I am quite new in lisp programing, so maybe this a stupid question, but anyway I have a list of numbers like ( 6000 6100 6200 6200 7200 etc.) and I want to find and replace the second same value ( to add 1200 to the second same value) so the result should be (6000 6100 6200 7400 7200). Can you help me with this? Thank you very much.
I don't see why the falk downvote the question, honestly. While it is not an interesting question, it is a clear one. For the pedants out there, let me generalize it in an attempt to make it more interesting. I am new to lisp, and I need to apply a function to a list. The output of the function depends not only on the current element but also on the preceding and/or subsequent elements, what is the best construct to use?
I think the easiest is to use loop:
(defun nahrad (list change)
(loop for f = nil then s
for s in list
collect (if (eql f s) (+ change s) s)))
(defparameter *test* '(6000 6100 6200 6200 7200))
(nahrad *test* 1200)
Related
I am trying to do Recursive method to find max value in list.
Can anyone explain where I made the mistake on this code and how to approach it next time.
(defun f3 (i)
(setq x (cond (> (car (I)) (cdr (car (I))))
(f3 (cdr (I)))))
)
(f3 '(33 11 44 2) )
also I tried this following method and didn't work:
(defun f3 (i)
(cond ((null I )nil )
(setq x (car (i))
(f3(cdr (i)))
(return-from max x)
)
Thanks a lot for any help. I am coming from java if that helps.
If you're working in Common Lisp, then you do this:
(defun max-item (list)
(loop for item in list
maximizing item))
That's it. The maximizing item clause of loop determines the highest item value seen, and implicitly establishes that as the result value of loop when it terminates.
Note that if list is empty, then this returns nil. If you want some other behavior, you have to work that in:
(if list
(loop for item in list
maximizing item))
(... handle empty here ...))
If the number of elements in the list is known to be small, below your Lisp implementation's limit on the number of arguments that can be passed to a function, you can simply apply the list to the max function:
(defun max-item (list)
(apply #'max list))
If list is empty, then max is misused: it requires one or more arguments. An error condition will likely be signaled. If that doesn't work in your situation, you need to add code to supply the desired behavior.
If the list is expected to be large, so that this approach is to be avoided, you can use reduce, treating max as a binary function:
(defun max-item (list)
(reduce #'max list))
Same remarks regarding empty list. These expressions are so small, many programmers will avoid writing a function and just use them directly.
Regarding recursion, you wouldn't use recursion to solve this problem in production code, only as a homework exercise for learning about recursion.
You are trying to compute the maximum value of a list, so please name your function maximum and your parameter list, not f3 or i. You can't name the function max without having to consider how to avoid shadowing the standard max function, so it is best for now to ignore package issues and use a different name.
There is a corner case to consider when the list is empty, as there is no meaningful value to return. You have to decide if you return nil or signal an error, for example.
The skeleton is thus:
(defun maximum (list)
(if (null list)
...
...))
Notice how closing parentheses are never preceded by spaces (or newlines), and opening parentheses are never followed by spaces (or newlines). Please note also that indentation increases with the current depth . This is the basic rules for Lisp formatting, please try following them for other developers.
(setq x <value>)
You are assigning an unknown place x, you should instead bind a fresh variable if you want to have a temporary variable, something like:
(let ((x <value>))
<body>)
With the above expression, x is bound to <value> inside <body> (one or more expressions), and only there.
(car (i))
Unlike in Java, parentheses are not used to group expressions for readability or to force some evaluation order, in Lisp they enclose compound forms. Here above, in a normal evaluation context (not a macro or binding), (i) means call function i, and this function is unrelated to your local variable i (just like in Java, where you can write int f = f(2) with f denoting both a variable and a method).
If you want to take the car of i, write (car i).
You seem to be using cond as some kind of if:
(cond (<test> <then>) <else>) ;; WRONG
You can have an if as follows:
(if <test> <then> <else>)
For example:
(if (> u v) u v) ;; evaluates to either `u` or `v`, whichever is greater
The cond syntax is a bit more complex but you don't need it yet.
You cannot return-from a block that was undeclared, you probably renamed the function to f3 without renaming that part, or copied that from somewhere else, but in any case return-from is only needed when you have a bigger function and probably a lot more side-effects. Here the computation can be written in a more functionnal way. There is an implicit return in Lisp-like languages, unlike Java, for example below the last (but also single) expression in add evaluates to the function's return value:
(defun add-3 (x)
(+ x 3))
Start with smaller examples and test often, fix any error the compiler or interpreter prints before trying to do more complex things. Also, have a look at the available online resources to learn more about the language: https://common-lisp.net/documentation
Although the other answers are right: you definitely need to learn more CL syntax and you probably would not solve this problem recursively in idiomatic CL (whatever 'idiomatic CL' is), here's how to actually do it, because thinking about how to solve these problems recursively is useful.
