Merging Two Matrixes... in LISP - lisp

(defun merge-matrix (matrix-1 matrix-2)
(if (not (or (eql (matrix-rows matrix-1) (matrix-rows matrix-2)) (null matrix-1) (null matrix-2))) (error "Invalid dimensions."))
(cond
((null matrix-1) (copy-tree matrix-2))
((null matrix-2) (copy-tree matrix-1))
(t (let ((result (copy-tree matrix-1)))
(dotimes (i (matrix-rows matrix-1))
(setf (nth i result) (nconc (nth i result) (nth i matrix-2))))
result))))
(merge-matrix '((3 1) (1 3)) '((4 2) (1 1)))
*** - EVAL: variable NULL has no value
I receive an error like that how I can fix the problem, thanks

The OP's code works for me. However I felt motivated to improve it and
I implemented the same idea (but a bit more powerful).
The semantics are the same as Matlab's vertcat.
The function appends all arguments into one big matrix.
Note that due to the declarations my code should be super efficient.
(deftype mat ()
"Non-square matrices. Last index is columns, i.e. row-major order."
`(simple-array single-float 2))
(defun are-all-elements-typep (type ls)
(reduce #'(lambda (b x) (and b (typep x type)))
ls))
(defun are-all-matrix-heights-equalp (ls)
(let ((first-height (array-dimension (first ls) 0)))
(reduce #'(lambda (b x) (and b
(= first-height
(array-dimension x 0))))
ls)))
(defun vertcat (&rest rest)
(declare (type cons rest))
(unless (are-all-elements-typep 'mat rest)
(break "At least one of the arguments isn't a matrix."))
(unless (are-all-matrix-heights-equalp rest)
(break "All Matrices must have the same number of rows."))
(let* ((height (array-dimension (first rest) 0))
(widths (mapcar #'(lambda (mat) (array-dimension mat 1)) rest))
(result (make-array (list height
(reduce #'+ widths))
:element-type 'single-float))
(current-width 0))
(dotimes (m (length rest))
(let ((e (elt rest m)))
(destructuring-bind (y x) (array-dimensions e)
(dotimes (j y)
(dotimes (i x)
(setf (aref result j (+ current-width i))
(aref e j i))))
(incf current-width (elt widths m)))))
(the mat result)))
#+nil
(let ((a (make-array '(2 3)
:initial-contents '((1s0 2s0 3s0)
(2s0 4s0 5s0))
:element-type 'single-float))
(b (make-array '(2 2)
:initial-contents '((6s0 7s0)
(9s0 8s0))
:element-type 'single-float)))
(vertcat a b a))
;=> #2A ((1.0 2.0 3.0 6.0 7.0 1.0 2.0 3.0) (2.0 4.0 5.0 9.0 8.0 2.0 4.0 5.0))

The error message you're getting suggests that lisp is trying to treat one of your calls to null as a variable. I was able to replicate this behavior by defining matrix-rows like Frank Shearar did and deleting the parentheses around the ((null matrix-1) (copy-tree matrix-2)) s-expression, for example. I'd suggest you check your parentheses, either manually or using something like SLIME, which gave me a warning when I tried to compile the function.

Related

Did anybody write when-let-cond?

