I'm new to Clojure. I was wondering how I could optimize an algorithm to count the number of inversions in a list. From what I understand, Clojure doesn't do tail call optimization unless specifically asked to? How do you get it to do this?
This first attempt with a mutated variable has a runtime of about 3.5s. But my second attempt was a functional version and it takes about 1m15s! and both require growing the stack size quite a bit (like -Xss12m).
How would I go about getting better performance?
I'd prefer to not have mutable variables (like the functional one) if possible. You can create the array file by typing something like seq 100000 | sort -R > IntArray.txt.
The first attempt w/ mutable variable:
(use 'clojure.java.io)
(def inversions 0)
(defn merge_and_count' [left right left_len]
(if (empty? right) left
(if (empty? left) right
(if (<= (first left) (first right))
(cons (first left) (merge_and_count' (rest left) right (- left_len 1)))
(let [_ (def inversions (+ inversions left_len))]
(cons (first right) (merge_and_count' left (rest right) left_len)))
))))
(defn inversion_count [list]
(if (or (empty? list) (nil? (next list))) list
(let [mid (quot (count list) 2)]
(merge_and_count' (inversion_count (take mid list))
(inversion_count (drop mid list)) mid)
)))
(defn parse-int [s]
(Integer. (re-find #"\d+" s )))
(defn get-lines [fname]
(with-open [r (reader fname)]
(doall (map parse-int (line-seq r)))))
(let [list (get-lines "IntArray.txt")
_ (inversion_count list)]
(print inversions))
My second attempt to be purely functional (no mutability):
(use 'clojure.java.io)
(defn merge_and_count' [left right inversions]
(if (empty? right) (list left inversions)
(if (empty? left) (list right inversions)
(if (<= (first left) (first right))
(let [result (merge_and_count' (rest left) right inversions)]
(list (cons (first left) (first result)) (second result)))
(let [result (merge_and_count' left (rest right) (+ inversions (count left)))]
(list (cons (first right) (first result)) (second result)))
))))
(defn inversion_count [list' list_len]
(if (or (empty? list') (nil? (next list'))) (list list' 0)
(let [mid (quot list_len 2)
left (inversion_count (take mid list') mid)
right (inversion_count (drop mid list') (- list_len mid))]
(merge_and_count' (first left) (first right) (+ (second left) (second right)))
)))
(defn parse-int [s]
(Integer. (re-find #"\d+" s )))
(defn get-lines [fname]
(with-open [r (reader fname)]
(doall (map parse-int (line-seq r)))))
(let [list (get-lines "IntArray.txt")
result (inversion_count list 100000)]
(print (second result)))
The stack overflows due to the recursion in merge-and-count. I tried this approach, and for 100000 items, it came back instantly.
(defn merge_and_count [left right inversions]
(loop [l left r right inv inversions result []]
(cond (and (empty? r) (empty? l)) [result inv]
(empty? r) [(apply conj result l) inv]
(empty? l) [(apply conj result r) inv]
(<= (first l) (first r)) (recur (rest l) r inv (conj result (first l)))
:else (recur l (rest r) (+ inv (count l)) (conj result (first r))))))
You need to replace this code with code from your second approach.
Related
I need to iterate through a list with sublists in Racket using list iteration and filtering, one of the lists is a nested list, I tried using "list?" and "car" to iterate inside but of course that would only apply to the first value of the sublist.
Is there a way to iterate through the whole nested list using list iteration and filtering?
(define (count-evens lst)
(length
(filter
(lambda (x)
(cond
[(and (list? x)
(and (number? (car x))
(eq? (modulo (car x) 2) 0)))
#t]
[(and (number? x)
(eq? (modulo x 2) 0))
#t]
[else
#f]))
lst)))
(count-evens '(1 2 5 4 (8 4 (b (10 3 3))) 3))
=> 3
Should return => 5
I would use a recursive function to do this but the assignment doesn't allow it.
"...assignment doesn't allow [recursive functions]"
Not sure what is allowed for this assignment, but
in ye olden days we processed recursive data structures with stacks...
