for example, what should I input is here finfirst #'oddp '(1 2 3), and it should find the first odd number return to the list, so what I think I need to do is to write a function just have one argument which is list, but I only know to find the first element in the list, so how can I use the condition in my code
(defun finfirst(list)(cond((null list) nil)
if I finish this, then it will told me that I need two argument， I just don't know what should I do for this function, just give me some hint for that
If you just want the functionality you described, you can use the function find-if, eg:
(find-if #'oddp '(1 2 3))
If you want to implement it yourself, you could do something like this:
(defun finfirst (function list)
(cond
((null list) nil)
((funcall function (first list)) (first list))
(t (finfirst function (rest list)))))
Then use it like this:
(finfirst #'oddp '(1 2 3))
Related
I'm trying to write a macro that recursively 'unwraps' the values of a list into individual values. For example, unwrap '(1 2 3)) would produce (separately):
(car '(1 2 3)) ; 1
(car (cdr '(1 2 3)) ; 2
(car (cdr (cdr '(1 2 3)) ; 3
This would be used like (+ (unwrap '(1 2 3))) ==> (+ 1 2 3). (I know you can use the standard libary apply for this, but I'm trying to write my own version).
My problem is that in writing the macro, I don't know how to recursively process the list argument passed into the unwrap macro. Here's what I've tried:
(define-syntax (unwrap stx)
(syntax-case stx ()
[(_ lst)
#`(begin (car lst) ; error at the last step since (car '()) is invalid
(unwrap (cdr lst)))]))
or
(define-syntax (unwrap stx)
(syntax-case stx ()
[(_ lst)
#`(if (null? lst)
'()
(car (unwrap (cdr lst))))])) ; infinite loops
These are both obviously wrong, but essentially I don't know write a macro that check the value of the input list (or matches against an empty list pattern) and returns something different in either scenario. It's also possible I'm approaching this entirely wrong, so any help would be greatly appreciated. Thank you!
You want
(+ (unwrap '(1 2 3))) ==> (+ 1 2 3)
but this is impossible. You are asking for unwrap to macroexpand to three disconnected forms, which is not allowed. Any macro must expand into exactly one form. So as you say in your question, for the example you gave, apply is the right answer.
For other possible usages, the answer is to back up a step and ask why you think this unwrap function would be useful: what cases do you hope to use it in? Then design a different solution, that works within the constraints of the language.
I am trying to evaluate each atom of a list and see if it's equal to the number provided and remove if its not but I am running into a slight problem.
I wrote the following code:
(defun equal1(V L)
(cond((= (length L) 0))
(T (cond( (not(= V (car(equal1 V (cdr L))))) (cdr L) )))
)
)
(equal1 5 '(1 2 3 4 5))
I obtain the following error
Error: Cannot take CAR of T.
If I add (write "hello") for the action if true, the following error is obtained:
Error: Cannot take CAR of "hello".
I'm still quite new to LISP and was wondering what exactly is going on and how could I fix this so I could evaluate each atom properly and remove it if its not, thus the cdr L for the action.
car and cdr are accessors of objects of type cons. Since t and "hello" are not cons you get an error message.
To fix it you need to know what types your function returns and not car unless you know that it's a cons
EDIT
First off ident and clean up the code.. The nested cond are uneccesary since cond is a if-elseif-else structure by default:
(defun remove-number (number list)
(cond ((= (length list) 0)
t)
((not (= number (car (remove-number number (cdr list)))))
(cdr list))))
(t
nil)))
I want you to notice I've added the default behaviour of returning t when a consequent is not given as we know = returns either t or nil so it returns t when the length is 0 in this case.
I've added the default case where none of the two previous predicates were truthy and it defaults to returning nil.
I've named it according to the functions used. = can only be used for numeric arguments and thus this will never work on symbols, strings, etc. You need to use equal if you were after values that look the same.
Looking at this now we can see that the functions return value is not very easy to reason about. We know that t, nil and list or any part of the tail of list are possible and thus doing car might not work or in the case of (car nil) it may not produce a number.
A better approach to doing this would be:
check if the list is empty, then return nil
check if the first element has the same numeric value as number, then recurse with rest of the list (skipping the element)
default case should make cons a list with the first element and the result fo the recursion with the rest of the list.
The code would look something like this:
(defun remove-number (number list)
(cond ((endp list) '())
((= (car list) number) (remove-number ...))
(t (cons ...))))
