I set myself to the task to write a Common Lisp function that concatenates two lists without using append.
Common Lisp input (concat-lists '(1 2 3) '(4 5 6)) should return (1 2 3 4 5 6)
Even though my solution seems to work it looks overtly complicated
(defun concat-lists(seq1 seq2)
(cond ((not (null seq1)) (cons (car seq1) (concat-lists (cdr seq1) seq2)))
(T (cond ((not (null seq2)) (cons (car seq2) (concat-lists seq1 (cdr seq2))))
(T nil)))))
What I'm looking for is a more elegant solution using reduce where I use seq1 as initial value and then apply a function to each element of seq2, thereby appending each value of the list to seq1. Somehow I always get stuck when trying....
Any help or input is much appreciated. Thanks!
CL-USER 39 > (reduce #'cons
'(1 2 3 4 5)
:initial-value '(a b c d e)
:from-end t)
(1 2 3 4 5 A B C D E)
The solution of Rainer Joswig is really elegant and simple, and respects your request of using reduce.
If you want to see also a recursive, simple solution, then here is the classical one:
(defun concat-lists (seq1 seq2)
(if (null seq1)
seq2
(cons (car seq1) (concat-lists (cdr seq1) seq2))))
(concat-lists '(1 2 3) '(4 5 6))
(1 2 3 4 5 6)
I do understand your requirement for 'reduce'. and here other options:
CL also has 'concatenante'
(concatenate 'list '(1 2 3) '(4 5 6))
There is also the other less complicated (IMHO), and not as elegant.
(defun concat-lists (list1 list2)
(let ((a (copy-list list1))
(b (copy-list list2)))
(rplacd (last a) b)
a))
or
(defun concat-lists (list1 list2)
(let ((a (copy-list list1))
(b (copy-list list2)))
(nconc a b)))
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))
How to convert a flat list into an arbitrarily complex tree-like structure? First, a simple example, convert '(1 2 3 4) into '(1 (2 (3 (4)))). I know how to do it with classical recursion:
(defun nestify (xs)
(if (null xs)
(list)
(list (car xs) (nestify (cdr xs)))))
Now, what if the nested structure is arbitrarily complex? For example, I want to convert '(1 2 3 4 5 6 7 8) into '(1 (2 3) (4 (5 6) 7) 8). How can I write a general function that is able to convert a flat list in any such nested structure? I can think of giving a template with dummy values. For example:
* (nestify '(1 2 3 4 5 6 7 8) '(t (t t) (t (t t) t) t))
'(1 (2 3) (4 (5 6) 7) 8)
My first attempt using recursion and custom tree size finding function:
(defun length* (tr)
"Count number of elements in a tree."
(cond ((null tr) 0)
((atom tr) 1)
(t (+ (length* (car tr))
(length* (cdr tr))))))
(defun tree-substitute (xs tpl)
"(tree-substitute '(1 2 3) '(t (t) t)) -> '(1 (2) 3)"
(cond ((null tpl) nil)
((atom (car tpl))
(cons (car xs) (tree (cdr xs) (cdr tpl))))
(t (cons (tree xs (car tpl))
(tree (nthcdr (length* (car tpl)) xs) (cdr tpl))))))
Is there any way to do this better, in a more elegant and concise way? For example, the function converting a list into a tree might not use the template, although I can't think of the method. Can I abstract away recursion and other details and have a neat reduce or some other high-level function?
Turning (1 2 3 4) into (1 (2 (3 (4)))) actually isn't quite as simple as you might hope, if you're using reduce. You need to specify :from-end t if you want to process 4 first, and the reduction function is either called with 3 and 4, if no :initial-value is specified, or with 4 and the initial value, if one is. That means you can use something like this, where the function checks for the special initial case:
(reduce (lambda (x y)
(if y
(list x y)
(list x)))
'(1 2 3 4)
:from-end t
:initial-value nil)
;=> (1 (2 (3 (4))))
A solution that involves a template is much more interesting, in my opinion. It's easy enough to define a maptree function that maps a function over a tree and returns a new tree with the function results:
(defun maptree (function tree)
"Return a tree with the same structure as TREE, but
whose elements are the result of calling FUNCTION with
the element from TREE. Because TREE is treated as an
arbitrarily nested structure, any occurrence of NIL is
treated as an empty tree."
(cond
((null tree) tree)
((atom tree) (funcall function tree))
((cons (maptree function (car tree))
(maptree function (cdr tree))))))
(maptree '1+ '(1 2 (3 (4 5)) (6 7)))
;=> (2 3 (4 (5 6)) (7 8))
Given the maptree function, it's not hard to call it with a function that provides an element from a list of elements, until that list of element is exhausted. This provides a definition of substitute-into:
(defun substitute-into (items tree)
"Return a tree like TREE, but in which the elements
of TREE are replaced with elements drawn from ITEMS.
