What I have to do is removing some elements from the list,the 1st,2nd,4th,8th,elements on positions power of 2.I figured out that the easyest way for me to solve this is to construct how the result list should look like without destroying the original list.Here's my code but it doesn't work yet,I'm getting a type error.I'm using contor to know with which element of the list I'm working with an counter to specify only the position from which the elements should be removed.My question is what am I doing wrong and how can it be fixed?
(defun remo(l)
(defparameter e ())
(setq contor 0)
(setq counter 0)
(dolist (elem l) (
(cond
(
((or (< (expt 2 contor) counter) (> (expt 2 contor) counter))
((push elem e) (setq contor (+ 1 contor))))
))
(setq counter (+1 counter))
)
)
(print e)
)
(defun remo (l)
(do ((power-of-2 1)
(counter 1 (1+ counter))
(result ())
(sublist l (cdr sublist)))
((null sublist) (nreverse result))
(if (= counter power-of-2)
(setq power-of-2 (* 2 power-of-2))
(push (car sublist) result))))
(remo '(1 2 3 4 5 6 7 8 9 10))
=> (3 5 6 7 9 10)
I already improved another of your attempts at https://stackoverflow.com/a/20711170/31615, but since you stated the real problem here, I propose the following solution:
(defun remove-if-index-power-of-2 (list)
(loop :for element :in list
:for index :upfrom 1 ; correct for language: "1st" is index 0
:unless (power-of-2-p index)
:collect element))
(defun power-of-2-p (number)
"Determines whether number, which is assumed to be a nonnegative
integer, is a power of 2 by counting the bits."
(declare (type (integer 0 *) number))
(= 1 (logcount number)))
Related
I have a list who's length is divisible by two, and I'm looking for something similar to the answer to this question:
(loop for (a b) on lst while b
collect (+ a b))
However there is overlap between elements:
(1 2 3 4 5) -> (3 5 7 9)
adding 1 and 2 and then 2 and 3 etc.
Where as I have a list like (1 2 3 4) and am looking for something like
((1 2) (3 4))
as output. Is there a way to make loop step correctly over the list?
Another solution.
Something like this should work:
(let ((list '(1 2 3 4)))
(loop :for (a b) :on list :by #'cddr :while b
:collect (cons a b)))
Also a more verbose variant:
(let ((list '(1 2 3 4)))
(loop :for a :in list :by #'cddr
:for b :in (cdr list) :by #'cddr
:collect (cons a b)))
Another approach using the SERIES package.
See also the user manual from Richard C. Waters.
Setup
(ql:quickload :series)
(defpackage :stackoverflow (:use :series :cl))
(in-package :stackoverflow)
Code
(defun pairs (list)
(collect 'list
(mapping (((odd even) (chunk 2 2 (scan 'list list))))
(list odd even))))
scan the content of list as a "serie"
chunk it with M=2 and N=2:
This function has the effect of breaking up the input series items
into (possibly overlapping) chunks of length m. The starting positions
of successive chunks differ by n. The inputs m and n must both be
positive integers.
More precisely, (chunk 2 2 (scan '(1 2 3 4))) produces #Z(1 3) and #Z(2 4)
mapping in parallel over each odd and even element of those series, produce a series of couples, as done by (list odd even).
finally, collect the result, as a list.
