...just like packages do.
I use Emacs (maybe, it can offer some kind of solution).
For example (defun the-very-very-long-but-good-name () ...) is not to useful later in code. But the name like Fn-15 or the first letters abbreviation is not useful too.
Is it possible either to have an alias like for packages or to access the documentation string while trying to recall the function's name?
In other words, is it possible for functions to mix somehow self-documenting and short names?
You want defalias. (defalias 'newname 'oldname) will preserve documentation and even show "newname is an alias for `oldname'" when its documentation is requested.
You could use setf to assign the function to the function cell of another, for example:
(defmacro alias (new-name prev-name)
`(setf (symbol-function ,new-name) (symbol-function ,prev-name)))
from 《On Lisp》？Here is the code:
(defmacro alias (new-name prev-name)
`(defmacro ,new-name (&rest args)
`(,',prev-name ,#args)))
; use: (alias df defun)
(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)))
(defmacro aliasx (&rest names)
`(alias
,#(mapcar #'(lambda (pair)
`(alias ,#pair))
(group names 2))))
; use: (aliasx df1 defun
; df2 defun
; df3 defun)
If it's all the typing which makes continual use of long names undesirable, then yes, emacs can help. Check out abbrev-mode. Also well thought-of in this context is hippie-expand.
If it's a question of readability, that's harder.
If your problem is that you can't remember a very long function name, but you remember PART of the name, that's what "apropos" is for. In my Emacs, I have "C-h a" bound to "hyper-apropos". You enter a substring of the symbol you're looking for, and it lists all the matches.
I dont know Emacs, but wouldn't (define shortname longnamefunctionblahblah) work?
You could simply have a function that just calls another function.
you can use (defmacro ...) to alias a function
Related
I am trying to make a list of callback functions, which could look like this:
(("command1" . 'callback1)
("command2" . 'callback2)
etc)
I'd like it if I could could do something like:
(define-callback callback1 "command1" args
(whatever the function does))
Rather than
(defun callback1 (args)
(whatever the function does))
(add-to-list 'callback-info ("command1" . 'callback1))
Is there a convenient way of doing this, e.g., with macros?
This is a good example of a place where it's nice to use a two-layered approach, with an explicit function-based layer, and then a prettier macro layer on top of that.
Note the following assumes Common Lisp: it looks just possible from your question that you are asking about elisp, in which case something like this can be made to work but it's all much more painful.
First of all, we'll keep callbacks in an alist called *callbacks*:
(defvar *callbacks* '())
Here's a function which clears the alist of callbacks
(defun initialize-callbacks ()
(setf *callbacks* '())
(values)
Here is the function that installs a callback. It does this by searching the list to see if there is a callback with the given name, and if there is then replacing it, and otherwise installing a new one. Like all the functions in the functional layer lets us specify the test function which will let us know if two callback names are the same: by default this is #'eql which will work for symbols and numbers, but not for strings. Symbols are probably a better choice for the names of callbacks than strings, but we'll cope with that below.
(defun install-callback (name function &key (test #'eql))
(let ((found (assoc name *callbacks* :test test)))
(if found
(setf (cdr found) function)
(push (cons name function) *callbacks*)))
name)
Here is a function to find a callback, returning the function object, or nil if there is no callback with that name.
(defun find-callback (name &key (test #'eql))
(cdr (assoc name *callbacks* :test test)))
And a function to remove a named callback. This doesn't tell you if it did anything: perhaps it should.
(defun remove-callback (name &key (test #'eql))
(setf *callbacks* (delete name *callbacks* :key #'car :test test))
name)
Now comes the macro layer. The syntax of this is going to be (define-callback name arguments ...), so it looks a bit like a function definition.
There are three things to know about this macro.
It is a bit clever: because you can know at macro-expansion time what sort of thing the name of the callback is, you can decide then and there what test to use when installing the callback, and it does this. If the name is a symbol it also wraps a block named by the symbol around the body of the function definition, so it smells a bit more like a function defined by defun: in particular you can use return-from in the body. It does not do this if the name is not a symbol.
