How to specify arguments for substitute in mapcar? - lisp

For example, I have a list of "strings" : ("First" "second" "third.").
And I need to replace all "s" to "a" -> (("Firat")...etc
I've found very beautiful function called substitute, but it works only for one sequence.
Can I use substitute in mapcar ? Something like this:
(mapcar 'substitute #\d #\o '(("test") ("wow") ("waow")))

You can wrap a LAMBDA around the function:
(mapcar (lambda (string) (substitute #\d #\o string)) '("test" "wow" "waow"))
;=> ("test" "wdw" "wadw")
Or you can use a helper function called CURRY (which is not part of the Common Lisp standard, but is available in Alexandria for example).
(ql:quickload :alexandria)
(use-package :alexandria)
(mapcar (curry #'substitute #\d #\o) '("test" "wow" "waow"))
;=> ("test" "wdw" "wadw")
For multiple strings in sublists, you can use nested mapcar/lambda:
(mapcar (lambda (sentence)
(mapcar (lambda (string)
(substitute #\d #\o string))
sentence))
'(("I" "like" "my" "dog.") ("My" "dog" "likes" "me" "too.")))
;=> (("I" "like" "my" "ddg.") ("My" "ddg" "likes" "me" "tdd."))
The inner LAMBDA could be changed to a CURRY too:
(mapcar (lambda (sentence)
(mapcar (curry #'substitute #\d #\o) sentence))
'(("I" "like" "my" "dog.") ("My" "dog" "likes" "me" "too.")))
Or even:
(mapcar (curry #'mapcar (curry #'substitute #\d #\o))
'(("I" "like" "my" "dog.") ("My" "dog" "likes" "me" "too.")))

I'm adding to jkiiski's very nice answer one final possibility to specify arguments to substitute in this case, though it's probably not what you need:
(mapcar #'substitute '(#\x #\x #\x) '(#\a #\b #\c) '("abcd" "abcd" "abcd"))
;; => ("xbcd" "axcd" "abxd")
Here for every i up to the lowest length of any passed list, mapcar passes the ith element of the first list as first argument, the ith element of the second as second argument and the i th element as third argument to substitute. This works because mapcar takes one or any number of lists.
Using this only makes sense if you want to perform different substitutions on (almost) each of the strings.

Beyond a flat list one could consider recursion, to handle the general nesting case:
(labels ((my-sub (s)
(typecase s
(string (substitute #\d #\o s))
(list (mapcar #'my-sub s))
(otherwise s))))
(my-sub '(("I" "like" "my" "dog.")
("My" ("dog" "likes" "me") "too."))))

Related

Racket: How do you use the value of an argument as a symbol?

I'm new to Racket, and am struggling to find the right words to explain what I'm trying to do. The best I can come up with is this: How do I use the value of an argument as a symbol, without calling the function which has the same name as the value of the argument, and without just quoting the name of the argument?
Here is a minimal example to show this.
There is a function called ul which does something.
There is another function called make-list which has a single parameter, list-type.
There is a third function, make-ul, which calls make-list with an argument ul.
In the true branch, the function ul is applied. Good so far.
In the false branch, I want to be able to use the value of the argument, which in this case is 'ul, as a symbol. In effect, this line of code would run (λ args (list* 'ul args)). How do I achieve this?
#lang racket
(define (ul . xs) `(div ,#xs))
(define (make-list list-type)
(if (equal? 'something 'something-else)
(λ args (apply list-type args)) ; true branch: call the function "ul"
(λ args (list* list-type args)))) ; false branch: how to use `ul without calling the "ul" function?
(define make-ul (make-list ul))
If your goal is to write good Racket code, then the best answer is don't.
Add a separate argument to take the symbol, like this:
;; make-list : Procedure Symbol -> Any ... -> Any
(define (make-list proc tag-sym)
(if ....
proc ;; (λ args (apply proc args)) simplifies to just proc
(λ args (list* tag-sym args))))
(define make-ul (make-list ul 'ul))
Or if the branch is independent of args, you could just have a single argument of type (U Procedure Symbol) instead:
;; make-list : (U Procedure Symbol) -> Any ... -> Any
(define (make-list list-type)
(cond [(procedure? list-type)
;; (λ args (apply list-type args))
list-type]
[(symbol? list-type)
(λ args (list* list-type args))]))
(define make-ul (make-list ul))
(define make-li (make-list 'li))
If you want to explore Racket features, there are two main ways to do it.
You can make a macro that takes a procedure name and uses it both as a reference and quotes it as a symbol. For example, using the first version of make-list above:
(define-syntax-rule (make-list* proc-name) (make-list proc-name (quote proc-name)))
(define make-ul (make-list* ul)) ;; => (define make-ul (make-list ul 'ul))
You can call object-name on a procedure to ask what Racket thinks its name is. For example, (object-name list) returns 'list, and (object-name ul) should return 'ul. But (object-name make-ul) will not return 'make-ul, because Racket tracks names at compile time, not at run time. (I would advise against writing code that depends on object-name; it makes your code very fragile.)