If there are more elements in TREE than there are in
ITEMS, the original elements of TREE remain in the result,
but a new tree structure is still constructed."
(maptree #'(lambda (x)
(if (endp items) x
(pop items)))
tree))
(substitute-into '(1 2 3 4 5) '(t (u (v)) (w x)))
;=> (1 (2 (3)) (4 5))
(substitute-into '(1 2 3 4 5) '(t u (v w x) y z))
;=> (1 2 (3 4 5) Y Z)
See Also
The maptree above is actually just a special case of a more general reduce, or fold, function for trees. Have a look at Using reduce over a tree in Lisp for some more information about how you can fold over trees. In this case, you could use my tree-reduce function from my answer to that question:
(defun tree-reduce (node-fn leaf-fn tree)
(if (consp tree)
(funcall node-fn
(tree-reduce node-fn leaf-fn (car tree))
(tree-reduce node-fn leaf-fn (cdr tree)))
(funcall leaf-fn
tree)))
and define maptree in terms of it:
(defun maptree (function tree)
(tree-reduce 'cons function tree))
My attempt:
(defun mimicry (source pattern)
(labels ((rec (pattern)
(mapcar (lambda (x)
(if (atom x)
(pop source)
(rec x)))
pattern)))
(rec pattern)))
Test:
CL-USER> (mimicry '(1 2 3 4 5) '(t (u (v)) (w x)))
(1 (2 (3)) (4 5))
i am trying to write a function in lisp which have 2 parameters one function F and one list L
if i place '> in place of F and list L is '(1 2 3 4 5) it will return 5 as 5 is biggest.
and if we put '< then it compares all list elements and gives the smallest one as output.
and so on.
we can even put custom written function in place of F for comparison.
i wish i could provide more sample code but i am really stuck at the start.
(DEFUN givex (F L)
(cond
(F (car L) (car (cdr L))
;after this i got stuck
)
)
another attemp to write this function
(defun best(F list)
(if (null (rest list)) (first list)
(funcall F (first List) (best (F list)))))
You are almost there, just the else clause returns the f's return value instead of the the best element:
(defun best (F list)
(let ((first (first list))
(rest (rest list)))
(if (null rest)
first
(let ((best (best f rest)))
(if (funcall F first best)
best
first)))))
Examples:
(best #'< '(1 2 3))
==> 3
(best #'> '(1 2 3))
==> 1
Note that this recursive implementation is not tail-recursive, so it is not the most efficient one. You might prefer this instead:
(defun best (f list)
(reduce (lambda (a b) (if (funcall f a b) b a)) list))
Or, better yet,
(defmacro fmax (f)
`(lambda (a b) (if (,f a b) b a)))
(reduce (fmax <) '(1 2 3))
==> 1
(reduce (fmax >) '(1 -2 3 -4) :key #'abs)
==> 1
(reduce (fmax <) '(1 -2 3 -4) :key #'abs)
==> 4
How to calculate the difference between two sets in Emacs Lisp? The sets should be lists.
The programm should be very simple and short, or else I won't understand it. I'm a newbee.
Thx
There is a set-difference function in the Common Lisp extensions:
elisp> (require 'cl-lib)
cl-lib
elisp> (cl-set-difference '(1 2 3) '(2 3 4))
(1)
When I write Elisp code that has lots of list data transformations, I use dash library, because it has loads of functions to work with lists. Set difference can be done with -difference:
(require 'dash)
(-difference '(1 2 3 4) '(3 4 5 6)) ;; => '(1 2)
Disclaimer: this is not an efficient way to do it in eLisp. An efficient way is through a hash-table with a hash function, but since you asked about lists, then here it is:
(defun custom-set-difference (a b)
(remove-if
#'(lambda (x) (and (member x a) (member x b)))
(append a b)))
(custom-set-difference '(1 2 3 4 5) '(2 4 6))
(1 3 5 6)
(defun another-set-difference (a b)
(if (null a) b
(let (removed)
(labels ((find-and-remove
(c)
(cond
((null c) nil)
((equal (car c) (car a))
(setq removed t) (cdr c))
(t (cons (car c) (find-and-remove (cdr c)))))))
(setf b (find-and-remove b))
(if removed
(another-set-difference (cdr a) b)
(cons (car a) (another-set-difference (cdr a) b)))))))
(another-set-difference '(1 2 3 4 5) '(2 4 6))
(1 3 5 6)
The second is slightly more efficient, because it will remove the elements as it makes consequent checks, but the first is shorter and more straight-forward.
Also note that lists are not good representation of sets because they naturally allow repetition. Hash maps are better for that purpose.
Here is a simple & short definition, which should be easy to understand. It is essentially the same as the set-difference function in the Common Lisp library for Emacs, but without any treatment of a TEST argument.