There are a couple of things you could do to improve this function.
Firstly, let's indent it properly
(defun equal1 (V L)
(cond
((= (length L) 0))
(T (cond
((not (= V (car (equal1 V (cdr L))))) (cdr L))))))
Rather than saying (= (length l) 0), you can use (zerop (length l)). A minor sylistic point. Worse is that branch returns no value. If the list L is empty what should we return?
The issue with the function is in the T branch of the first cond.
What we want to do is
remove any list item that is the same value as V
keep any item that is not = to V
The function should return a list.
The expression
(cond
((not (= V (car (equal1 V (cdr L))))) (cdr L)))
is trying (I think) to deal with both conditions 1 and 2. However it's clearly not working.
We have to recall that items are in a list and the result of the equal function needs to be a list. In the expression above the result of the function will be a boolean and hence the result of the function call will be boolean.
The function needs to step along each element of the list and when it sees a matching value, skip it, otherwise use the cons function to build the filtered output list.
Here is a skeleton to help you out. Notice we don't need the embedded cond and just have 3 conditions to deal with - list empty, filter a value out, or continue to build the list.
(defun equal-2 (v l)
(cond
((zerop (length L)) nil)
((= v (car l)) <something goes here>) ;skip or filter the value
(t (cons (car l) <something goes here>)))) ;build the output list
Of course, this being Common Lisp, there is a built-in function that does this. You can look into remove-if...
I want to make a function that checks if an element is a member of a list. The list can contain other lists.
This is what I came with so far:
(defun subl(l)
(if (numberp l)
(if (= l 10)
(princ "Found"))
(mapcar 'subl l)))
Now the number I am searching for is hard-coded and it is 10. I would like to write it somehow so the function takes another parameter(the number I am searching for) and returns true or 1 when it finds it. The main problem is that I can't see a way to control mapcar. mapcar executes subl on each element of l, if l si a list. But how can I controll the returned values of each call?
I would like to check the return value of each subl call and if one of it is true or 1 to return true or 1 till the last recursive call. So in the end subl returns true or one if the element is contained in the list or nil otherwise.
Any idea?
This procedure below should process as you have described;
(defun member-nested (el l)"whether el is a member of l, el can be atom or cons,
l can be list of atoms or not"
(cond
((null l) nil)
((equal el (car l)) t)
((consp (car l)) (or (member-nested el (car l))
(member-nested el (cdr l))))
(t (member-nested el (cdr l)))))
mapcar is a very generic primitive to map a function over a list. You can use one of the built-in combinators which are much more closely suited with what you're trying to do. Look into the member function.
Your function seems to play the role of main function and helper at the same time. That makes your code a lot more difficult to understand than it has to be..
So imagine you split the two:
;; a predicate to check if an element is 10
(defun number10p (l)
(and (numberp l)
(= l 10)))
;; the utility function to search for 10 amongst elements
(defun sublistp (haystack)
(mapcar #'number10p haystack)))
But here when you do (sublistp '(5 10 15 20)) you'll get (nil t nil nil) back. Thats because mapcar makes a list of every result. For me it seems you are describing some since it stops at the first true value.
(defun sublistp (haystack)
(some #'number10p haystack)))
(sublistp '(5 10 15 20)) ; ==> t
Now to make it work for any data type we change the predicate and make it as a local function where we have the argument we are searching for:
(defun sublistp (needle haystack)
(flet ((needlep (x)
(equal x needle)))
(some #'needlep haystack)))
(sublistp '(a b) '(a b c (a b) d e f)) ; ==> t
You can also do this with an anonymous predicate like this:
(defun sublistp (needle haystack)
(some #'(lambda (x)
(equal x needle))
haystack))
An implementation of this is the member function, except it returns the match as truth value. That's ok since anything but nil is true in CL:
(member 10 '(5 10 15 20)) ; ==> (10 15 20)
EDIT
You commented on a different answer that you are required to use mapcar in that case use it together with append to get a list of all matches and check if the list has greater than 0 elements:
(defun sublistp (needle haystack)
(flet ((needle-check (x)
(if (equal x needle) '(t) nil)))
(< 0 (length
(apply #'append
(mapcar #'needle-check haystack))))))
How it works is that for each match you get a list of one element and for every non match you get an empty list. When appending the lists you'll get the empty list when there is not match. For all other results you have a match. This is not a very efficient implementation.
