I am writing a function, try-weak-cues, to select a response from a large set of responses. This program is essentially a conversation with the user.
(define try-weak-cues
(lambda (sentence context)
(define helper
(lambda(list-of-pairs)
(define helper2
(lambda(list-of-pairs context)
(cond((null? list-of-pairs)
(cond((null? list-of-pairs) '())
((any-good-fragments?(cue-part(car list-of-pairs))sentence) (helper2(cdr(car list-of-pairs))context))
(else(helper(cdr list-of-pairs)))))))))
(helper *weak-cues*))))
Below is the list of responses that the function is supposed to pull from:
(define *weak-cues*
'( ( ((who) (whos) (who is))
((first-base)
((thats right) (exactly) (you got it)
(right on) (now youve got it)))
((second-base third-base)
((no whos on first) (whos on first) (first base))) )
( ((what) (whats) (what is))
((first-base third-base)
((hes on second) (i told you whats on second)))
((second-base)
((right) (sure) (you got it right))) )
( ((whats the name))
((first-base third-base)
((no whats the name of the guy on second)
(whats the name of the second baseman)))
((second-base)
((now youre talking) (you got it))))
))
The error:
define: bad syntax (multiple expressions after identifier) in: (define
helper (lambda (list-of-pairs) (define helper2 (lambda (list-of-pairs
context) (cond ((null? list-of-pairs) (cond ((null? list-of-pairs)
(quote ())) ((any-good-fragments? (cue-part (car list-of-pairs))
sentence) (helper2 (cdr (car list-of-pairs)) context)) (else (helper
(cdr list-of-pairs))))))))) (helper weak-cues))
The problem is that, when you define an internal procedure inside the body of a lambda, you have to write an expression after the definition, typically calling the internal procedure. For example, this is wrong:
(define f
(lambda (x)
(define g
(lambda (y)
<body 1>))))
But this is correct, notice how after the inner lambda definition there's a <body 2> part:
(define f
(lambda (x)
(define g
(lambda (y)
<body 1>))
<body 2>))
Also, your code will be much easier to debug if you avoid this style of procedure definition:
(define f
(lambda (x)
<body>))
It'll be shorter and clearer like this:
(define (f x)
<body>)
Now, back to your code. After fixing the formatting and switching to the shorter procedure definition syntax, your code will look like this:
(define (try-weak-cues sentence context)
(define (helper list-of-pairs)
(define (helper2 list-of-pairs context)
(cond ((null? list-of-pairs)
(cond ((null? list-of-pairs) '())
((any-good-fragments? (cue-part (car list-of-pairs)) sentence)
(helper2 (cdr (car list-of-pairs)) context))
(else (helper (cdr list-of-pairs)))))))
<missing body>)
(helper *weak-cues*))
Now it's clear that the body of helper is missing, after the inner definition of helper2 you didn't write anything. Possibly you intended to call helper2 at that point, but your code is confusing enough as it is and I can't guess what to write in the missing body, it's up to you.
Related
So I am new to Scheme and have encountered a problem. What I am trying to do is the following. It's a rather simple problem nevertheless I am receiving several errors:
I try to sum up the elements of lists (which only consists of numbers). If the total amount is even, the procedure should return <'divisible_by_2>.
If the total amount is odd, it should return <'not_divisible_by_2>.
The initial sstep was to build a procedure that sums up the lists. This one works. The second step was to build an if function which takes the sum of lists and returns <'divisible_by_2> if the sum is even and <'not_divisible_by_2> if it is odd.