(defun set-diff (list1 list2 &optional key)
"Combine LIST1 and LIST2 using a set-difference operation.
Optional arg KEY is a function used to extract the part of each list
item to compare.
The result list contains all items that appear in LIST1 but not LIST2.
This is non-destructive; it makes a copy of the data if necessary, to
avoid corrupting the original LIST1 and LIST2."
(if (or (null list1) (null list2))
list1
(let ((keyed-list2 (and key (mapcar key list2)))
(result ()))
(while list1
(unless (if key
(member (funcall key (car list1)) keyed-list2)
(member (car list1) list2))
(setq result (cons (car list1) result)))
(setq list1 (cdr list1)))
result)))
GNU Emacs Lisp Reference Manual, Sets and Lists suggests using cl-lib's
cl-set-difference LIST1 LIST2 &key :test :test-not :key
(require 'cl-lib)
(cl-set-difference '(1 2 3) '(2 3 4))
(1)
The following Emacs Lisp function takes a list of lists and returns a list in which the items of the inner lists have been concatenated to one big list. It is pretty straight-forward and I am convinced something like this must already be part of the standard function library.
(defun flatten (LIST)
(if LIST
(append (car LIST) (flatten (cdr LIST)))
nil))
I am looking for a function that will take a single list of lists as its argument and then append all the inner lists.
(flatten '((a b) (c d)))
will give
(a b c d)
Does anyone know whether this function is already built in, and if so, under which name?
Thanks!
You're either looking for append:
(defun flatten (list-of-lists)
(apply #'append list-of-lists))
If (and only if) you know that you'll always have a list of lists.
Otherwise:
(defun flatten (list)
(mapcan (lambda (x) (if (listp x) x nil)) list))
Emacs 27.1 has flatten-tree:
(flatten-tree '((a b) (c d)))
(a b c d)
See: https://www.gnu.org/software/emacs/manual/html_node/elisp/Building-Lists.html
I stepped into this only recently whilst looking for something different; there is something that might not have been put into evidence by the test data utilized to check the function, depending on whether the original question was meant to refer to generic lists (i.e.: list of list of list of list of...) or just to two-level lists (as in the example).
The solution based on append works fine only with two-level lists, and there is a further issue with the solution based on mapcan.
Basically, the general solution has to be recursive both on car and cdr, as in the flatten defun below.
(setq l '((((1 2) 3) 4) (5 6 7)))
(defun flatten(x)
(cond ((null x) nil)
((listp x) (append (flatten (car x)) (flatten (cdr x))))
(t (list x))))
(defun flatten2(l)
(if l (append (car l) (flatten2 (cdr l))) nil))
(defun flatten3(l)
(mapcan (lambda(x) (if (listp x) x nil)) l))
(flatten l)
(1 2 3 4 5 6 7)
(apply #'append l)
(((1 2) 3) 4 5 6 7)
(flatten2 l)
(((1 2) 3) 4 5 6 7)
The further issue is with the usage of mapcan in flatten3: as mapcan hides an nconc inside, the user must remember that it alters its argument.
l
((((1 2) 3) 4) (5 6 7))
(flatten3 l)
(((1 2) 3) 4 5 6 7)
l
((((1 2) 3) 4 5 6 7) (5 6 7))
Dash is a modern list library for Emacs, and has flatten. It's the second most downloaded package on Melpa, after magit. From the readme:
-flatten (l): Takes a nested list l and returns its contents as a single, flat list.
(-flatten '((1))) ;; => '(1)
(-flatten '((1 (2 3) (((4 (5))))))) ;; => '(1 2 3 4 5)
(-flatten '(1 2 (3 . 4))) ;; => '(1 2 (3 . 4))
-flatten-n (num list): Flatten num levels of a nested list.
(-flatten-n 1 '((1 2) ((3 4) ((5 6))))) ;; => '(1 2 (3 4) ((5 6)))
(-flatten-n 2 '((1 2) ((3 4) ((5 6))))) ;; => '(1 2 3 4 (5 6))
(-flatten-n 3 '((1 2) ((3 4) ((5 6))))) ;; => '(1 2 3 4 5 6)
This package was started 2012-09.
I realize that the original question was "what is the built in function". It appears that there is none. The other solutions do not actually flatten all lists that I tested. This function appears to work. I'm posting it here because this was the first place Google hit when I did my search.
(defun flatten (LIST)
"flattens LIST"
(cond
((atom LIST) (list LIST))
((null (cdr LIST)) (flatten (car LIST)))
(t (append (flatten (car LIST)) (flatten (cdr LIST))))))
e.g.
(flatten (list "a" (list "b" "c" nil) (list (list "d" "e") "f")))
("a" "b" "c" nil "d" "e" "f")
Have a look at nconc