It is not quite clever enough: in particular it does not deal with docstrings in any useful way (it ought to pull them out of the block I think). I am not sure this matters.
The switch to decide the test uses expressions like '#'eql which reads as (quote (function eql)): that is to avoid wiring in functions into the expansion because functions are not externalisable objects in CL. However I am not sure I have got this right: I think what is there is safe but it may not be needed.
So, here it is
(defmacro define-callback (name arguments &body body)
`(install-callback ',name
,(if (symbolp name)
`(lambda ,arguments
(block ,name
,#body))
`(lambda ,arguments
,#body))
:test ,(typecase name
(string '#'string=)
(symbol '#'eql)
(number '#'=)
(t '#'equal))))
And finally here are two different callbacks being defined:
(define-callback "foo" (x)
(+ x 3))
(define-callback foo (x)
(return-from foo (+ x 1)))
These lists are called assoc lists in Lisp.
CL-USER 120 > (defvar *foo* '(("c1" . c1) ("c2" . c2)))
*FOO*
CL-USER 121 > (setf *foo* (acons "c0" `c1 *foo*))
(("c0" . C1) ("c1" . C1) ("c2" . C2))
CL-USER 122 > (assoc "c1" *foo* :test #'equal)
("c1" . C1)
You can write macros for that, but why? Macros are advanced Lisp and you might want to get the basics right, first.
Some issues with you example you might want to check out:
what are assoc lists?
what are useful key types in assoc lists?
why you don't need to quote symbols in data lists
variables are not quoted
data lists need to be quoted
You can just as easy create such lists for callbacks without macros. We can imagine a function create-callback, which would be used like this:
(create-callback 'callback1 "command1"
(lambda (arg)
(whatever the function does)))
Now, why would you use a macro instead of a plain function?
In the end, assisted by the responders above, I got it down to something like:
(defmacro mk-make-command (name &rest body)
(let ((func-sym (intern (format "mk-cmd-%s" name))))
(mk-register-command name func-sym)
`(defun ,func-sym (args &rest rest)
(progn
,#body))))
i'm new to common lisp and therefore my problem could be very easy, but i didn't find anything, maybe i used the wrong search terms.
i've got the following problem:
i have a function that does a special addition on an arbitrary number of parameters. the next step would be to apply that function to an arbitrary number of lists of same size, the result would be an list of that size.
it works if i call
(mapcar #'addition list1 list2 ...)
but if i have to define a function
(defun list-add (list &rest lists)
(mapcar #'addition list lists))
it won't work, because &rest lists now is a list of lists. the function addition needs to be called with all parameters as sequence, so a recursive call is not possible.
does anyone have a solution?
See APPLY.
Also note the value of CALL-ARGUMENTS-LIMIT.
The obvious solution would be:
(defun list-add (&rest lists)
(apply #'mapcar #'addition lists))
I'm not sure I got the question correctly but try
(defun list-add (list &rest lists)
(mapcar (lambda (l) (apply #'addition list l))
lists))
I'm not sure if this is necessarily better or worse than the answers already supplied, but here's what I came up with:
(defun list-add (first-required-list &rest other-lists)
(let ((all-lists (cons first-required-list
other-lists)))
(reduce (lambda (left-list right-list)
(mapcar #'addition left-list right-list))
all-lists)))
I'm trying to write a function which checks if every element in the list x has property a, so I wrote:
(defun check (a x)
(if (listp x)
(eval (cons 'and (mapcar #'a x)))))
but it doesn't work. (Basically I want a to be the name of a function, say blablabla, and in the body of the check-function, by #'a I want to mean the function blablabla, instead of a function called a.) Now the code above doesn't work. I think in Lisp one should be able to plug in functions. How can I fix it?
(It is literally my first day on lisp, so it might be a stupid question ;)
and BTW I'm using Lispworks 6.0 personal version.)
There is no need to use the sharp-quote syntax here. Its purpose is to use a function name in a variable position, but a is a variable already. Just write a instead of #'a.