What I wrote so far:
(define (divisible_or_not list-sum lst)
(if (odd? list-sum lst)
(lambda (list-sum lst)
(cond
((null? lst)
0)
((pair? (car lst))
(+(list-sum (car lst)) (list-sum (cdr lst)))
(else
(+ (car lst) (list-sum (cdr lst)))
)
)
)
('divisible_by_2)
('not_divisible_by_2)
)
)
)
Version 2.0 (lst=tree; tree-count=sum-lst):
(define (divisible-or-not tree)
(define (tree-count tree)
(cond
((null? tree)
0)
((pair? (car tree))
(+(tree-count (car tree)) (tree-count (cdr tree)))
(else
(+ (car tree) (tree-count (cdr tree))))))
(if (odd? tree-count tree)
('divisible-by-2)
('not-divisible-by-2))))
Your code, properly indented, looks like so:
(define (divisible_or_not list-sum lst)
(if (odd? list-sum lst)
(lambda (list-sum lst)
(cond
((null? lst)
0)
((pair? (car lst))
(+(list-sum (car lst)) (list-sum (cdr lst)))
(else
(+ (car lst) (list-sum (cdr lst))))))
('divisible_by_2)
('not_divisible_by_2))))
The structure of your program looks like this:
(if ...
(lambda (...) ...))
In other words, if your test succeeds, you return an anonymous function, and if the test fails, you return nothing (in Scheme, the value is undefined in that case).
Inside your lambda, the code is a list of three expressions, a cond, the form ('divisible_by_2) and the form ('not_divisible_by_2).
First of all, do not use underscores for separating words in Lisp/Scheme, use dashes, like so: divisible-by-2.
Secondly, only the last expression's value is returned from the lambda, so the intermediate cond, since it has no side-effect, is basically doing work for nothing. The second form, ('divisible_by_2), looks like a function call but is going to give you an error. If you want to return a symbol, just quote it, without parentheses: 'divisible-by-2.
Since you already have an intermediate function, you can associate it to a name:
(define tree-count (sum tree)
(cond ...))
I named it tree-count because you also recurse into the car of your lists.
Once you have this function, you only need to apply it:
(if (even? (tree-count tree))
'divisible-by-2
'not-divisible-by-2)
I need to write the lisp macro in scheme (please on hygienic macros and syntax-rules etc) that will have function call and Alist as argument
I want function and macro that call that function to have syntax like this:
(foo '(10 (a (lambda () (display "10")) b (lambda () (display "20"))))
or macro without quotes.
My last code is working, but not sure if this is how you suppose to write function/macro like this. It seems that I need double backquote but don't know how to write it. (I'm right now reading On Lips by Paul Graham and he said that double backquote is very hard and only need by macros defining macros, but It seems that this is what I need).
(define (foo expr)
`(list ,(car expr)
(,(string->symbol "quasiquote") ,(pair-map (lambda (a b)
(cons (symbol->string a)
(list 'unquote b)))
(cadr expr)))))
(define-macro (bar expr)
(foo expr))
(define xx (bar (10 (a 20 b (lambda () (display "x") (newline))))))
;; (list 10 `((a . ,20) (b . ,(lambda () (display "x") (newline))))
(define bfn (cdr (assoc "b" (cadr xx)))))
(bfn)
;; "x"
and here is definition of pair-map
(define (pair-map fn seq-list)
"(seq-map fn list)
Function call fn argument for pairs in a list and return combined list with
values returned from function fn. It work like the map but take two items from list"
(let iter ((seq-list seq-list) (result '()))
(if (null? seq-list)
result
(if (and (pair? seq-list) (pair? (cdr seq-list)))
(let* ((first (car seq-list))
(second (cadr seq-list))
(value (fn first second)))
(if (null? value)
(iter (cddr seq-list) result)
(iter (cddr seq-list) (cons value result))))))))
with (string->symbol "quasiquote") I was able not to use double backquote, can this be written with double backquote/quasiquote? How this should look like?
I'm asking if this can be written different way so I can fix few issues in my own lisp interpreter (not sure if is working correctly but it seems that this final version works the same in guile).
I came up with shorter quasiquote version, but still it require inserting symbols:
(define (foo expr)
`(list ,(car expr)
(,'quasiquote ,(pair-map (lambda (a b)
`(,(symbol->string a) . (,'unquote ,b)))
(cadr expr)))))
I want to solve a lisp function that returns a NUMBER(count) of numbers which are greater than the first number in the list.The list is a linear list of numbers.