I was thinking about a cond with a twist
(let ((a 0))
(let* ((result nil))
(tagbody
(let ((b1 (+ 0 a)))
(when (eq b1 1)
(print "1")
(setf result b1)
(go finish)))
(let ((b2 (+ 0 a)))
(when (eq b2 2)
(print "2")
(setf result b2)
(go finish)))
(when T
(print "else")
(setf result a))
(format t "=== ~A~%" a)
finish)
result))
where when test-form is wrapped in let. On one hand this seems to fit into a problem I am working on, but also seems overcomplicated. Can it be simplified with a macro? What would be the best way to simplify it if I had lots of test-forms?
Part of the problem in trying to do it that way is restricting the let blocks to only one test-form and its body.
But I wonder if I am going down the wrong path. Playing with an imaginary variant of when-let suggests there is no benefit of going down this path.
Trying cond
The version using cond appears to be more compact.
(let ((a 3))
(let* ((b1 (+ 0 a))
(b2 (+ 0 a)))
(cond
((eq b1 1)
(print "1")
b1)
((eq b2 2)
(print "2")
b2)
(T (print "else")
a))))
All boils down to the variables defined in the let* which in real life example would be used to avoid calculating the same value twice and improve readability. What should I do?
I'd prefer to think more in terms of blocks and returning values from them, instead working with goto and variables. If one really needs separate let-bound variables and their own scope:
(prog ((a 0))
(let ((b1 (+ 0 a)))
(when (eql b1 1)
(print "1")
(return b1)))
(let ((b2 (+ 0 a)))
(when (eql b2 2)
(print "2")
(return b2)))
(return
(progn
(print "else")
(return a))))
Somebody did now. I wanted it to be compatible with cond which raises a trouble: if you want the binding clauses to be like
(cond/binding
...
((var expr) <use var>)
...)
But you want to allow just general test clauses, then a function with one argument is ambiguous: should
(cond/binding
...
((car x) ...)
...)
call car or bind car? To make this work then you need to bind a useless variable in that case:
(cond/binding
...
((useless (car x)) <useless not used here>)
...)
And that means you either need to insert ignore or ignorable declarations all over the place, or live with compiler warnings.
So, well, I decided it would be better to go the other way: you have to say when you want to bind a variable. And you do that by a clause like:
(cond/binding
...
((bind var expr) <var is bound here>)
...)
And note that bind is magic in the syntax (so this means you can't call a function called bind, but that's OK as I already use bind as a keyword in other macros.
The macro also tries hard (well, hard given I basically just typed it in and it's had no testing) to actually behave like cond: returning multiple values, for instance.
So this:
(cond/binding
((f x y z) t)
((bind x 3) (print x) (values x t))
(t (values nil nil))
(1))
expands to
(block #:cond/binding
(when (f x y z)
(return-from #:cond/binding (progn t)))
(let ((x 3))
(when x
(return-from #:cond/binding
(progn (print x) (values x t)))))
(when t
(return-from #:cond/binding (progn (values nil nil))))
(let ((r 1))
(when r
(return-from #:cond/binding r))))
(where all the blocks are the same block).
So, here:
(defmacro cond/binding (&body clauses)
;; Like COND but it can bind variables. All clauses are (should be)
;; like COND, except that a clause of the form ((bind var <expr>)
;; ...) will bind a variable. Note that bind has to be literally
;; the symbol BIND: it's magic in the syntax.
(let ((bn (make-symbol "COND/BINDING")))
`(block ,bn
,#(mapcar
(lambda (clause)
(unless (consp clause)
(error "bad clause ~S" clause))
(case (length clause)
(1
`(let ((r ,(car clause)))
(when r (return-from ,bn r))))
(otherwise
(destructuring-bind (test/binding &body forms) clause
(typecase test/binding
(cons
(case (car test/binding)
((bind)
(unless (and (= (length test/binding) 3)
(symbolp (second test/binding)))
(error "bad binding clause ~S" test/binding))
(destructuring-bind (var expr) (rest test/binding)
`(let ((,var ,expr))
(when ,var
(return-from ,bn
(progn ,#forms))))))
(otherwise
`(when ,test/binding
(return-from ,bn
(progn ,#forms))))))
(t
`(when ,test/binding
(return-from ,bn
(progn ,#forms)))))))))
clauses))))
Caveat emptor.
If I understand you problem correctly, then you can use or and rely on the fact that when is evaluated to nil if the condition is not true, e.g.,
(defun example (a)
(or
(let ((b1 (+ 0 a)))
(when (eql b1 1)
(print "1")
b1))
(let ((b2 (+ 0 a)))
(when (eql b2 2)
(print "2")
b2))
(progn
(print "else")
a)))
Using macrolet is the best solution so far. That allows me to bypass the limitations of when-let and not all bindins in the let form have to evaluate to true.
(let ((a 3))
(let ((result nil))
(macrolet ((ret-go (res)
`(progn
(setf result ,res)
(go finish))))
(tagbody
(let ((b1 (+ 0 a)))
(when (eq b1 1)
(print "1")
(ret-go b1)))
(let ((b2 (+ 0 a)))
(when (eq b2 2)
(print "2")
(ret-go b2)))
(when T
(print "else")
(setf result a))
(format t "=== ~A~%" a)
finish)
result)))

LISP function which, given a number and a list, returns the first even number greater than n

I'm having trouble finding my error.
This keeps returning nil:
(even-greater-n 5 '(1 2 3 4 5 6 7))
(defun even-greater-n (n L)
(cond ((null L) nil)
((and (> (car L) n) (evenp n)) (car L))
(t (even-greater-n n (cdr L)))))
Your error
You are passing to evenp n
instead of (car L).
Iteration
This is relatively easy to implement using
loop:
(defun even-greater (n l)
(loop for k in l
when (and (< n k)
(evenp k))
return k))
(even-greater 5 '(1 2 3 4 5 6 7 8))
==> 6
Recursion
If you are required to use recursion, you can do it too:
(defun even-greater (n l)
(cond ((endp l) nil)
((and (< n (first l))
(evenp (first l)))
(first l))
(t (even-greater n (rest l)))))
(even-greater 3 '(1 2 3 4 5 6 7 8))
==> 4
Library
And, of course, Lisp has a very powerful library, including
find-if:
(defun even-greater (n l)
(find-if (lambda (k)
(and (< n k)
(evenp k)))
l))
(even-greater 2 '(1 2 3 4 5 6 7 8))
==> 4
You must look for (car L) is even or not.
Using find-if and a single, open-coded lambda function:
(defun even-greater (n list)
(find-if (lambda (item) (and (> item n) (evenp item))) list))
Using functional combinators:
;; Combine multiple functions with AND:
;; Returns a function of one-argument which
;; passes that argument to the functions in the list,
;; one by one. If any function returns nil, it stops
;; and returns nil. Otherwise it returns the value
;; returned by the last function:
(defun andf (&rest functions)
(lambda (arg)
(let (res)
(dolist (f functions res)
(unless (setf res (funcall f arg))
(return))))))
;; Returns a one-argument function which tests
;; whether its argument is greater than quant.
(defun greater (quant)
(lambda (arg) (> arg quant)))
;; "find it, if it is greater than n, and even"
(defun even-greater (n list)
(find-if (andf (greater n) #'evenp) list))

