I could figure out some way to do this myself but I have a feeling there's a simpler, perhaps built-in way to do this. I want to see if any two lists share an element. These are the two lists I'm dealing with at the moment:
((0 1 2) (3 4 5) (6 7 8) (0 3 6) (1 3 7) (2 4 8) (0 4 8) (2 4 6))
((0 1 7) (0 1 6) (0 1 3) (0 3 7) (0 3 6) (0 6 7) (1 3 7) (1 3 6) (1 6 7) (3 6 7))
Since both lists contain (1 3 7), I'd like a comparison of the lists to return T.
Is there a better way to do this than just setting up a couple DOLISTs?
How about INTERSECTION?
(defvar a '((0 1 2) (3 4 5) (6 7 8) (0 3 6) (1 3 7) (2 4 8) (0 4 8) (2 4 6)))
=> A
(defvar b '((0 1 7) (0 1 6) (0 1 3) (0 3 7) (0 3 6) (0 6 7) (1 3 7) (1 3 6) (1 6 7) (3 6 7)))
=> B
(intersection a b :test 'equal)
=> ((1 3 7) (0 3 6))
Related
I would like to ask how can I merge 2 different lists of numbers to a new list keeping the "common points" between them in Common Lisp.
Example
list1: (1 2 3 2 2 )
List2: (1/2 1/2 1 2 2 1 2 1)
Result:(1/2 1/2 1 1 1 2 1 1 1 1)
I hope the image below can give an exact description of the problem.
The lists are numbers because it must compare the different units of the two series and further combine the points of start of each number of both series into a new serie.
Image_1. I think this image is the best way to describe the problem.
Based on your description, I wrote two mutually-recursive functions MRG and SPLIT:
MRG iterates over the first list, all calls SPLIT for each element
SPLIT tries to collect from the second list enough elements for which the sum is equal to the current element in the first list. If the element in the second list is too large, it is split and the remaining is reinjected into the second list. SPLIT also calls MRG when it has finished processing the current element in the first list.
Here is a trace of execution showing how the result is computed.
0: (MRG (1 2 3 2 2) (1/2 1/2 1 2 2 1 2 1))
1: (SPLIT 1 (1/2 1/2 1 2 2 1 2 1) (2 3 2 2))
2: (SPLIT 1/2 (1/2 1 2 2 1 2 1) (2 3 2 2))
3: (SPLIT 0 (1 2 2 1 2 1) (2 3 2 2))
4: (MRG (2 3 2 2) (1 2 2 1 2 1))
5: (SPLIT 2 (1 2 2 1 2 1) (3 2 2))
6: (SPLIT 1 (2 2 1 2 1) (3 2 2))
7: (SPLIT 0 (1 2 1 2 1) (3 2 2))
8: (MRG (3 2 2) (1 2 1 2 1))
9: (SPLIT 3 (1 2 1 2 1) (2 2))
10: (SPLIT 2 (2 1 2 1) (2 2))
11: (SPLIT 0 (1 2 1) (2 2))
12: (MRG (2 2) (1 2 1))
13: (SPLIT 2 (1 2 1) (2))
14: (SPLIT 1 (2 1) (2))
15: (SPLIT 0 (1 1) (2))
16: (MRG (2) (1 1))
17: (SPLIT 2 (1 1) NIL)
18: (SPLIT 1 (1) NIL)
19: (SPLIT 0 NIL NIL)
20: (MRG NIL NIL)
20: MRG returned NIL
19: SPLIT returned NIL
18: SPLIT returned (1)
17: SPLIT returned (1 1)
16: MRG returned (1 1)
15: SPLIT returned (1 1)
14: SPLIT returned (1 1 1)
13: SPLIT returned (1 1 1 1)
12: MRG returned (1 1 1 1)
11: SPLIT returned (1 1 1 1)
10: SPLIT returned (2 1 1 1 1)
9: SPLIT returned (1 2 1 1 1 1)
8: MRG returned (1 2 1 1 1 1)
7: SPLIT returned (1 2 1 1 1 1)
6: SPLIT returned (1 1 2 1 1 1 1)
5: SPLIT returned (1 1 1 2 1 1 1 1)
4: MRG returned (1 1 1 2 1 1 1 1)
3: SPLIT returned (1 1 1 2 1 1 1 1)
2: SPLIT returned (1/2 1 1 1 2 1 1 1 1)
1: SPLIT returned (1/2 1/2 1 1 1 2 1 1 1 1)
0: MRG returned (1/2 1/2 1 1 1 2 1 1 1 1)
I made no attempt to optimize the code, I just tried to do something that works correctly in a way that can produce a useful trace. But this looks like something for which a loop might work too.