Compilation
All the intermediate "series" are compiled away thanks to a stream-fusion mechanism. Here is the macro expansion when pointing at collect:
(LET* ((#:OUT-1120 LIST))
(LET (#:ELEMENTS-1117
(#:LISTPTR-1118 #:OUT-1120)
(#:COUNT-1113 0)
#:CHUNK-1114
#:CHUNK-1115
#:ITEMS-1123
(#:LASTCONS-1106 (LIST NIL))
#:LST-1107)
(DECLARE (TYPE LIST #:LISTPTR-1118)
(TYPE FIXNUM #:COUNT-1113)
(TYPE CONS #:LASTCONS-1106)
(TYPE LIST #:LST-1107))
(SETQ #:COUNT-1113 1)
(SETQ #:LST-1107 #:LASTCONS-1106)
(TAGBODY
#:LL-1124
(IF (ENDP #:LISTPTR-1118)
(GO SERIES::END))
(SETQ #:ELEMENTS-1117 (CAR #:LISTPTR-1118))
(SETQ #:LISTPTR-1118 (CDR #:LISTPTR-1118))
(SETQ #:CHUNK-1114 #:CHUNK-1115)
(SETQ #:CHUNK-1115 #:ELEMENTS-1117)
(COND ((PLUSP #:COUNT-1113) (DECF #:COUNT-1113) (GO #:LL-1124))
(T (SETQ #:COUNT-1113 1)))
(SETQ #:ITEMS-1123
((LAMBDA (ODD EVEN) (LIST ODD EVEN)) #:CHUNK-1114 #:CHUNK-1115))
(SETQ #:LASTCONS-1106
(SETF (CDR #:LASTCONS-1106) (CONS #:ITEMS-1123 NIL)))
(GO #:LL-1124)
SERIES::END)
(CDR #:LST-1107)))
CL-USER 156 > (loop with list = '(1 2 3 4)
while list
collect (loop repeat 2
while list
collect (pop list)))
((1 2) (3 4))
or
CL-USER 166 > (loop with list = '(1 2 3 4 5 6)
while (and list (cdr list))
collect (loop repeat 2 collect (pop list)))
((1 2) (3 4) (5 6))
CL-USER 167 > (loop with list = '(1 2 3 4 5 6 7)
while (and list (cdr list))
collect (loop repeat 2 collect (pop list)))
((1 2) (3 4) (5 6))
I'm trying to get the largest sublist from a list using Common Lisp.
(defun maxlist (list)
(setq maxlen (loop for x in list maximize (list-length x)))
(loop for x in list (when (equalp maxlen (list-length x)) (return-from maxlist x)))
)
The idea is to iterate through the list twice: the first loop gets the size of the largest sublist and the second one retrieves the required list. But for some reason I keep getting an error in the return-from line. What am I missing?
Main problem with loop
There are a few problems here. First, you can write the loop as the following. There are return-from and while forms in Common Lisp, but loop defines its own little language that also recognizes while and return, so you can just use those:
(loop for x in list
when (equalp maxlen (list-length x))
return x)
A loop like this can actually be written more concisely with find though. It's just
(find maxlen list :key list-length :test 'equalp)
Note, however, that list-length should always return a number or nil, so equalp is overkill. You can just use eql, and that's the default for find, so you can even write
(find maxlen list :key list-length)
list-length and maximize
list-length is a lot like length, except that if a list has circular structure, it returns nil, whereas it's an error to call length with an improper list. But if you're using (loop ... maximize ...), you can't have nil values, so the only case that list-length handles that length wouldn't is one that will still give you an error. E.g.,
CL-USER> (loop for x in '(4 3 nil) maximize x)
; Evaluation aborted on #<TYPE-ERROR expected-type: REAL datum: NIL>.
(Actually, length works with other types of sequences too, so list-length would error if you passed a vector, but length wouldn't.) So, if you know that they're all proper lists, you can just
(loop for x in list
maximizing (length x))
If they're not all necessarily proper lists (so that you do need list-length), then you need to guard like:
(loop for x in list
for len = (list-length x)
unless (null len) maximize len)
A more efficient argmax
However, right now you're making two passes over the list, and you're computing the length of each sublist twice. Once is when you compute the maximum length, and the other is when you go to find one with the maximum value. If you do this in one pass, you'll save time. argmax doesn't have an obvious elegant solution, but here are implementations based on reduce, loop, and do*.