(define (count-evens lst)
(define (lst-at stack) ;; (car stack) = index in deepest sub-list
;; produce list cursor within lst indexed by stack
(do ([stack (reverse stack) (cdr stack)]
[cursor (list lst) (list-tail (car cursor) (car stack))])
((null? stack) cursor)))
(do ([stack (list 0)
(cond
[(null? (lst-at stack))
(cdr stack)] ;; pop stack
[(pair? (car (lst-at stack)))
(cons 0 stack)] ;; push stack
[else ;; step through current (sub)list
(cons (+ 1 (car stack)) (cdr stack))])]
[count 0
(let ([item (car (lst-at stack))])
(if (and (number? item) (even? item)) (+ 1 count) count))])
((null? (lst-at stack)) count)))
> (count-evens '(1 2 5 4 (8 4 (b (10 3 3))) 3)) ;=>
5
I'm a newbie to scheme programming and I was writing small codes when I encountered the following issue and couldn't reason about it satisfactorily.
(define (at_i lst i)
(if (eq? i 0)
(car lst)
(at_i (cdr lst)
(- i 1) )))
Evaluation of (at_i '(1 2 3 4) 0) returns 1.
Now lets define same procedure with a lambda syntax.
(define (at_i lst i)
(lambda (vec x)
(if (eq? x 0)
(car vec)
(at_i (cdr vec)
(- x 1) )))
lst i)
But now Evaluation of (at_i '(1 2 3 4) 0) returns 0, which is not in the list and in general, it returns element at index-1.
I don't understand why this is happening.
Note: I just realized its not returning element at index - 1 but the index itself. The reason for this has been well explained below by #Renzo. Thanks!
First, you should indent properly the code if you intend to learn the language, since code indentation is very important to understand programs in Scheme and Lisp-like languages.
For instance, your function correctly indented is:
(define (at_i lst i)
(lambda (vec x)
(if (eq? x 0)
(car vec)
(at_i (cdr vec) (- x 1))))
lst
i)
From this you can see that you are defining the function at_i exactly as before in terms of a function with two parameters (lst and i), and whose body is constitued by three expressions: the first lambda (vec x) ... (- x 1)))), which is an (anonymous) function (lambda) which is not called or applied, the second, which is the first parameter lst, and finally the third which is the second parameter i. So, when the function at_i is called with two arguments, the result is the evaluation of the three expressions in sequence, the first two values are discarded (the function and the value of lst), and the result is the value of the second parameter i. This is reason for which the result of (at_i '(1 2 3 4) 0) is 0, since it is the value of i.
A proper redefinition of the function in lambda form would be the following, for instance:
(define at_i
(lambda (vec x)
(if (eq? x 0)
(car vec)
(at_i (cdr vec) (- x 1)))))
(at_i '(1 2 3 4) 0) ;; => 1
in which you can see that the name at_i, through the define, is associated to a two parameter function which calculates correctly the result.
eq? is memory object equality. Only some Scheme implementations interpret (eq? 5 5) as #t. Use = for numbers, eqv? for values, and equal? for collections.
(define (index i xs) ; `index` is a partial function,
(if (= i 0) ; `i` may go out of range
(car xs)
(index (- i 1) ; Tail recursion
(cdr xs) )))
Your second function returns the index because you missed parenthesis around the lambda's application. It should be
(define (index i xs)
((lambda (i' xs')
(if (= i' 0)
(car xs')
(index (- i' 1) ; Not tail-recursive
(cdr xs') )))
i xs))
But this is verbose and differs semantically from the first formulation.
You say you are defining the "same procedure with a lambda syntax", but you are not. That would be (define at_i (lambda lst i) ...). Instead, you are effectively saying (define (at_i lst i) 1 2 3), and that is 3, of course.
In your particular case, you defined the procedure at_i to return (lambda (vec x) ...), lst and i. Now, if you call (at_i '(1 2 3 4) 0), the procedure will return 0, since that is the value of i at this point.
I've got some function func and want to apply it on a list lst, so I used map but I need to have the first and last element of the list evaluated with some other function func2.
So basically I want this:
(map (lambda (x)
(cond [(isBeginningOfList? lst) (func2 x)]
[(isEndOfList? lst) (func2 x)]
[else (func x)]))
lst)
Obviously this doesn't work.