unexpected result from listp

I have a struct spider:
(defstruct spider omegas values k)
And an instance spider:
(set '*spider* (make-spider
:omegas '()
:values (list *input*)
:k '(#'omegashift #'dec #'dupval '((0 . #'dec) (1 . #'inc) (2 . #'dec)))))
But when I run the expression: (listp (car (spider-k *spider*)) on Emacs and SBCL (and SLIME is involved but I'm not sure what it is.)
The REPL returns T. This is obviously confusing as (car (spider-k *spider*) correctly returns #'OMEGASHIFT and (listp (function OMEGASHIFT)) properly returns NIL.
Why is (listp (car (spider-k *spider*)) true? Shouldn't it be false?
#'omegashift is a reader-macro that expands to the list (function omegashift).
When you evaluate (function omegashift) you get a function, but you're not evaluating it because you quoted the list. So you're just getting the list that the reader-macro expands to.
You'd see the same thing if you did (listp (car '('foo))). 'foo expands to the list (quote foo). This evaluates to the symbol foo, but the quote before the list prevents evaluation.
To get functions instead of lists, you need to evaluate the function expressions. You can do this by calling the function list rather than quoting a list.
(setq *spider* (make-spider
:omegas '()
:values (list *input*)
:k (list #'omegashift #'dec #'dupval (list (cons 0 #'dec) (cons 1 #'inc) (cons 2 #'dec)))))
You can also use backquote to simplify this:
(setq *spider* (make-spider
:omegas '()
:values (list *input*)
:k `(,#'omegashift ,#'dec ,#'dupval '((0 . ,#'dec) (1 . ,#'inc) (2 . ,#'dec)))))
Inside a backquoted expression, you use comma to mark the sub-expressions that you want to evaluate.
BTW, you should use setq to assign variables, not set with a quoted symbol. They're equivalent for global variables, but you can't use set with a local variable.