Iterative version (edit)
Here is a version without recursion along with debugging statements:
(defun mrg% (lx ly)
(with-list-collector (collect)
(flet ((collect (v)
"Add print statements to COLLECT"
(print (list :collect v))
(collect v)))
(dolist (x lx)
(loop
(print (list :split x ly))
(unless (plusp x)
(return))
(assert ly)
(let ((y (pop ly)))
(if (<= y x)
(decf x (collect y))
(return (push (- y (collect x)) ly)))))))))
With your example:
(mrg% '(1 2 3 2 2 )
'(1/2 1/2 1 2 2 1 2 1))
... prints:
(:SPLIT 1 (1/2 1/2 1 2 2 1 2 1))
(:COLLECT 1/2)
(:SPLIT 1/2 (1/2 1 2 2 1 2 1))
(:COLLECT 1/2)
(:SPLIT 0 (1 2 2 1 2 1))
(:SPLIT 2 (1 2 2 1 2 1))
(:COLLECT 1)
(:SPLIT 1 (2 2 1 2 1))
(:COLLECT 1)
(:SPLIT 3 (1 2 1 2 1))
(:COLLECT 1)
(:SPLIT 2 (2 1 2 1))
(:COLLECT 2)
(:SPLIT 0 (1 2 1))
(:SPLIT 2 (1 2 1))
(:COLLECT 1)
(:SPLIT 1 (2 1))
(:COLLECT 1)
(:SPLIT 2 (1 1))
(:COLLECT 1)
(:SPLIT 1 (1))
(:COLLECT 1)
(:SPLIT 0 NIL)
For completeness, here is the macro I am using:
(defmacro with-list-collector
((collector-name &optional name copy-p) &body body)
"Bind COLLECTOR-NAME as a local function to collect items in a list.
A call to (COLLECTOR-NAME VALUE) accumulates VALUE into a list, in the
same order as the calls are being made. The resulting list can be
accessed through the symbol NAME, if given, or as the return value of
WITH-LIST-COLLECTOR.
The return value of (COLLECTOR-NAME VALUE) is VALUE.
If COPY-P is T, each access to NAME performs a copy of the list under
construction. Otherwise, NAME refers to a list which last cons-cell is
modified after each call to COLLECTOR-NAME (except if it is NIL).
The return value of the whole form is the list being built, ONLY when
NAME is NIL. Otherwise, the return value is given by the last form of
BODY: it is assumed that the list will be accessed by NAME if
necessary, and that the interesting value is given by BODY."
(assert (or (not copy-p) name) ()
"A COPY argument is only valid when a NAME is given.")
(alexandria:with-gensyms (queue head value)
(let ((flet-expr `(flet ((,collector-name (,value)
(prog1 ,value
(setf ,queue
(setf (cdr ,queue)
(cons ,value nil))))))
(declare (inline ,collector-name))
,#body)))
`(let* ((,queue (cons nil nil))
(,head ,queue))
,(if name
`(symbol-macrolet
((,name ,(if copy-p
`(copy-seq (cdr ,head))
`(cdr ,head))))
,flet-expr)
;; anonymous list : return as result
`(progn ,flet-expr
(cdr ,head)))))))
It seems to me that the list elements are like pauses between beats. My algorithm would at each step look for the minimum pause, then reduce the remaining current pauses by that and advance the lists when their current pause is zero.