Getting a "bad binding form" error in LISP

I have to write a simple program in Lisp that multiplies a polynomial by some factor. In this example, I want to multiply (x + 5) * 5x. The answer should be 5x^2 + 25x.
When I put in ((1 1) (5 0)) (5 1)) I should get (5 2) (25 1). However, I'm getting various errors ranging from undefined operator TERM in (TERM) and bad binding form. I'm a novice at Lisp and trying to return a list as shown above. Below is my short block of code:
(defun get-coef (term)
(car term))
(defun get-power (term)
(cadr term))
(defun make-term (coef power)
(cons coef power))
(defun poly-eval (poly factor)
(if (null poly) 0
(let ((term (car poly))
(let (coef ((* (get-coef(term)) (get-coef(factor)))))
(power ((+ (cadr(term)) (cadr(factor)))))
(make-term (coef power))
(poly-eval (cdr poly) factor))))))
Any help is appreciated!!
Several problems with your code:
You are using (fun (arg1 arg2)) syntax. It should be (fun arg1 arg2). For example, you write (make-term (coef power)) but it should be (make-term coef power).
Your bindings in let are all over the place. The correct syntax is
(let ((v1 e1)
(v2 e2)
(v3 e3))
e0)
i.e. all the bindings are in one list, and each binding is a list of two elements. Note that the expressions that the variables are bound to (e1 etc.) are not wrapped in any extra layers of parentheses.
make-term doesn't use the same representation as get-power. In get-power you use cadr so you need to make sure make-term puts the power in the right position.
Your poly-eval doesn't actually combine (make-term coef power) with the recursive call to (poly-eval (cdr poly) factor), so it gets lost. You should cons the "here"-result to the "there"-result.
Your poly-eval returns 0 instead of the empty list for empty polynomials.
All in all, your code can be fixed as
(defun get-coef (term)
(car term))
(defun get-power (term)
(cadr term))
(defun make-term (coef power)
(list coef power))
(defun poly-eval (poly factor)
(if (null poly) nil
(let ((term (car poly)))
(let
((coef (* (get-coef term) (get-coef factor)))
(power (+ (get-power term) (get-power factor))))
(cons (make-term coef power)
(poly-eval (cdr poly) factor))))))
giving e.g.
(poly-eval '((1 1) (5 0)) '(5 1))
resulting in
((5 2) (25 1))
Your make-term uses CONS but your get-power takes the CADR:
(defun get-power (term) (cadr term))
(defun make-term (coef power) (cons coef power))
You prolly wanted (list coef power).
(cons 'c 'p) returns (c . p), not (c p).
Now your get-power goes for CADR, the CAR of the CDR, but the CDR is 'p.
Your inputs are lists of coeff and power eg (5 1), so it seems the only problem is in your make-term.
Or you can turn around and be consistent with (( 5 . 1)(5 . 0) and then change get power to be (cdr term).
Another way:
(defun mult(term factor)
(list (* (first term) (first factor)) (+ (second term) (second factor))))
(defun polyeval(poly factor)
(cond
((null poly) nil)
(t (cons (mult (first poly) factor) (polyeval (rest poly) factor)))))
Note: first=car, rest=cdr, second=cadr