(defun argmax (fn list &key (predicate '>) (key 'identity))
(destructuring-bind (first &rest rest) list
(car (reduce (lambda (maxxv x)
(destructuring-bind (maxx . maxv) maxxv
(declare (ignore maxx))
(let ((v (funcall fn (funcall key x))))
(if (funcall predicate v maxv)
(cons x v)
maxxv))))
rest
:initial-value (cons first (funcall fn (funcall key first)))))))
(defun argmax (function list &key (predicate '>) (key 'identity))
(loop
for x in list
for v = (funcall function (funcall key x))
for maxx = x then maxx
for maxv = v then maxv
when (funcall predicate v maxv)
do (setq maxx x
maxv v)
finally (return maxx)))
(defun argmax (function list &key (predicate '>) (key 'identity))
(do* ((x (pop list)
(pop list))
(v (funcall function (funcall key x))
(funcall function (funcall key x)))
(maxx x)
(maxv v))
((endp list) maxx)
(when (funcall predicate v maxv)
(setq maxx x
maxv v))))
They produce the same results:
CL-USER> (argmax 'length '((1 2 3) (4 5) (6 7 8 9)))
(6 7 8 9)
CL-USER> (argmax 'length '((1 2 3) (6 7 8 9) (4 5)))
(6 7 8 9)
CL-USER> (argmax 'length '((6 7 8 9) (1 2 3) (4 5)))
(6 7 8 9)
Short variant
CL-USER> (defparameter *test* '((1 2 3) (4 5) (6 7 8 9)))
*TEST*
CL-USER> (car (sort *test* '> :key #'length))
(6 7 8 9)
Paul Graham's most
Please, consider also Paul Graham's most function:
(defun most (fn lst)
(if (null lst)
(values nil nil)
(let* ((wins (car lst))
(max (funcall fn wins)))
(dolist (obj (cdr lst))
(let ((score (funcall fn obj)))
(when (> score max)
(setq wins obj
max score))))
(values wins max))))
This is the result of test (it also returns value that's returned by supplied function for the 'best' element):
CL-USER> (most #'length *test*)
(6 7 8 9)
4
extreme utility
After a while I came up with idea of extreme utility, partly based on Paul Graham's functions. It's efficient and pretty universal:
(declaim (inline use-key))
(defun use-key (key arg)
(if key (funcall key arg) arg))
(defun extreme (fn lst &key key)
(let* ((win (car lst))
(rec (use-key key win)))
(dolist (obj (cdr lst))
(let ((test (use-key key obj)))
(when (funcall fn test rec)
(setq win obj rec test))))
(values win rec)))
It takes comparison predicate fn, list of elements and (optionally) key parameter. Object with the extreme value of specified quality can be easily found:
CL-USER> (extreme #'> '(4 9 2 1 5 6))
9
9
CL-USER> (extreme #'< '(4 9 2 1 5 6))
1
1
CL-USER> (extreme #'> '((1 2 3) (4 5) (6 7 8 9)) :key #'length)
(6 7 8 9)
4
CL-USER> (extreme #'> '((1 2 3) (4 5) (6 7 8 9)) :key #'cadr)
(6 7 8 9)
7
Note that this thing is called extremum in alexandria. It can work with sequences too.
Using recursion:
(defun maxim-list (l)
(flet ((max-list (a b) (if (> (length a) (length b)) a b)))
(if (null l)
nil
(max-list (car l) (maxim-list (cdr l))))))
The max-list internal function gets the longest of two list. maxim-list is getting the longest of the first list and the maxim-list of the rest.
I have a sorted list of integers, (1 2 4 5 6 6 7 8 10 10 10). I want to group them all, so that I get ((1) (2) (4) (5) (6 6) (7) (8) (10 10 10)).
So far I have this, which works:
(let ((current-group (list)) (groups (list)))
(dolist (n *sorted*)
(when (and (not (null current-group)) (not (eql (first current-group) n)))
(push current-group groups)
(setf current-group (list)))
(push n current-group))
(push current-group groups)
(nreverse groups))
But I'm sure there must be a much more LISPy way to do this. Any ideas?
Not that bad. I would write it this way:
(defun group (list)
(flet ((take-same (item)
(loop while (and list (eql (first list) item))
collect (pop list))))
(loop while list
collect (take-same (first list)))))
CL-USER 1 > (group '(1 2 4 5 6 6 7 8 10 10 10))
((1) (2) (4) (5) (6 6) (7) (8) (10 10 10))
There's already an accepted answer, but I think it's worth looking at another way of decomposing this problem, although the approach here is essentially the same). First, let's define cut that takes a list and a predicate, and returns the prefix and suffix of the list, where the suffix begins with the first element of the list that satisfies the predicate, and the prefix is everything before that that didn't:
(defun cut (list predicate)
"Returns two values: the prefix of the list
containing elements that do no satisfy predicate,
and the suffix beginning with an element that
satisfies predicate."
(do ((tail list (rest tail))
(prefix '() (list* (first tail) prefix)))
((or (funcall predicate (first tail))
(endp tail))
(values (nreverse prefix) tail))))
(cut '(1 1 1 2 2 3 3 4 5) 'evenp)
;=> (1 1 1) (2 2 3 3 4 5)
(let ((l '(1 1 2 3 4 4 3)))
(cut l (lambda (x) (not (eql x (first l))))))
;=> (1 1), (2 3 4 4 3)
Then, using cut, we can move down the an input list taking prefixes and suffixes with a predicate that's checking whether an element is not eql to the first element of the list. That is, beginning with (1 1 1 2 3 3) you'd cut with the predicate checking for "not eql to 1", to get (1 1 1) and (2 3 3). You'd add the first to the list of groups, and the second becomes the new tail.