Call several functions with the same value

I have various functions and I want to call each function with the same value. For instance,
I have these functions:
(defun OP1 (arg) ( + 1 arg) )
(defun OP2 (arg) ( + 2 arg) )
(defun OP3 (arg) ( + 3 arg) )
And a list containing the name of each function:
(defconstant *OPERATORS* '(OP1 OP2 OP3))
So far, I'm trying:
(defun TEST (argument) (dolist (n *OPERATORS*) (n argument) ) )
I've tried using eval, mapcar, and apply, but these haven't worked.
This is just a simplified example; the program that I'm writing has eight functions that are needed to expand nodes in a search tree, but for the moment, this example should suffice.
Other answers have provided some idiomatic solutions with mapcar. One pointed out that you might want a list of functions (which *operators* isn't) instead of a list of symbols (which *operators* is), but it's OK in Common Lisp to funcall a symbol. It's probably more common to use some kind of mapping construction (e.g., mapcar) for this, but since you've provided code using dolist, I think it's worth looking at how you can do this iteratively, too. Let's cover the (probably more idiomatic) solution with mapping first, though.
Mapping
You have a fixed argument, argument, and you want to be able to take a function function and call it with that `argument. We can abstract this as a function:
(lambda (function)
(funcall function argument))
Now, we want to call this function with each of the operations that you've defined. This is simple to do with mapcar:
(defun test (argument)
(mapcar (lambda (function)
(funcall function argument))
*operators*))
Instead of operators, you could also write '(op1 op2 op3) or (list 'op1 'op2 'op3), which are lists of symbols, or (list #'op1 #'op2 #'op3) which is a list of functions. All of these work because funcall takes a function designator as its first argument, and a function designator is
an object that denotes a function and that is one of: a symbol (denoting the function named by that symbol in the global environment), or a function (denoting itself).
Iteratively
You can do this using dolist. The [documentation for actually shows that dolist has a few more tricks up its sleeve. The full syntax is from the documentation
dolist (var list-form [result-form]) declaration* {tag | statement}*
We don't need to worry about declarations here, and we won't be using any tags, but notice that optional result-form. You can specify a form to produce the value that dolist returns; you don't have to accept its default nil. The common idiom for collecting values into a list in an iterative loop is to push each value into a new list, and then return the reverse of that list. Since the new list doesn't share structure with anything else, we usually reverse it destructively using nreverse. Your loop would become
(defun test (argument)
(let ((results '()))
(dolist (op *operators* (nreverse results))
(push (funcall op argument) results))))
Stylistically, I don't like that let that just introduces a single value, and would probably use an &aux variable in the function (but this is a matter of taste, not correctness):
(defun test (argument &aux (results '()))
(dolist (op *operators* (nreverse results))
(push (funcall op argument) results)))
You could also conveniently use loop for this:
(defun test2 (argument)
(loop for op in *operators*
collect (funcall op argument)))
You can also do somewhat succinctly, but perhaps less readably, using do:
(defun test3a (argument)
(do ((results '() (list* (funcall (first operators) argument) results))
(operators *operators* (rest operators)))
((endp operators) (nreverse results))))
This says that on the first iteration, results and operators are initialized with '() and *operators*, respectively. The loop terminates when operators is the empty list, and whenever it terminates, the return value is (nreverse results). On successive iterations, results is a assigned new value, (list* (funcall (first operators) argument) results), which is just like pushing the next value onto results, and operators is updated to (rest operators).
FUNCALL works with symbols.
From the department of silly tricks.
(defconstant *operators* '(op1 op2 o3))
(defun test (&rest arg)
(setf (cdr arg) arg)
(mapcar #'funcall *operators* arg))
There's a library, which is almost mandatory in any anywhat complex project: Alexandria. It has many useful functions, and there's also something that would make your code prettier / less verbose and more conscious.
Say, you wanted to call a number of functions with the same value. Here's how you'd do it:
(ql:quickload "alexandria")
(use-package :alexandria)
(defun example-rcurry (value)
"Calls `listp', `string' and `numberp' with VALUE and returns
a list of results"
(let ((predicates '(listp stringp numberp)))
(mapcar (rcurry #'funcall value) predicates)))
(example-rcurry 42) ;; (NIL NIL T)
(example-rcurry "42") ;; (NIL T NIL)
(defun example-compose (value)
"Calls `complexp' with the result of calling `sqrt'
with the result of calling `parse-integer' on VALUE"
(let ((predicates '(complexp sqrt parse-integer)))
(funcall (apply #'compose predicates) value)))
(example-compose "0") ;; NIL
(example-compose "-1") ;; T
Functions rcurry and compose are from Alexandria package.

Function name and dynamic binding in Common Lisp

I'm reading Peter Norvig's Paradigms of AI. In chapter 6.2, the author uses code like below (not the original code, I picked out the troubling part):
Code Snippet:
(progv '(op arg) '(1+ 1)
(eval '(op arg)))
As the author's original intent, this code should return 2, but in sbcl 1.1.1, the interpreter is apparently not looking up op in the environment, throwing out op: undefined function.
Is this implementation specific? Since the code must have been tested on some other lisp.
p.s Original code
You probably mean
(progv '(op arg) '(1+ 1)
(eval '(funcall op arg)))
Edit(2013-08-21):
PAIP was written in pre-ANSI-Common-Lisp era, so it's possible the code
there contains a few noncompliances wrt the standard. We can make
the examples work with the following revision:
(defun match-if (pattern input bindings)
"Test an arbitrary expression involving variables.
The pattern looks like ((?if code) . rest)."
(and (eval (reduce (lambda (code binding)
(destructuring-bind (var . val) binding
(subst val var code)))
bindings :initial-value (second (first pattern))))
(pat-match (rest pattern) input bindings)))
;; CL-USER> (pat-match '(?x ?op ?y is ?z (?if (eql (?op ?x ?y) ?z))) '(3 + 4 is 7))
;; ((?Z . 7) (?Y . 4) (?OP . +) (?X . 3) (T . T))
;; CL-USER> (pat-match '(?x ?op ?y (?if (?op ?x ?y))) '(3 > 4))
;; NIL
Elements in first positions are not looked up as values, but as functions and there is no concept of dynamic binding in the function namespace.
I'd say after a quick look that the original code was designed to evaluate in a context like
(progv '(x y) '(12 34)
(eval '(> (+ x y) 99)))
i.e. evaluating a formula providing substitution for variables, not for function names.
The other answers so far are right, in that the actual form being evaluated is not the variables being bound by progv (simply (op arg)), but none have mentioned what is being evaluated. In fact, the comments in the code you linked to provide a (very) short explanation (this is the only code in that file that uses progv):
(defun match-if (pattern input bindings)
"Test an arbitrary expression involving variables.
The pattern looks like ((?if code) . rest)."
;; *** fix, rjf 10/1/92 (used to eval binding values)
(and (progv (mapcar #'car bindings)
(mapcar #'cdr bindings)
(eval (second (first pattern))))
(pat-match (rest pattern) input bindings)))
The idea is that a call to match-if gets called like
(match-if '((?if code) . rest) input ((v1 val1) (v2 val2) ...))
and eval is called with (second (first pattern)), which the value of code. However, eval is called within the progv that binds v1, v2, &c., to the corresponding val1, val2, &c., so that if any of those variables appear free in code, then they are bound when code is evaluated.
Problem
The problem that I see here is that, by the code we can't tell if the value is to be saved as the variable's symbol-value or symbol-function. Thus when you put a + as a value to some corresponding variable, say v, then it'll always be saved as the symbol-value of var, not it's symbol-function.
Therefore when you'll try to use it as, say (v 1 2) , it won't work. Because there is no function named v in the functions' namespace(see this).
So, what to do?
A probable solution can be explicit checking for the value that is to be bound to a variable. If the value is a function, then it should be bound to the variable's function value. This checking can be done via fboundp.
So, we can make a macro functioner and a modified version of match-if. functioner checks if the value is a function, and sets it aptly. match-if does the dynamic local bindings, and allows other code in the scope of the bound variables.
(defmacro functioner (var val)
`(if (and (symbolp ',val)
(fboundp ',val))
(setf (symbol-function ',var) #',val)
(setf ,var ,val)))
(defun match-if (pattern input bindings)
(eval `(and (let ,(mapcar #'(lambda (x) (list (car x))) bindings)
(declare (special ,# (mapcar #'car bindings)))
(loop for i in ',bindings
do (eval `(functioner ,(first i) ,(rest i))))
(eval (second (first ',pattern))))
(pat-match (rest ',pattern) ',input ',bindings))))