I need a function that will take in a list of characters and numbers, and then return the numbers added up (ignoring the characters). This is what I have so far:
(define (adder lst)
(cond
((null? lst)
0)
((number? (car lst))
(+(adder (car lst)) (adder (cdr lst))))
((char? (car lst))
((adder(cdr lst))))
))
(display (adder '(asd12sdf)))
Running it on codepad.org just displays void. I know the code is wrong because it looks wrong, but I have no idea how to fix it... How do I have the function keep track of the first number it finds and add it to the next one it finds, while skipping all characters?
In your second cond case, there's no reason to run adder on (car lst). Just adding (car list) itself to the recursive step should work.
For the last line, don't test (char? (car lst)). Just make the last line the else clause, meaning that anything BUT a number will go to the else line.
The reason you're getting void is because your input doesn't satisfy any of the cond conditions, and you have no else, so the answer is nothing (i.e. (void)).
The last mistake is in the input you're giving it. '(asd12sdf) is literally a list with one symbol named "asd12sdf". I think you want to give it '(a s d 1 2 s d f) (a list of 6 symbols and 2 numbers) which should result in 3. Notice that there's a very important difference between the symbol 'a and the character #\a.
It looks like you have the logic down, so your problem doesn't seem to be functional languages, just Scheme's syntax.
Edit: and in the last line, you have ((adder(cdr lst))) which has one too many parens wrapped around it. That will cause Scheme to attempt to evaluate the result of adder (which is a number) as a procedure (error!).
You should observe that this function is more or less sum which can be defined simply by using fold.
(define (adder lst)
(fold + 0 lst))
What does fold do? Basically, it's defined like so:
(define (fold f initial lst)
(if (null? lst)
initial
(fold f (f (car lst) initial) (cdr lst))))
(In other words, it calls f, a function of 2 arguments, on each element of lst, using the car of the lst as the first argument, and the accumulated result as the second argument to f.)
The issue here which you need to address is that + doesn't know how to operate on non-numeric values. No problem, you've already dealt with that. What happens if it's a character instead? Well, you're not adding anything to the total value, so replace it with a 0. Therefore, your solution is as simple as:
(define (adder lst)
(fold your-new-protected-+ 0 lst))
In Common Lisp:
(reduce #'+ '(1 #\a #\b 2 1 2 #\c #\d 4)
:key (lambda (item) (if (numberp item) item 0)))
or
(loop for item in '(1 #\a #\b 2 1 2 #\c #\d 4)
when (numberp item) sum item)
The Common Lisp HyperSpec says in the funcall entry that
(funcall function arg1 arg2 ...)
== (apply function arg1 arg2 ... nil)
== (apply function (list arg1 arg2 ...))
Since they are somehow equivalent, when would you use apply, and when funcall?
You should use funcall if you have one or more separate arguments and apply if you have your arguments in a list
(defun passargs (&rest args) (apply #'myfun args))
or
(defun passargs (a b) (funcall #'myfun a b))
apply is useful when the argument list is known only at runtime, especially when the arguments are read dynamically as a list. You can still use funcall here but you have to unpack the individual arguments from the list, which is inconvenient. You can also use apply like funcall by passing in the individual arguments. The only thing it requires is that the last argument must be a list:
> (funcall #'+ 1 2)
3
> (apply #'+ 1 2 ())
3
Well I think a good rule of thumb would be: use apply when you can't use funcall: the latter is clearer but is also less general than apply in that it doesn't allow you to call a function whose number of arguments is only known at runtime.
Of course it is only good practice and you could systematically do this the ugly way (systematically using apply), but as you've probably noticed, using the ugly way when a very similar but cleaner way is available is not very common-lisp-y.
Example of function that needs apply instead of funcall:
could you implement map in such a way that (map #'+ '(1 2) '(2 3)) and (map #'+ '(1 2) '(2 3) '(3 4)) both work (which is the case with the standard function) without using apply (or eval, which is cheating)?
EDIT: as has also been pointed out, it would be silly to write:(funcall func (first list) (second list) (third list) etc.) instead of (apply func list).
Apply function is curring the result, like it returns a function that applies to next argument, to next argument.
It is important subject on functional programming languages.
(mapcar 'list '((1 2)(3 4)))
(((1 2)) ((3 4)))
(funcall 'mapcar 'list '((1 2)(3 4)))
(((1 2)) ((3 4)))
(apply 'mapcar 'list '((1 2)(3 4)))
((1 3) (2 4))