You don't need eval you can use apply.
To the problem: You need funcall because you provide a as argument. (Edit: Not in this case.) By quoting you just refer to the function a not the a in this function.
(defun check (a xs)
(if (listp xs)
(every #'identity (mapcar a
xs))))
Better, use loop:
(defun check (a xs)
(if (listp xs)
(loop for x in xs
always (funcall a x))))
Best, use every:
(defun check (a xs)
(if (listp xs)
(every a xs)))
Here is how I would write something like your check function. I tried to give it a more descriptive name.
(defun are-all-elements-fullfilling-fun-p (fun ls)
(every #'identity (mapcar fun ls)))
Edit: Note that a shorter and better definition is
(defun are-all-elements-fullfilling-fun-p (fun ls)
(every fun ls)))
Now let's say we want to call it with this function. Note that I tend to use declarations when possible. I quite often screw something up and debugging is easy if the compiler can figure the error out. Also the code will run faster.
(defun is-even-p (n)
(declare (type number n))
(the boolean (= 0 (mod n 2))))
You have to place the #' here:
(are-all-elements-fullfilling-fun-p #'is-even-p '(1 2 3 4))
(are-all-elements-fullfilling-fun-p #'is-even-p '(38 2 4))
The only thing I don't like about Emacs is the lack of namespaces, so I'm wondering if I can implement them on my own.
This is my first attempt, and it's obvious that I can't just replace every match of a name with its prefixed version, but what should I check? I can check for bindings with (let) then mark the entire subtree, but what if somebody creates a (my-let) function that uses let? Is my effort destined to fail? :(
Also, why are my defuns failing to define the function? Do I have to run something similar to intern-symbol on every new token?
Thanks!
Since this is the first google result for elisp namespaces...
There's a minimalist implementation of namespaces called fakespace which you can get on elpa, which does basic encapsulation. I'm working on something ambitious myself, which you can check out here.
To handle things like my-let or my-defun, you need to macroexpand those definitions, e.g. with macroexpand-all.
For the failure to define the functions, you need to use intern instead of make-symbol (because make-symbol always creates a new distinct fresh uninterned symbol).
Adding namespaces will take more than prefixing the identifiers with the namespace names. The interpreter has to be able to tell the namespaces. Some tinkering must go into the interpreter as well. That might need to go through a thorough discussion at gnu.emacs.sources and/or #emacs at irc.freenode.org.
This is a fixed version of the code from #vpit3833 to provide namespace support (using the hint from #Stefan). It’s too good to leave around half-fixed :)
;; Simple namespace definitions for easier elisp writing and clean
;; access from outside. Pythonesque elisp :)
;;
;; thanks to vpit3833 → http://6e5e5ae9206fa093.paste.se/
(defmacro namespace (prefix &rest sexps)
(let* ((naive-dfs-map
(lambda (fun tree)
(mapcar (lambda (n) (if (listp n) (funcall naive-dfs-map fun n)
(funcall fun n))) tree)))
(to-rewrite (loop for sexp in sexps
when (member (car sexp)
'(defvar defmacro defun))
collect (cadr sexp)))
(fixed-sexps (funcall naive-dfs-map
(lambda (n) (if (member n to-rewrite)
(intern
(format "%s-%s" prefix n)) n))
sexps)))
`(progn ,#fixed-sexps)))
;; (namespace test
;; (defun three () 3)
;; (defun four () (let ((three 4)) three))
;; (defun + (&rest args) (apply #'- args)))
;; (test-+ 1 2 3)
(provide 'namespace)
I am working on a genetic programming hobby project.
I have a function/macro setup that, when evaluated in a setq/setf form, will generate a list that will look something like this.
(setq trees (make-trees 2))
==> (+ x (abs x))
Then it will get bound out to a lambda function #<FUNCTION :LAMBDA (X) ... > via strategic use of functions/macros
However, I want to get a bit more effective with this than manually assigning to variables, so I wrote something like this:
(setq sample
(let* ((trees (make-trees 2))
(tree-bindings (bind-trees trees))
(evaluated-trees (eval-fitness tree-bindings))))
(list (trees tree-bindings evaluated-trees)))
However, I get EVAL: trees has no value when I place this in a let form. My suspicion is that the macro expansions don't get fully performed in a LET as compared to a SETF, but that doesn't make sense to me.