(defun foo (lst)
(cond ((null lst) 0)
(car = k)
((> (car lst) k)
(1+ (foo (cdr lst))))
(T (foo (cdr lst)))))
My problem is that I cannot keep the first element and compare it with the others.
Let's take apart your problem:
You have a set of numbers. Really, you have a “special” first number, and then the rest of them. Specifically, you probably want only real numbers, because “less than” does not make sense in terms of complex (imaginary) numbers.
You can use first to get the first number from the list, and rest for the others.
Of these, you want to count any that are not greater than the first.
So let's start with sort of pseudocode
(defun count-numbers-greater-than-first (list)
;; split out first and rest
;; call the real count function
)
Well, we know now that we can use first and rest (also, as you used, historically car and cdr), so:
(defun count-numbers-greater-than-first (list)
(count-numbers-greater-than (first list) (rest list))
You already probably know that > is used to test whether real numbers are greater than one another.
A quick look at the CLHS reveals a nice function called count-if
(defun count-numbers-not-greater-than (reference other-numbers)
(count-if ??? other-numbers))
The ??? needs to be an object of function type, or the name of a function. We need to “curry” the reference (first number) into that function. This means we want to create a new function, that is only used for one run through the count-if, that already has “closed over” the value of reference.
If we knew that number would always be, say, 100, that function would look like this:
(defun greater-than-100 (number)
(> number 100))
That function could then get used in the count-if:
(defun count-numbers-greater-than (reference other-numbers)
(count-if (function greater-than-100)
other-numbers))
(defun count-numbers-greater-than (reference other-numbers)
(count-if #'greater-than-100 other-numbers))
But that doesn't solve the problem of getting the reference number “curried” into the function.
Without reaching for Alexandria (I'll explain in a moment), you can use a lambda form to create a new, anonymous function right here. Since reference is available within count-numbers-not-greater-than, you can use its value within that lambda. Let's convert for 100 first:
(defun count-numbers-greater-than (reference other-numbers)
(count-if (lambda (number) (> number 100))
other-numbers))
Now we can use reference:
(defun count-numbers-greater-than (reference other-numbers)
(count-if (lambda (number) (> number reference))
other-numbers))
And, in fact, you could even merge this back into the other function, if you wanted:
(defun count-numbers-greater-than-first (list)
(count-if (lambda (number) (> number (first list)))
(rest list)))
That Alexandria thing
But, what about Alexandria? Alexandria is a collection of super-useful utility functions that's available in Quicklisp or elsewhere.
(ql:quickload "alexandria")
(use-package #:alexandria)
Of course, you'd normally use it in your own defpackage
(defpackage my-cool-program
(:use :common-lisp :alexandria))
Two of the things it provides are curry and rcurry functions. It turns out, that lambda function in there is a really common case. You have an existing function — here, > — that you want to call with the same value over and over, and also some unknown value that you want to pass in each time.
These end up looking a lot like this:
(lambda (x) (foo known x))
You can use curry to write the same thing more concisely:
(curry #'foo known)
It also work with any number of arguments. RCurry does the same, but it puts the unknown values “x” at the left, and your known values at the right.
(lambda (x) (foo x known)) = (rcurry #'foo known)
So another way to write the count-if is:
(defun count-numbers-greater-than-first (list)
(count-if (rcurry #'> (first list))
(rest list)))
* (count-numbers-greater-than-first '(10 9 8 7 11 12))
2
Your function indented correctly looks like this:
(defun foo (lst)
(cond ((null lst) 0)
(car = k) ; strange cond term
((> (car lst) k)
(1+ (foo (cdr lst))))
(T (foo (cdr lst)))))
I have commented the second term in your cond. It is quite strange. It first evaluates the variable car (not the function #'car). If car is not nil it first evaluates the variable = (not the function #'=) and since it is not the last consequent expression in the cond term it throws that away and returns the last which is k.