How to transform this function into macro?

I have difficulties understanding the new macro system of Scheme. Somewhere along the path I began to write my "macro" as a function first, and then later apply it as a macro.
So my mission is to turn the following structure:
;; highlight-rules: rule id, color and the regexp matches
(define highlight-rules
`((important ,(with-esc "[1;33m") ("foo"
"bob"))
(unimportant ,(with-esc "[1;30m") ("case of unimport"))
(urgent ,(with-esc "[1;31m") ("urgents"))))
Into this kind of cond series with match strings compiled to regexpes:
;; just an example. `line` is an argument bound by the function application
(cond
((string-match (regexp ".*sudo:session.*") line)
(with-color *important* line))
(else line))
I have written a function that seems to do the trick:
;; (cdar highlight-rules) -> (colorstring list-of-rules)
(define (parse-highlight-rules rules)
;; aux function to do one 'class' of patterns
(define (class-of-rules colorstr rulelist)
(map (lambda (rule)
`((string-match ,(regexp rule)) (with-color ,colorstr line)))
rulelist))
(define (do-loop accumulator rules)
(let* ((highlight-group (cdar rules))
(colorstr (car highlight-group))
(grouprules (cadr highlight-group))
(acc* (append (class-of-rules colorstr grouprules) accumulator))
(rest (cdr rules)))
(if (null? rest)
acc*
(do-loop acc* rest))))
; wrap the list in cond.
`(apply cond ,(do-loop '() rules)))
With given highlight-rules the function returns correct-looking list (well apart from applying the apply -- in clojure one would use splicing):
CSI> (parse-highlight-rules highlight-rules)
(apply cond (((string-match #<regexp>) (with-color "\x1b[1;31m" line))
((string-match #<regexp>) (with-color "\x1b[1;30m" line))
((string-match #<regexp>) (with-color #0="\x1b[1;33m" line))
((string-match #<regexp>) (with-color #0# line))))
But how to proceed with this? I've been stuck with this for a while. Chicken Scheme is my dialect.
The easiest way of transforming your function into a macro is by using Chicken's explicit-renaming macro facility, which works similarly to Clojure's defmacro (except that an explicit-renaming macro takes some additional arguments that can be used to preserve hygiene).
Splicing works basically the same way as it does in Clojure. The syntax is ,#. Therefore, the following should work:
(define-for-syntax (parse-highlight-rules rules)
;; ... insert missing code here ...
`(cond ,#(do-loop '() rules)))
(define-syntax highlight
(er-macro-transformer
(lambda (form rename compare)
(parse-highlight-rules (cdr form)))))