Occurrence of symbol A found anywhere in L. LISP

Question

Here is my function

(defun freq (symbol_A List_L)
    (cond ((atom (car List_L)) 
        (cond ((eq (car List_L) symbol_A) t (+ 1 (freq symbol_A (cdr List_L))))
            (t 0))) 
    (T (freq symbol_A (cdr List_L))))
)

I am getting an error variable ATOM has no value. Here is what I am testing with

(freq  'c '((a c) c e)) --> 2
(freq  'f '(((s) o ) d)) --> 0
(freq  'f '(((f) f) f f)) --> 4

Can not understand where is my error.

I also tried this:

(defun freq (a L)
  (cond
   ((null L) 0)
   ((equal a (car L)) (+ 1 (freq a (cdr L))))
   (t (freq a (cdr L)))))

you really need to indent and format your code. Writing/reading unformatted Lisp code is really a pain. — Rainer Joswig, Oct 28 '17 at 07:29

Rainer Joswig · Answer 1 · 2017-10-28T06:28:52.400

1

There is a syntax error in your code.

Indent and format your code

First you need to format your code:

(defun freq (A L)
  (cond (atom (car L))             ; first clause
        (eq A (car (car L )))      ; second clause
        (T (+ 1 (freq A (cdr L)))) ; third clause
        (T (freq A (cdr L)))))     ; fourth clause

A useful COND:

(cond ((> i 0) :positive)
      ((< i 0) :negative)
      (t       :equal))

You have written something like:

(cond (> i 0) :positive
      (< i 0) :negative
      (t      :equal))

Here a Lisp interpreter would complain that > has no value. It's because it is used as a variable. (> i 0) is seen as the first clause and the test form is the variable >. Since > usually is not a variable, this is an error.

A compiler might even reject the whole form at compile time, since there are clauses which are not lists.

As you can see the parentheses around a clause are missing in your code. Check the syntax of cond...

cond {clause}* => result*

A cond has zero or more clauses.

clause::= (test-form form*)

Each clause is a list (!!) starting with a test-form and followed by zero or more forms.

It makes also no sense to have two clause with T as the test form. The first one will always be take and the second one will always be ignored.

edited Oct 28 '17 at 06:28

answered Oct 28 '17 at 06:07

Rainer Joswig

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@MaksimKniazev:; about what? That ATOM is used as a variable, because you have a syntax error in your code? – Rainer Joswig Oct 28 '17 at 06:23
yes I got the logic but I am lost on this parenthesis. Have you tried to fix it and run to see if for sure that's the problem? – Yaroslav Dukal Oct 28 '17 at 06:48
@Maksim: I don't need to try that. It's obvious that your usage of COND is missing parentheses in the clauses and has a double T clause. You need to fix that. – Rainer Joswig Oct 28 '17 at 06:51
I edited my answer, I am no longer getting an error, but am getting wrong output... – Yaroslav Dukal Oct 28 '17 at 07:16
If I replace 1 for 2 then it will be correct, but I am not sure if that's the right way to do it... – Yaroslav Dukal Oct 28 '17 at 07:24

uselpa · Accepted Answer · 2017-10-28T12:28:27.537

1

Knowing that nil is an atom, you can simply recurse over the car and cdr of each cons cell. When you hit an atom, add 1 if it matches or 0 if it doesn't.

(defun freq (sym tree)
  (if (atom tree)
      (if (eq sym tree) 1 0)
      (+ (freq sym (car tree)) (freq sym (cdr tree)))))

edited Oct 28 '17 at 12:28

answered Oct 28 '17 at 10:33

uselpa

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Perfect, thanks! It would be better to know how to do it using primitive functions (without elt) – Yaroslav Dukal Oct 28 '17 at 11:13
`elt` is just a variable here, you can call it however you want. It's not the standard function. – uselpa Oct 28 '17 at 12:26
I've renamed it to `sym` to avoid confusion. – uselpa Oct 28 '17 at 12:28
Got it! Thanks again – Yaroslav Dukal Oct 28 '17 at 20:21

Occurrence of symbol A found anywhere in L. LISP

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