What is the cause of this issue?
--- edit: yanked my code and put the whole file in a pastebin ---
Supposing that I decide that a setq isn't going to do it for me and I write a simple function to do it:
(defun generate-sample ()
(let ((twiggs (make-trees 2)))
(let ((tree-bindings (bind-trees twiggs)))
(let ((evaluated-trees (eval-fitness tree-bindings)))
(list twiggs tree-bindings evaluated-trees)))))
This yields an explosion of ...help file error messages (??!?)... and "eval: variable twiggs has no value", which stems from the bind-trees definition on SLIME inspection.
I am reasonably sure that I've completely hosed my macros. http://pastebin.org/673619
(Setq make-trees 2) sets the value of the variable make-trees to 2, then returns 2.
I do not see a reason for a macro in what you describe. Is it true that your make-trees creates a single random tree, which can be interpreted as a program? Just define this as a function with defun. I am thinking of something like this:
(defun make-tree (node-number)
(if (= node-number 1)
(make-leaf)
(cons (get-random-operator)
(mapcar #'make-tree
(random-partition (- node-number 1))))))
Let and setq do totally different things. Setq assigns a value to an existing variable, while let creates a new lexical scope with a number of lexical bindings.
I think that you should present more of your code; currently, your question does not make a lot of sense.
Update:
I will fix your snippet's indentation to make things clearer:
(setq sample
(let* ((trees (make-trees 2))
(tree-bindings (bind-trees trees))
(evaluated-trees (eval-fitness tree-bindings))))
(list (trees tree-bindings evaluated-trees)))
Now, as written before, let* establishes lexical bindings. These
are only in scope within its body:
(setq sample
(let* ((trees (make-trees 2))
(tree-bindings (bind-trees trees))
(evaluated-trees (eval-fitness tree-bindings)))
;; here trees, tree-bindings, and evaluated-trees are bound
) ; end of let* body
;; here trees, tree-bindings, and evaluated trees are not in scope anymore
(list (trees tree-bindings evaluated-trees)))
That last line is spurious, too. If those names were bound, it would
return a list of one element, which would be the result of evaluating
the function trees with tree-bindings and evaluated-trees as
arguments.
You might get what you want like this:
(setq sample
(let* ((trees (make-trees 2))
(tree-bindings (bind-trees trees))
(evaluated-trees (eval-fitness tree-bindings)))
(list trees tree-bindings evaluated-trees)))
Another update:
The purpose of macros is to eliminate repeated code when that elimination is not possible with functions. One frequent application is when dealing with places, and you also need them to define new control constructs. As long as you do not see that something cannot work as a function, do not use a macro for it.
Here is some code that might help you:
(defun make-tree-lambda (depth)
(list 'lambda '(x)
(new-tree depth)))
(defun make-tree-function (lambda-tree)
(eval lambda-tree))
(defun eval-fitness (lambda-form-list input-output-list)
"Determines how well the lambda forms approach the wanted function
by comparing their output with the wanted output in the supplied test
cases. Returns a list of mean quadratic error sums."
(mapcar (lambda (lambda-form)
(let* ((actual-results (mapcar (make-tree-function lambda-form)
(mapcar #'first input-output-list)))
(differences (mapcar #'-
actual-results
(mapcar #'second input-output-list)))
(squared-differences (mapcar #'square
differences)))
(/ (reduce #'+ squared-differences)
(length squared-differences))))
lambda-form-list))
(defun tree-fitness (tree-list input-output-list)
"Creates a list of lists, each inner list is (tree fitness). Input
is a list of trees, and a list of test cases."
(mapcar (lambda (tree fitness)
(list tree fitness))
tree-list
(eval-fitness (mapcar #'make-tree-lambda tree-list)
input-output-list)))