First of all let's write a function max/2 which returns the maximum of two numbers. This is pretty easy:
(defun max/2 (a b)
(if (> a b) a b))
Now the trick is this: assume you have some idea of what the maximum of a list of numbers is: call this guess m. Then:
if the list is empty, the maximum is m;
otherwise the list has a first element, so pick a new m which is the maximum of the first element of the list and the current m, and recurse on the rest of the list.
So, we can write this function, which I'll call max/carrying (because it 'carries' the m):
(defun max/carrying (m list)
(if (null list)
m
(max/carrying (max/2 (first list) m)
(rest list))))
And this is now almost all we need. The trick is then to write a little shim around max/carrying which bootstraps it:
to compute the maximum of a list:
if the list is empty it has no maximum, and this is an error;
otherwise the result is max/carrying of the first element of the list and the rest of the list.
I won't write that, but it's pretty easy (to signal an error, the function you want is error).
is there a function like find in lisp that returns true instead of the element we're trying to find?
example:
I want it to do
(find 'x '(a c x)) = t
not
(find 'x '(a c x)) = x
Also, the reason I am asking is because I am trying to reach the deepest element in a list. My plan was to flatten the list every time i recursively called it.
I would then stop the recursive call when
(mapcar 'atom list)
would tell me every atom in there is true.
Do you believe this is a good approach to this problem?
There's no such function, but it can't be easier to write one:
(defun find-t (&rest args)
(when (apply #'find args)
t))
Also instead of (mapcar 'atom list) you can use (every #`(eql t %) list), i.e. check that every item in list is exactly t. (Here #`() is syntactic sugar for one-argument lambdas, that I use.)
But overall it's unclear, what you're trying to achieve with all this. Can you elaborate on what you're trying to do?
Is there anyone who can help me to write a function in common LISP that counts the numbers in a list?
The code that I have written is below, but it does not work!
(defun count-numbers(lst)
(let(result()))
(dolist(number lst)
(push number result))
(length result))
For example, when I enter this query "(count'(r 4 f d w 2 3 4 1 z))", I must get 5.
Since it's homework, I'll just give some pointers. First: simplicity. If you are new to Common-Lisp, just use its basic features. For example: recursion. In pure functional style. Think about something like this:
(defun count (list counter)
;; something
)
we first check list. If it's empty, we already checked all the elements, so we return counter. If list is not empty, we
take its first element
we check if it's a number
it's a number! We recursively call count on the rest of the list and with counter = counter + 1
it's not a number! We recursively call count on the rest of the list with counter the same as before
Use (numberp n). It returns T if n is a number, NIL if not.
(defun count-numbers (lst)
(let (result ()))
(dolist (number lst)
(push number result))
(length result))
Check the indentation. Is that what you wanted? Maybe not.
Then you also push all elements to the result list? Is that what you want?
Here is a list of functions on numbers. http://www.lispworks.com/documentation/HyperSpec/Body/c_number.htm
Maybe there is one you need?
This is a good introductory Lisp book for download: http://www.cs.cmu.edu/~dst/LispBook/
I'd say there are a variety of ways to solve this, one would be an imperative loop, like the mostly correct solution already written, a recursive counting function (which is probably the worst way since there's no guarantee in CL that you won't blow the stack), or the functional approach you would probably actually use in production. The last one would be this:
(defun count-numbers (list) (count-if #'numberp list))
I have to reverse the elements of a simple (single-dimension) list. I know there's a built-in reverse function but I can't use it for this.
Here's my attempt:
(defun LISTREVERSE (LISTR)
(cond
((< (length LISTR) 2) LISTR) ; listr is 1 atom or smaller
(t (cons (LISTREVERSE (cdr LISTR)) (car LISTR))) ; move first to the end
)
)
Output pretty close, but is wrong.
[88]> (LISTREVERSE '(0 1 2 3))
((((3) . 2) . 1) . 0)
So I tried to use append instead of cons:
(t (append (LISTREVERSE (cdr LISTR)) (car LISTR)))
But got this error:
*** - APPEND: A proper list must not end with 2
Any help?
I can give you a couple of pointers, because this looks like homework:
The base case of the recursion is when the list is empty (null), and not when there are less than two elements in the list
Consider defining a helper function with an extra parameter, an "accumulator" initialized in the empty list. For each element in the original list, cons it at the head of the accumulator. When the input list is empty, return the accumulator
As an aside note, the above solution is tail-recursive.
As a follow-up to Óscar López (and fighting the temptation to just write a different solution down):
Using both append and length makes the posted solution just about the least efficient way of reversing a list. Check out the documentation on cons and null for some better ideas on how to implement this.