How can I achieve this functionality?
Can I somehow get a key of each list entry? Like lambda(key,val) and then compare (equal? key 0) / (equal? key (length lst))?
There's for/list with in-indexed and that does what you describe:
(define (f lst f1 f2)
(define last (sub1 (length lst)))
(for/list (((e i) (in-indexed lst)))
(if (< 0 i last)
(f1 e)
(f2 e))))
then
> (f '(1 2 3 4 5) sub1 add1)
'(2 1 2 3 6)
You can use a map on all the elements except the first and the last one and treat those two separately. In this way you avoid those comparisons which you would do for every element.
(define special-map
(λ (lst f1 f2)
(append (list (f1 (car lst)))
(map f2 (drop-right (cdr lst) 1))
(list (f1 (last lst))))))
Example
Let's try to increment the first and the last elements and decrement all the others.
> (special-map '(1 2 3 4 5) add1 sub1)
'(2 1 2 3 6)
Later edit
I changed (take (cdr lst) (- (length lst) 2)) with (drop-right (cdr lst) 1).
I am working with a score in Lilypond that has a lot of repetitions, where basically every bar has to be repeated a certain number of times. I would like to be able to write above every bar the number of times it should be repeat, similar to the score below (which was not created in Lilypond):
It would be great to be able to have some brackets above the bar and also to have the "3x" centralized, just like in the example above. So far, the only (temporary) solution I was able to come up with in Lilypond was to add repeat bars and then simply write "3x" above the first note of every bar (since I could not have it centralized on the bar either). It does not look very good, but gets the job done. This temporary solution looks like this:
Any suggestions of how to make this example look more similar to the first inn Lilypond would be extremely welcome!
This is a workaround for this problem:
\version "2.19.15"
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% COPY ALL THIS BELOW %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% repeatBracket snippet
% will add .----Nx----. above a bar, where N is the number of repetitions
% based on the wonderful spanner by David Nalesnik (see: http://lists.gnu.org/archive/html/lilypond-user/2014-10/msg00446.html )
#(define (test-stencil grob text)
(let* ((orig (ly:grob-original grob))
(siblings (ly:spanner-broken-into orig)) ; have we been split?
(refp (ly:grob-system grob))
(left-bound (ly:spanner-bound grob LEFT))
(right-bound (ly:spanner-bound grob RIGHT))
(elts-L (ly:grob-array->list (ly:grob-object left-bound 'elements)))
(elts-R (ly:grob-array->list (ly:grob-object right-bound 'elements)))
(break-alignment-L
(filter
(lambda (elt) (grob::has-interface elt 'break-alignment-interface))
elts-L))
(break-alignment-R
(filter
(lambda (elt) (grob::has-interface elt 'break-alignment-interface))
elts-R))
(break-alignment-L-ext (ly:grob-extent (car break-alignment-L) refp X))
(break-alignment-R-ext (ly:grob-extent (car break-alignment-R) refp X))
(num
(markup text))
(num
(if (or (null? siblings)
(eq? grob (car siblings)))
num
(make-parenthesize-markup num)))
(num (grob-interpret-markup grob num))
(num-stil-ext-X (ly:stencil-extent num X))
(num-stil-ext-Y (ly:stencil-extent num Y))
(num (ly:stencil-aligned-to num X CENTER))
(num
(ly:stencil-translate-axis
num
(+ (interval-length break-alignment-L-ext)
(* 0.5
(- (car break-alignment-R-ext)
(cdr break-alignment-L-ext))))
X))
(bracket-L
(markup
#:path
0.1 ; line-thickness
`((moveto 0.5 ,(* 0.5 (interval-length num-stil-ext-Y)))
(lineto ,(* 0.5
(- (car break-alignment-R-ext)
(cdr break-alignment-L-ext)
(interval-length num-stil-ext-X)))
,(* 0.5 (interval-length num-stil-ext-Y)))
(closepath)
(rlineto 0.0
,(if (or (null? siblings) (eq? grob (car siblings)))
-1.0 0.0)))))
(bracket-R
(markup
#:path
0.1
`((moveto ,(* 0.5
(- (car break-alignment-R-ext)
(cdr break-alignment-L-ext)
(interval-length num-stil-ext-X)))
,(* 0.5 (interval-length num-stil-ext-Y)))
(lineto 0.5
,(* 0.5 (interval-length num-stil-ext-Y)))
(closepath)
(rlineto 0.0
,(if (or (null? siblings) (eq? grob (last siblings)))
-1.0 0.0)))))
(bracket-L (grob-interpret-markup grob bracket-L))
(bracket-R (grob-interpret-markup grob bracket-R))
(num (ly:stencil-combine-at-edge num X LEFT bracket-L 0.4))
(num (ly:stencil-combine-at-edge num X RIGHT bracket-R 0.4)))
num))
#(define-public (Measure_attached_spanner_engraver context)
(let ((span '())
(finished '())
(event-start '())
(event-stop '()))
(make-engraver
(listeners ((measure-counter-event engraver event)
(if (= START (ly:event-property event 'span-direction))
(set! event-start event)
(set! event-stop event))))
((process-music trans)
(if (ly:stream-event? event-stop)
(if (null? span)
(ly:warning "You're trying to end a measure-attached spanner but you haven't started one.")