(defun group (list)
(do ((group '()) ; group's initial value doesn't get used
(results '() (list* group results))) ; empty, but add a group at each iteration
((endp list) (nreverse results)) ; return (reversed) results when list is gone
(multiple-value-setq (group list) ; update group and list with the prefix
(cut list ; and suffix from cutting list on the
(lambda (x) ; predicate "not eql to (first list)".
(not (eql x (first list))))))))
(group '(1 1 2 3 3 3))
;=> ((1 1) (2) (3 3 3))
On implementing cut
I tried to make that cut relatively efficient, insofar as it only makes one pass through the list. Since member returns the entire tail of the list that begins with the found element, you can actually use member with :test-not to get the tail that you want:
(let ((list '(1 1 1 2 2 3)))
(member (first list) list :test-not 'eql))
;=> (2 2 3)
Then, you can use ldiff to return the prefix that comes before that tail:
(let* ((list '(1 1 1 2 2 3))
(tail (member (first list) list :test-not 'eql)))
(ldiff list tail))
;=> (1 1 1)
It's a simple matter, then, to combine the approaches and to return the tail and the prefix as multiples values. This gives a version of cut that takes only the list as an argument, and might be easier to understand (but it's a bit less efficient).
(defun cut (list)
(let ((tail (member (first list) list :test-not 'eql)))
(values (ldiff list tail) tail)))
(cut '(1 1 2 2 2 3 3 3))
;=> (1 1), (2 2 2 3 3)
I like to use reduce:
(defun group (lst)
(nreverse
(reduce (lambda (r e) (if (and (not (null r)) (eql e (caar r)))
(cons (cons e (car r)) (cdr r))
(cons (list e) r)))
lst
:initial-value nil)))
or using push:
(defun group (lst)
(nreverse
(reduce (lambda (r e)
(cond
((and (not (null r)) (eql e (caar r))) (push e (car r)) r)
(t (push (list e) r))))
lst
:initial-value nil)))
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)
Is there a way to do this:
(defvar long-list ((1 1 1 1) (2 2 2 2) (3 3 3 3)
(4 4 4 4) (5 5 5 5) (6 6 6 6))
(format t "magic" long-list)
To output something like:
(1 1 1 1) (2 2 2 2) (3 3 3 3)
(4 4 4 4) (5 5 5 5) (6 6 6 6)
Where I would define the number of columns to print?
I know about (format t "~/my-function/" long-list) option, but maybe there's something built-in?
The reference is being highly unhelpful on this particular topic.
OK, sorry, I actually found it: http://www.lispworks.com/documentation/lw51/CLHS/Body/f_ppr_fi.htm#pprint-tabular but before I found it, I wrote this:
(defun pplist-as-string (stream fmt colon at)
(declare (ignore colon at))
(dolist (i fmt)
(princ i stream)))
(defun ppcolumns (stream fmt colon at cols)
(declare (ignore at colon))
(when (or (not cols) (< cols 1)) (setq cols 1))
(let* ((fmt-length (length fmt))
(column-height (floor fmt-length cols))
(remainder (mod fmt-length cols))
(printed 0)
columns
column-sizes)
(do ((c fmt (cdr c))
(j 0 (1+ j))
(r (if (zerop remainder) 0 1) (if (zerop remainder) 0 1))
(i 0 (1+ i)))
((null c))
(when (or (= j (+ r column-height)) (zerop i))
(setq columns (cons c columns)
column-sizes
(cons
(+ r column-height) column-sizes))
(unless (zerop remainder)
(unless (zerop i) (decf remainder)))
(setq j 0)))
(setq columns (reverse columns)
column-sizes (reverse column-sizes))
(when (/= fmt-length (* column-height cols))
(incf column-height))
(dotimes (i column-height)
(do ((c columns (cdr c))
(size column-sizes (cdr size)))
((or (null c)))
(when (> printed (1- fmt-length))
(return-from ppcolumns))
(when (< 0 (car size))
(pplist-as-string stream (caar c) nil nil)
(when (caar c) (incf printed))
(unless (null c) (princ #\ ))
(rplaca c (cdar c))))
(princ #\newline))))
which prints it in another direction. In case you would need it.