2

I'm currently trying to learn lisp and am using emacs on linux. To the best of my ability, I have written two functions.

Both functions first remove the first element of the list.

  • series adds all the elements in the given list.
  • parallel 1) takes the inverse of each number in the list, then 2) adds all the elements in the list, then 3) takes the inverse of the sum of the elements.

Code:

(defun series (lst)
    (apply #'+' (cdr lst)) )

(defun parallel (lst)
    (/ 1 (apply #'+' (apply #'/' (cdr 'lst ) ) ) ))

I can evaluate the function, but when I try to use the function, as below:

(series (list 3 3 4 5))

I get the error : value CDR is not of the expected type NUMBER. I see this, and I think, why is emacs treating cdr as a number rather than a function? I'm new to lisp and emacs, so I don't know to fix this error. Any help would be appreciated.

UPDATE

I've the problems in this code and I think it works...

(defun series (lst)
 (apply #'+ (cdr lst) ))

(defun parallel(lst)
(/ 1 (apply #'+ (mapcar #'/ (make-list (- (length lst) 1) :initial-element 1) (cdr lst)  ) )))

Hopefully, what I was trying to do is understood now.

Improve this question
4
  • 3
    The ' notation in Lisp is very different from quoted character strings. There is no closing quote. '<anything> is a shorthand which stands for the list structure (quote <anything>). This is useful because when (quote <anything>) is evaluated, the value is always object <anything> itself, whatever it is. The value of (quote (+ 2 2)) is the list (+ 2 2) rather than 4. Lisp has a double quote for string literals, like "abc", and character constants have a '#\' (hash backslash) syntax which has some variety in it such as '#\space' denoting space and '#\A' denoting upper case A. Commented May 4, 2012 at 1:52
  • @Kaz this should definitely go into your answer, not stay hidden in comments. :) Commented May 5, 2012 at 15:11
  • in Common LISP mapcar stops when shorter list ends. This means, circular list can be used to supply the default argument: (nth 5 (setq b '(1.0) b (rplacd b b))) ==> 1.0. Don't know about emacs-Lisp though. btw do you get same answer when using 1 and 1.0? Commented May 5, 2012 at 20:34
  • Hmmm....... I'm not familiar with circular lists, but the concept seems more intuitive. In this case, both lists next to mapcar are guaranteed to be the same length. I also saw '#1='(1 2 3 . #1#) as an example of a circular list. Does this work? Commented May 5, 2012 at 22:29

2 Answers 2

Reset to default
3

You have extra apostrophes that are confusing the LISP parser. The syntax to refer to the + function is just #'+; there's no "close quote".

Improve this answer
Sign up to request clarification or add additional context in comments.

3 Comments

Also, no quote on lst in (cdr 'lst).
Ahhh... I see now, I wrote it wrong. Removing the extra quotes (as in Kaz's comment & Reed's answer) fixed the problems in this code.
Yes, and note there can be whitespace between ' and whatever follows, so that 'A' B actually means the same thing as 'A 'B.
1

Re:

(defun parallel (lst)
    (/ 1 (apply #'+' (apply #'/' (cdr 'lst ) ) ) ))

Why are you trying to discard the first element. That is curious.

Note that the logic of this function won't work even if the quoting issues are solved.

(apply #'/ (cdr lst)) will pass the remainder of lst to the / function. This function will produce a number.

And so you then effectively (apply #'+ number), problem being that apply wants a list.

The parallel semantics you are hinting at will not happen.

Perhaps you want this:

 ;; divide the sum of the remaining elements, by the fraction of those elements
 (/ (apply #'+ (cdr lst)) (apply #'/ (cdr lst)))

Note that in Emacs Lisp, division of integers by integers yields an integer, and not a rational like in Common Lisp.

Improve this answer

5 Comments

the OP wants to "take[...] the inverse of each number in the list". That's (mapcar #'/ (cdr lst)). The names series and parallel probably refer to calculations of ohmic resistance in electrical circuits. :)
@WillNess that's exactly what I meant to do :)
@Kaz I've edited my answer to include the updated code, so what I was trying to accomplish can be understood.
@WillNess I got a mortarboard badge over in Electronics yesterday and it didn't occur to me. Say, I don't have an installation of Emacs handy right now, but looking at the GNU Emacs manual online, it appears that one-argument form of the / function is not documented! In Common Lisp (/ x) does means (/ 1 x), which is analogous to unary minus (- x) -> (- 0 x). Is this something that Just Works? If not, you have to do (mapcar (lambda (num) (/ 1 num)) list).
turns out my old emacs installation is gone too. wasn't a big fan of getting a pain in my left wrist every time I tried using it, after a day or two. ;)

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.