Secondly you write that you say you use the first element as comparison, however you call it k in your function but it is not defined anywhere. You need to do something before you do the recursion and thus you cannot let the actual function do the recursion since it will take the first element each time. Here is where labels can be used:
;; didn't call it foo since it's not very descriptive
(defun count-larger-than-first (list)
(let ((first (car list)))
(labels ((helper (list)
(cond ((null list) 0)
((> (car list) first)
(1+ (helper (cdr list))))
(t (helper (cdr list))))))
(helper (cdr list)))))
Of course. Since you now have the possibility to add more arguments I would have added an accumulator:
(defun count-larger-than-first (list)
(let ((first (car list)))
(labels ((helper (list acc)
(cond ((null list) acc)
((> (car list) first)
(helper (cdr list) (1+ acc)))
(t (helper (cdr list) acc)))))
(helper (cdr list) 0))))
And of course recursion might blow the stack so you should really write it without in Common Lisp:
(defun count-larger-than-first (list)
(let ((first (car list)))
(loop :for element :in (cdr list)
:counting (> element first))))
There are higher order functions that count too which might be more suitable:
(defun count-larger-than-first (list)
(let ((first (car list)))
(count-if (lambda (element) (> element first))
(cdr list))))
When I compile the following code, SBCL complains that g!-unit-value and g!-unit are undefined. I'm not sure how to debug this. As far as I can tell, flatten is failing.
When flatten reaches the unquoted part of defunits, it seems like the entire part is being treated as an atom. Does that sound correct?
The following uses code from the book Let over Lambda:
Paul Graham Utilities
(defun symb (&rest args)
(values (intern (apply #'mkstr args))))
(defun mkstr (&rest args)
(with-output-to-string (s)
(dolist (a args) (princ a s))))
(defun group (source n)
(if (zerop n) (error "zero length"))
(labels ((rec (source acc)
(let ((rest (nthcdr n source)))
(if (consp rest)
(rec rest (cons (subseq source 0 n) acc))
(nreverse (cons source acc))))))
(if source (rec source nil) nil)))
(defun flatten (x)
(labels ((rec (x acc)
(cond ((null x) acc)
((atom x) (cons x acc))
(t (rec (car x) (rec (cdr x) acc))))))
(rec x nil)))
Let Over Lambda Utilities - Chapter 3
(defmacro defmacro/g! (name args &rest body)
(let ((g!-symbols (remove-duplicates
(remove-if-not #'g!-symbol-p
(flatten body)))))
`(defmacro ,name ,args
(let ,(mapcar
(lambda (g!-symbol)
`(,g!-symbol (gensym ,(subseq
(symbol-name g!-symbol)
2))))
g!-symbols)
,#body))))
(defun g!-symbol-p (symbol-to-test)
(and (symbolp symbol-to-test)
(> (length (symbol-name symbol-to-test)) 2)
(string= (symbol-name symbol-to-test)
"G!"
:start1 0
:end1 2)))
(defmacro defmacro! (name args &rest body)
(let* ((o!-symbols (remove-if-not #'o!-symbol-p args))
(g!-symbols (mapcar #'o!-symbol-to-g!-symbol o!-symbols)))
`(defmacro/g! ,name ,args
`(let ,(mapcar #'list (list ,#g!-symbols) (list ,#o!-symbols))
,(progn ,#body)))))
(defun o!-symbol-p (symbol-to-test)
(and (symbolp symbol-to-test)
(> (length (symbol-name symbol-to-test)) 2)
(string= (symbol-name symbol-to-test)
"O!"