To illustrate, I put a print instruction into the loop:
(defun merge-beats (&rest lists)
(do* ((minpause nil (reduce #'min (mapcar #'first pauses)))
(result () (cons minpause result))
(pauses lists
(remove nil
(mapcar (lambda (pause-list)
(let ((current-pause (- (first pause-list)
minpause)))
(if (zerop current-pause)
(rest pause-list)
(cons current-pause
(rest pause-list)))))
pauses)))
(- #1=(print (list :minpause minpause :result result :pauses pauses))
#1#))
((endp pauses) (nreverse result))))
CL-USER> (merge-beats '(1 2 3 2 2)
'(1/2 1/2 1 2 2 1 2 1))
(:MINPAUSE NIL :RESULT NIL :PAUSES ((1 2 3 2 2) (1/2 1/2 1 2 2 1 2 1)))
(:MINPAUSE 1/2 :RESULT (1/2) :PAUSES ((1/2 2 3 2 2) (1/2 1 2 2 1 2 1)))
(:MINPAUSE 1/2 :RESULT (1/2 1/2) :PAUSES ((2 3 2 2) (1 2 2 1 2 1)))
(:MINPAUSE 1 :RESULT (1 1/2 1/2) :PAUSES ((1 3 2 2) (2 2 1 2 1)))
(:MINPAUSE 1 :RESULT (1 1 1/2 1/2) :PAUSES ((3 2 2) (1 2 1 2 1)))
(:MINPAUSE 1 :RESULT (1 1 1 1/2 1/2) :PAUSES ((2 2 2) (2 1 2 1)))
(:MINPAUSE 2 :RESULT (2 1 1 1 1/2 1/2) :PAUSES ((2 2) (1 2 1)))
(:MINPAUSE 1 :RESULT (1 2 1 1 1 1/2 1/2) :PAUSES ((1 2) (2 1)))
(:MINPAUSE 1 :RESULT (1 1 2 1 1 1 1/2 1/2) :PAUSES ((2) (1 1)))
(:MINPAUSE 1 :RESULT (1 1 1 2 1 1 1 1/2 1/2) :PAUSES ((1) (1)))
(:MINPAUSE 1 :RESULT (1 1 1 1 2 1 1 1 1/2 1/2) :PAUSES NIL)
(1/2 1/2 1 1 1 2 1 1 1 1)
CL-USER>
i am programming in Common lisp, and i need an algorithm that delete sublist with (at least) two common elements.
i don't have any idea to how i can solve the problem. I thought to use this:
(defun remove-duplicates-list (list) (remove-duplicates list :test 'equal :key 'cdr))
but the cdr between the sublists is different, and i just can't figure out how i can count two elements and remove the sublist.
Also i need to don't remove the "original" sublist (in the example '(1 2 3 4)), bot only (2 3 4) and (1 3 4)
Example :
input: '( (1 2 3 4) (2 3 4) (5 6 7 8) (1 3 4) (9 9 9 9) )
output: '( (1 2 3 4) (5 6 7 8) (9 9 9 9) )
input: '( ((1 . 1) (2 . 2) (3 . 3) (4 . 4)) ((1 . 1) (2 . 2) (4 . 4)))
output : '((1 . 1) (2 . 2) (3 . 3) (4 . 4))
Thank you!
P.s sorry if, at first, i have not explained well my problem, i correct many mistake in the post. this is my first question on this community, so forgive please
Actually you could do it with remove-duplicates. You need to make a test function that returns true when at least two elements are similar. eg.
(defun two-similar-p (lst1 lst2)
...)
(two-similar-p '(1 2 3) '(1 4 5)) ; ==> nil
(two-similar-p '(1 2 5) '(1 4 5)) ; ==> t
Using a hash is the fastest and best time complexity while iterating one list (length other-list) times is possibly the easiest. Then you can solve your problem this way:
(defun remove-duplicates-list (list)
(remove-duplicates list :test #'two-similar-p :from-end t))
(remove-duplicates-list '((1 2 3 4) (2 3 4) (5 6 7 8) (1 3 4) (9 9 9 9)))
; ==> ((1 2 3 4) (5 6 7 8) (9 9 9 9))
I'd like to insert the result of an evaluated Clojure expression directly in my Emacs buffer, in pretty-printed form.