Please, please indent properly.
Tail recursion really is both more efficient and reasonably simple in this case. Try it if you haven't already. labels is the form you want to use to define local recursive functions.
It may be worth your while to flip through The Little Schemer. It'll give you a better feel for recursion in general.
It's ok what you did. You only missed the composition of the result list.
Think about it: You have to append the 1-element list of the CAR to the end of the list of the reversed CDR:
(defun LISTREVERSE (LISTR)
(cons
((< (length LISTR) 2) LISTR) ; listr is 1 atom or smaller
(t (append (LISTREVERSE (cdr LISTR)) (list (car LISTR))))))
(defun listreverse (list)
(let (result)
(dolist (item list result)
(push item result))))
Don't use recursion in Common Lisp when there is a simple iterative way to reach the same goal. Common Lisp does not make any guarantees about tail recursion, and your tail recursive function invocation may not be optimized to a jump at the discretion of the compiler.
push prepends the item to the result
dolist has an optional third argument which is the value returned. It is evaluated when the loop is exited.
Should I use
(apply + (filter prime? (range 1 20)))
or
(reduce + (filter prime? (range 1 20)))
Edit: This is the source for prime in clojure from optimizing toolkit.
(defn prime? [n]
(cond
(or (= n 2) (= n 3)) true
(or (divisible? n 2) (< n 2)) false
:else
(let [sqrt-n (Math/sqrt n)]
(loop [i 3]
(cond
(divisible? n i) false
(< sqrt-n i) true
:else (recur (+ i 2)))))))
If you are asking in terms of performance, the reduce is better by a little:
(time (dotimes [_ 1e6] (apply + (filter even? (range 1 20)))))
"Elapsed time: 9059.251 msecs"
nil
(time (dotimes [_ 1e6] (reduce + (filter even? (range 1 20)))))
"Elapsed time: 8420.323 msecs"
nil
About 7% difference in this case, but YMMV depending on the machine.
You haven't provided your source for the prime? function, so I have substituted even? as the predicate. Keep in mind that your runtime may be dominated by prime?, in which case the choice between reduce and apply matters even less.
If you are asking which is more "lispy" then I would say that the reduce implementation is preferrable, as what you are doing is a reduce/fold in the functional programming sense.
I would think that reduce would be preferable when it is available, because apply uses the list as arguments to the function, but when you have a large number -- say, a million -- elements in the list, you will construct a function call with a million arguments! That might cause some problems with some implementations of Lisp.
(reduce op ...) is the norm and (apply op ...) the exception (notably for str and concat).
I would expect apply to realize a lazy list which could be ugly, and you never want to assume your list is not lazy, 'cause you could suddenly find yourself getting smacked with massive memory useage.
Reduce is going to grab them 1 by one and roll the results together into a single whole, never taking the whole list in at once.
i am going to play devil's advocate and argue for apply.
reduce is clojure's take on fold (more exactly foldl), the left-fold, and is usually defined with an initial element, because a fold operation has two parts:
an initial (or "zero") value
an operation for combining two values
so to find the sum of a set of numbers the natural way to use + is as (fold + 0 values) or, in clojure, (reduce + 0 values).
this explicitly shows the result for an empty list, which is important because it is not obvious to me that + returns 0 in this case - after all, + is a binary operator (all that fold needs or assumes).
now, in practice, it turns out that clojure's + is defined as more than a binary operator. it will take many or, even, zero values. cool. but if we're using this "extra" information it's friendly to signal that to the reader. (apply + values) does this - it says "i am using + in a strange way, as more than a binary operator". and that helps people (me, at least) understand the code.
[it's interesting to ask why apply feels clearer. and i think it's partly that you are saying to the reader: "look, + was designed to accept multiple values (that's what apply is used for), and so the language implementation will include the case of zero values." that implicit argument is not present with reduce applied to a single list.]
alternatively, (reduce + 0 values) is also fine. but (reduce + values) triggers an instinctive reaction in me: "huh, does + provide a zero?".
and if you disagree, then please, before you downvote or post a reply, are you sure about what (reduce * values) will return for an empty list?
I agree with Andrew Cooke. But I would like to add that since it already knows how to work and handle multiple arguments, we are essentially re-telling the '+' operator how to work by using it in the 'reduce', in my opinion.
Also, if I told you do to add a bunch of numbers together, you would probably use the '+' operator by itself, and not use 'reduce' at all.
(+ 1 2 3 etc...)
'apply' is just saying, I want to do the same thing, but use what's inside this list for the arguments. There is no extra reasoning required and you are using it as if the numbers didn't exist in a list to begin with, like you would if you were writing it out yourself.
(Reduce is so heavily used that using/reasoning about it in this case isn't a big deal at all, I think either way is fine, but personally prefer apply)