:start1 0
:end1 2)))
(defun o!-symbol-to-g!-symbol (o!-symbol)
(symb "G!" (subseq (symbol-name o!-symbol) 2)))
Let Over Lambda - Chapter 5
(defun defunits-chaining (u units prev)
(if (member u prev)
(error "~{ ~a~^ depends on~}"
(cons u prev)))
(let ((spec (find u units :key #'car)))
(if (null spec)
(error "Unknown unit ~a" u)
(let ((chain (second spec)))
(if (listp chain)
(* (car chain)
(defunits-chaining
(second chain)
units
(cons u prev)))
chain)))))
(defmacro! defunits (quantity base-unit &rest units)
`(defmacro ,(symb 'unit-of- quantity)
(,g!-unit-value ,g!-unit)
`(* ,,g!-unit-value
,(case ,g!-unit
((,base-unit) 1)
,#(mapcar (lambda (x)
`((,(car x))
,(defunits-chaining
(car x)
(cons
`(,base-unit 1)
(group units 2))
nil)))
(group units 2))))))
This is kind of tricky:
Problem: you assume that backquote/comma expressions are plain lists.
You need to ask yourself this question:
What is the representation of a backquote/comma expression?
Is it a list?
Actually the full representation is unspecified. See here: CLHS: Section 2.4.6.1 Notes about Backquote
We are using SBCL. See this:
* (setf *print-pretty* nil)
NIL
* '`(a ,b)
(SB-INT:QUASIQUOTE (A #S(SB-IMPL::COMMA :EXPR B :KIND 0)))
So a comma expression is represented by a structure of type SB-IMPL::COMMA. The SBCL developers thought that this representation helps when such backquote lists need to be printed by the pretty printer.
Since your flatten treats structures as atoms, it won't look inside...
But this is the specific representation of SBCL. Clozure CL does something else and LispWorks again does something else.
Clozure CL:
? '`(a ,b)
(LIST* 'A (LIST B))
LispWorks:
CL-USER 87 > '`(a ,b)
(SYSTEM::BQ-LIST (QUOTE A) B)
Debugging
Since you found out that somehow flatten was involved, the next debugging steps are:
First: trace the function flatten and see with which data it is called and what it returns.
Since we are not sure what the data actually is, one can INSPECT it.
A debugging example using SBCL:
* (defun flatten (x)
(inspect x)
(labels ((rec (x acc)
(cond ((null x) acc)
((atom x) (cons x acc))
(t (rec (car x) (rec (cdr x) acc))))))
(rec x nil)))
STYLE-WARNING: redefining COMMON-LISP-USER::FLATTEN in DEFUN
FLATTEN
Above calls INSPECT on the argument data. In Common Lisp, the Inspector usually is something where one can interactively inspect data structures.
As an example we are calling flatten with a backquote expression:
* (flatten '`(a ,b))
The object is a proper list of length 2.
0. 0: SB-INT:QUASIQUOTE
1. 1: (A ,B)
We are in the interactive Inspector. The commands now available:
> help
help for INSPECT:
Q, E - Quit the inspector.
<integer> - Inspect the numbered slot.
R - Redisplay current inspected object.
U - Move upward/backward to previous inspected object.
?, H, Help - Show this help.
<other> - Evaluate the input as an expression.
Within the inspector, the special variable SB-EXT:*INSPECTED* is bound
to the current inspected object, so that it can be referred to in
evaluated expressions.
So the command 1 walks into the data structure, here a list.
> 1
The object is a proper list of length 2.
0. 0: A
1. 1: ,B
Walk in further:
> 1
The object is a STRUCTURE-OBJECT of type SB-IMPL::COMMA.
0. EXPR: B
1. KIND: 0
Here the Inspector tells us that the object is a structure of a certain type. That's what we wanted to know.
We now leave the Inspector using the command q and the flatten function continues and returns a value:
> q
(SB-INT:QUASIQUOTE A ,B)
For anyone else who is trying to get defmacro! to work on SBCL, a temporary solution to this problem is to grope inside the unquote structure during the flatten procedure recursively flatten its contents:
(defun flatten (x)
(labels ((flatten-recursively (x flattening-list)
(cond ((null x) flattening-list)
((eq (type-of x) 'SB-IMPL::COMMA) (flatten-recursively (sb-impl::comma-expr x) flattening-list))
((atom x) (cons x flattening-list))
(t (flatten-recursively (car x) (flatten-recursively (cdr x) flattening-list))))))
(flatten-recursively x nil)))
But this is horribly platform dependant. If I find a better way, I'll post it.
