2 4 6))
  12. ;12
  13. (sumlist '(2 4 6 10 12 14 16))
  14. ;64
  15.  
  16. ;;multiply all items in list
  17. (defun multlist (l)
  18.   ( cond
  19.     ( (null l) 1)
  20.     ( t (* (car l) (multlist (cdr l))))
  21.     )
  22.   )
  23. ;;demo multlist
  24. (multlist '(2 4 6))
  25. ;48
  26. (multlist '(2 4 6 10 12 14 16))
  27. ;1290240
  28.  
  29. ;;filer argument and stop after first argument and return the tail
  30. (defun rember (x l)
  31.   ( cond
  32.     ( (null l) l)
  33.     ( (eq x (car l))(cdr l))
  34.     ( t ( cons (car l) (rember x (cdr l))))
  35.     )
  36.   )
  37. ;;demo rember 1
  38. (rember 2 '(2 3 4))
  39. ;(3 4)
  40. (rember 2 '(3 2 4 5))
  41. ;(3 4 5)
  42. (rember 2 '(3 2 4 2 5))
  43. ;(3 4 2 5)
  44.  
  45. ;;recursively filter arguement out of  all input
  46. (defun rember (x l)
  47.   ( cond
  48.     ( (null l) l)
  49.     ( (eq x (car l))(rember x (cdr l)))
  50.     ( t ( cons (car l) (rember x (cdr l))))
  51.     )
  52.   )
  53. ;;demo rember 2
  54. (rember 2 '(2 3 4))
  55. ;(3 4)
  56. (rember 2 '(3 2 4 5))
  57. ;(3 4 5)
  58. (rember 2 '(3 2 4 2 5))
  59. ;(3 4 5)
  60.  
  61.  
  62. ;;recursively take head of inner lists
  63. (defun firsts (l)
  64.   (cond
  65.    ( (null l)())
  66.    ( t (cons (car (car l))(firsts (cdr l))))
  67.    )
  68.   )
  69. ;;demo firsts
  70. (firsts '( (4 5) (6 3)(8 3 1)))
  71. ;(4 6 8)
  72.  
  73. ;;replace second param with first param
  74. (defun replace (new old l)
  75.   (cond
  76.    ( (null l) l)
  77.    ( (eq old (car l)) (cons new( cdr l)))
  78.    ( t ( (cons (car l) ( replace new old (cdr l)))))
  79.    )
  80.   )
  81. ;;demo replace
  82. (replace 3 6 '(6 5 7))
  83. ;(3 5 7)
  84.  
  85. ;;recursion replace all second param with first param
  86. (defun replaceall (new old l)
  87.   (cond
  88.    ( (null l) l)
  89.    ( (eq old (car l)) (cons new (replaceall new old (cdr l))))
  90.    ( t (cons (car l) (replaceall new old (cdr l))))
  91.   )
  92. )
  93. ;;demo replaceall
  94. (replaceall 3 6 '(6 5 7 6))
  95. ;(3 5 7 3)
  96.  
  97. ;; insert new old etc
  98. (defun insertR (new old l)
  99.   (cond
  100.    ( (null l) l)
  101.    ( (eq old (car l)) (cons (car l) (cons new (cdr l))))
  102.    ( t (cons (car l) (insertR new old (cdr l))))
  103.    )
  104.   )
  105. ;;demo insertR
  106. (insertR 3 6 '(6 7 8))
  107. ;(6 3 7 8)
  108. (insertR 3 6 '(1 4 6 7 6))
  109. ;(1 4 6 3 7 6)
  110.  
  111. ;;insert all occurences
  112. ;; insert new old etc
  113. (defun insertallR (new old l)
  114.   (cond
  115.    ( (null l) l)
  116.    ( (eq old (car l)) (cons (car l) (cons new (insertallR new old (cdr l)))))
  117.    ( t (cons (car l) (insertallR new old (cdr l))))
  118.    )
  119.   )
  120. ;;demo insertallR
  121. (insertallR 3 6 '(6 7 8))
  122. ;(6 3 7 8)
  123. (insertallR 3 6 '(1 4 6 7 6))
  124. ;(1 4 6 3 7 6 3)

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