K Isn't Lisp - Hackdot

Re:K Isn't Lisp by tmtvl on Tuesday February 13, 2024 @12:56PM (#39360557)

Doing the naive thing in CL without thinking about things like 'efficiency' or 'maintainability':

  (let ((test-input '((1 "one") (1 "two") (1 "three") (1 "one") (1 "four") (1 "two"))))
    (mapcar (lambda (string)
              (list (count string test-input :test #'string= :key #'second)
                    string))
      (remove-duplicates (mapcar #'second test-input)
                         :test #'string= :from-end t)))

I will admit it's not as short as:

  v:((1;"one");(1;"two");(1;"three");(1;"one");(1;"four");(1;"two"))
  +(#:'=v[;1];?v[;1])

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Re:K Isn't Lisp by taeric on Tuesday February 13, 2024 @01:31PM (#39360982)

My naive looked more like: (with apologies on likely messing up spacing in hn.)

    (let ((in '((1 "one") (1 "two") (1 "three") (1 "one") (1 "four") (1 "two"))))
           (loop with hash = (make-hash-table :test #'equal)
                 for v in in
                 do (incf (gethash (cadr v) hash 0))
                 finally (return (loop for key being the hash-keys of hash
                                       collect (list (gethash key hash) key)))))

[ Parent | on HN ]

Re:K Isn't Lisp by SeanLuke on Tuesday February 13, 2024 @01:25PM (#39360907)

Now you have me thinking about how short I can make that. My revision:

    (defun doit (elts)
        (mapcar (lambda (elt) `(,(count elt elts :test 'equal :key 'second) ,elt))
        (delete-duplicates (mapcar 'second elts) :test 'equal)))))

or maybe

    (defun doit (elts)
        (loop for elt in (delete-duplicates (mapcar 'second elts) :test 'equal)
        collect `(,(count elt elts :test 'equal :key 'second) ,elt)))

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Re:K Isn't Lisp by tmtvl on Tuesday February 13, 2024 @01:40PM (#39361115)

As much as Lisp isn't meant for code golf, I think the loop variation could be used as a (rather long) one-liner.

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