In case anyone's still interested in this one, here are my three cents. My objection to the above modification of flatten is that it might be more naturally useful as it were originally, while the problem with representations of unquote is rather endemic to defmacro/g!. I came up with a not-too-pretty modification of defmacro/g! using features to decide what to do. Namely, when dealing with non-SBCL implementations (#-sbcl) we proceed as before, while in the case of SBCL (#+sbcl) we dig into the sb-impl::comma structure, use its expr attribute when necessary and use equalp in remove-duplicates, as we are now dealing with structures, not symbols. Here's the code:
(defmacro defmacro/g! (name args &rest body)
(let ((syms (remove-duplicates
(remove-if-not #-sbcl #'g!-symbol-p
#+sbcl #'(lambda (s)
(and (sb-impl::comma-p s)
(g!-symbol-p (sb-impl::comma-expr s))))
(flatten body))
:test #-sbcl #'eql #+sbcl #'equalp)))
`(defmacro ,name ,args
(let ,(mapcar
(lambda (s)
`(#-sbcl ,s #+sbcl ,(sb-impl::comma-expr s)
(gensym ,(subseq
#-sbcl
(symbol-name s)
#+sbcl
(symbol-name (sb-impl::comma-expr s))
2))))
syms)
,#body))))
It works with SBCL. I have yet to test it thoroughly on other implementations.
I want to ask why this function doesn't work...
(defun nenum(ls)
(cond
((null ls) nil)
((listp car(ls)) (nenum (rest ls)))
((numberp car(ls)) (nenum (rest ls)))
(t (cons (car ls) (nenum (rest ls))))))
Example: (nenum '(l 1 i (b) (5) s -2 p)) --> (l i s p)
Thank you!
Looking at the predicate you have in one of your cond terms:
(listp car (ls))
Thus apply the function listp with the two arguments car and the result of calling the function ls with no arguments. car and ls both need to be free variables and listp needs to be a different function than the one defined in CLHS since it only takes one argument.
Perhaps you have though you were writing Algol? An Algol function call look like operator(operand) but not CL. CL is a LISP dialect and we have this form on our function calls:
(operand operator)
If we nest we do the same:
(operand (operand operator))
You got it right in the alternative (cons (car ls) (nenum (rest ls)))
Replace car(ls) with (car ls).
Here's a much easier way to write that function:
(defun nenum (list)
(remove-if (lambda (item)
(or (listp item)
(numberp item)))
list))
Note that NIL doesn't need its own test because listp covers it.
There's no need to write a function like this from scratch. Common Lisp already provides remove-if, and you can give it a predicate that matches numbers and non-atoms:
CL-USER> (remove-if #'(lambda (x)
(or (numberp x)
(not (atom x))))
'(l 1 i (b) (5) s -2 p))
;=> (L I S P)
Or, to make it even clearer that you're keeping non-numeric atoms, you can use remove-if-not with a predicate that checks for numeric atoms:
CL-USER> (remove-if-not #'(lambda (x)
(and (atom x)
(not (numberp x))))
'(l 1 i (b) (5) s -2 p))
;=> (L I S P)
Note that the empty list, which is often written as (), is just the symbol nil. As such, it too is a non-numeric atom. If you'd want to keep other symbols, e.g.,
CL-USER> (remove-if-not #'(lambda (x)
(and (atom x)
(not (numberp x))))
'(li (b) -1 (5) sp))
;=> (LI SP)
then you'll probably want to keep nil as well:
CL-USER> (remove-if-not #'(lambda (x)
(and (atom x)
(not (numberp x))))
'(van (b) () (5) a))
;=> (VAN NIL A)