Recursing Through Nested List LISP

How would I recurse through nested lists?
For example, given: '((A 1 2) (B 3 4))
How would I add 2 to the second element in each nested sublist?
(defun get-p0 (points)
(loop for x from 0 to
(- (list-length points) 1) do
(+ 2 (cadr (nth x points)))
)
)
I'm not really sure why (get-p0 '((A 1 2) (B 3 4))) returns NIL.
I'd go with something like this:
(loop for (letter x y) in '((A 1 2) (B 3 4))
collect (list letter (+ 2 x) y))
The reason: it's shorter and you don't measure the length of the list in order to iterate over it (why would you do that?)
Since you ask for a recursive solution:
(defun get-p0 (lst &optional (n 0))
(if (null lst)
nil
(let ((elt1 (first lst)) (eltn (cdr lst)))
(if (listp elt1)
(cons (get-p0 elt1) (get-p0 eltn))
(cons (if (= n 1) (+ elt1 2) elt1) (get-p0 eltn (+ n 1)))))))
so
? (get-p0 '((A 1 2) (B 3 4)))
((A 3 2) (B 5 4))
and it recurses further down if necessary:
? (get-p0 '((A 0 2) ((B -4 4) (C 10 4))))
((A 2 2) ((B -2 4) (C 12 4)))
The way you put it, you can consider the problem as a basic recursion pattern: you go through a list using recursion or iteration (mapcar, reduce, etc.; dolist, loop, etc.) and apply a function to its entries. Here is a functional solution:
(defun get-p0 (points)
(mapcar #'add-2 points))
where the auxiliary function can be defined as follows:
(defun add-2 (lst)
"Add 2 to the 2nd item"
(let ((res '()))
(do ((l lst (cdr l))
(i 1 (1+ i)))
((null l) (nreverse res))
(push (if (= 2 i)
(+ 2 (car l))
(car l))
res))))
As written your 'loop' use does not return anything; thus NIL is returned. As is your code is simply iterating over x and computing something; that something isn't stored anywhere.
So, how to get your desired result? Assuming you are willing to modify each point in points, this should work:
(defun get-p0 (points)
(loop for x from 0 to (- (list-length points) 1) do
(let ((point (nth x points)))
(setf (cadr point) (+ 2 (cadr point)))))
points)

Possible to have more than 2 values in equality checks? (> NUM1 NUM2) ELISP

The function "greaterthan", (< NUM1 NUM2), allows only for returning t/nil for comparing 2 values.
I would like to test (var1 > var2 < var3 < var4), is there any way to do that using only one function in lisp? If not, what is the best procedure?
The best procedure is not to bother: (and (< var2 var1) (< var2 var3) (< var3 var4)) is not harder to read that your ..>..<..<.. chain.
It makes sense to test for the ascending order:
(require 'cl)
(defun cl-< (&rest args)
(every '< args (cdr args))
These days I don't hesitate to (require 'cl) anymore, but if you do,
here is another variant:
(defun cl-< (arg &rest more-args)
(or (null more-args)
(and (< arg (first more-args))
(apply #'cl-< more-args))))
The following is a macro implementation for variadic <
(defmacro << (x y &rest args)
(if args
(if (or (symbolp y)
(numberp y))
`(and (< ,x ,y) (<< ,y ,#args))
(let ((ys (make-symbol "y")))
`(let (,ys)
(and (< ,x (setq ,ys ,y))
(<< ,ys ,#args)))))
`(< ,x ,y)))
for simple cases just expands to (and ...) chains
(<< x y z) ==> (and (< x y) (< y z))
where the expression is not a number and not a symbol expands to a more complex form to avoid multiple evaluations in presence of side effects
(<< (f x) (g y) (h z)) ==> (let ((gy)) (and (< (f x) (setq gy (g y)))
(< gy (h z))))
for example
(setq foo (list))
nil
(defun call (x) (push x foo) x)
call
(<< (call 1) (call 2) (call 5) (call 4) (call 0))
nil
foo
(4 5 2 1)
every function has been called once, except for 0 that didn't need to be called because of short circuiting (I'm not 100% sure if short circuiting is a really good idea or not... #'< in Common Lisp is a regular function with all arguments all evaluated exactly once in left-to-right order without short circuiting).
(defun << (arg1 arg2 arg3 arg4)
(when (and (< arg1 arg2) (< arg2 arg3) (< arg3 arg4)))
)
(<< 1 2 3 4)
Probably possible to extend with any amount of arguments, but such a general form would seem useful.
(defmacro << (&rest args)
(let ((first (car args))
(min (gensym))
(max (gensym))
(forms '(t)) iterator)
(setq args (reverse (cdr args))
iterator args)
`(let ((,min ,first) ,max)
,(or
(while iterator
(push `(setq ,min ,max) forms)
(push `(< ,min ,max) forms)
(push `(setq ,max ,(car iterator)) forms)
(setq iterator (cdr iterator))) `(and ,#forms)))))
(macroexpand '(<< 10 20 30 (+ 30 3) (* 10 4)))
(let ((G99730 10) G99731)
(and (setq G99731 20)
(< G99730 G99731)
(setq G99730 G99731)
(setq G99731 30)
(< G99730 G99731)
(setq G99730 G99731)
(setq G99731 (+ 30 3))
(< G99730 G99731)
(setq G99730 G99731)
(setq G99731 (* 10 4))
(< G99730 G99731)
(setq G99730 G99731) t))
This is the idea similar to 6502's, but it may create less code, in a less trivial situation, but it will create more code in a trivial situation.