For example, with something like:
;; [emacs lisp]
(insert (nrepl-dict-get (nrepl-sync-request:eval "(range 30)") "value"))
I get, in the buffer of interest,
;;=>
(0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29)
In the past, I've let Clojure pretty-print things for me, as so:
(nrepl-dict-get
(nrepl-sync-request:eval
(format "(clojure.core/let [x %s] (with-out-str (clojure.pprint/pprint x)))"
"(range 30)"))
"value")
;;=>
"(0\n 1\n 2\n 3\n 4\n 5\n 6\n 7\n 8\n 9\n 10\n 11\n 12\n 13\n 14\n 15\n 16\n 17\n 18\n 19\n 20\n 21\n 22\n 23\n 24\n 25\n 26\n 27\n 28\n 29)\n"
However, the " and \n are being inserted escaped; I want them to be inserted unescaped. In other words, I want the pretty-printed result to be inserted directly without escaping quotes or newlines. This used to work in earlier versions of Cider and cider-nrepl.
Wrapping:
(nrepl-dict-get
(nrepl-sync-request:eval
(format "(clojure.core/let [x %s] (with-out-str (clojure.pprint/pprint x)))"
"(range 30)"))
"value")
in read should solve this.
I've just added this feature to lispy (it's a Paredit-style
package that uses Cider for Clojure eval):
2E will to a pretty-printed eval-and-insert, while
E will keep doing a plain one.
Here's an example (| represents point):
|(for [x (range 8)] (range x))
After E:
|(for [x (range 8)] (range x))
(() (0) (0 1) (0 1 2) (0 1 2 3) (0 1 2 3 4) (0 1 2 3 4 5) (0 1 2 3 4 5 6))
After 2E:
|(for [x (range 8)] (range x))
(()
(0)
(0 1)
(0 1 2)
(0 1 2 3)
(0 1 2 3 4)
(0 1 2 3 4 5)
(0 1 2 3 4 5 6))
Of course you can still do EjM to accomplish the same thing:
(for [x (range 8)] (range x))
|(()
(0)
(0 1)
(0 1 2)
(0 1 2 3)
(0 1 2 3 4)
(0 1 2 3 4 5)
(0 1 2 3 4 5 6))
If I have a list like this
((0 1 2) (3 4 5) (6 7 8) (0 3 6) (1 3 7) (2 4 8) (0 4 8) (2 4 6))
And I want to find the index of (0 3 6), is there a built-in function to do this? POSITION doesn't seem to work when the search item is itself a list.
See hyperspec. POSITION can take a :test argument:
(position '(0 3 6)
'((0 1 2) (3 4 5) (6 7 8) (0 3 6) (1 3 7) (2 4 8) (0 4 8) (2 4 6))
:test #'equal))
3
The default test for POSITION (and other sequence operations) is EQL, by the way.
If I had a N lists each of length M, how could I write a nice clean function to return a single list of length M, where each element is the sum of the corresponding elements in the N lists?
(starting to learn lisp - go easy!)
This is a job for the map and apply functions. Here is a way to do it, with an EDIT suggested by Nathan Sanders:
(define (add-lists . more)
(apply map + more))
For a more matlab like syntax:
(define (piecewise func)
(lambda more
(apply map func more)))
(define pw piecewise)
((pw +) '(1 2 3 4 5) '(6 7 8 9 0))
((pw -) '(1 2 3 4 5) '(6 7 8 9 0))
((pw *) '(1 2 3 4 5) '(6 7 8 9 0))
((pw /) '(1 2 3 4 5) '(6 7 8 9 0.1))
outputs:
(7 9 11 13 5)
(-5 -5 -5 -5 5)
(6 14 24 36 0)
(1/6 2/7 3/8 4/9 50.0)
Just this works in MIT scheme.
(map + '(1 2 3) '(4 5 6) '(7 8 9))
;Value 28: (12 15 18)