(begin (set! finished span)
(ly:engraver-announce-end-grob trans finished event-start)
(set! span '())
(set! event-stop '()))))
(if (ly:stream-event? event-start)
(begin (set! span (ly:engraver-make-grob trans 'MeasureCounter event-start))
(set! event-start '()))))
((stop-translation-timestep trans)
(if (and (ly:spanner? span)
(null? (ly:spanner-bound span LEFT))
(moment<=? (ly:context-property context 'measurePosition) ZERO-MOMENT))
(ly:spanner-set-bound! span LEFT
(ly:context-property context 'currentCommandColumn)))
(if (and (ly:spanner? finished)
(moment<=? (ly:context-property context 'measurePosition) ZERO-MOMENT))
(begin
(if (null? (ly:spanner-bound finished RIGHT))
(ly:spanner-set-bound! finished RIGHT
(ly:context-property context 'currentCommandColumn)))
(set! finished '())
(set! event-start '())
(set! event-stop '()))))
((finalize trans)
(if (ly:spanner? finished)
(begin
(if (null? (ly:spanner-bound finished RIGHT))
(set! (ly:spanner-bound finished RIGHT)
(ly:context-property context 'currentCommandColumn)))
(set! finished '())))
(if (ly:spanner? span)
(begin
(ly:warning "I think there's a dangling measure-attached spanner :-(")
(ly:grob-suicide! span)
(set! span '())))))))
\layout {
\context {
\Staff
\consists #Measure_attached_spanner_engraver
\override MeasureCounter.font-encoding = #'latin1
\override MeasureCounter.font-size = 0
\override MeasureCounter.outside-staff-padding = 2
\override MeasureCounter.outside-staff-horizontal-padding = #0
}
}
repeatBracket = #(define-music-function
(parser location N note)
(number? ly:music?)
#{
\override Staff.MeasureCounter.stencil =
#(lambda (grob) (test-stencil grob #{ #(string-append(number->string N) "×") #} ))
\startMeasureCount
\repeat volta #N { $note }
\stopMeasureCount
#}
)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% ...UNTIL HERE %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{
\repeatBracket 7 {c'1}
\repeatBracket 32 {d' g}
\repeatBracket 14 {e' f g}
\repeatBracket 29 {f' a bes \break cis' e''}
\repeatBracket 1048 {g'1}
}
This code above gives the following result:
]
This solution was not created by myself, but sent to me by David Nalesnik, from lilypond-user mailing list. I just would like to share it here in case someone would need it as well. I've made just some very minor adjustments to what David sent me.
I just had a similar problem but preferred the style on your second example, with the 3x above the bar. The solution I found was:
f e d c |
\mark \markup {"3x"}\repeat volta 3 {c d e f}
f e d c |
generating
Maybe someone else has a use for it.
So I have to finish a project in Scheme and I'm pretty stuck. Basically, what the program does is open a file and output the statistics. Right now I am able to count the number of characters, but I also need to count the number of lines and words. I'm just trying to tackle this situation for now but eventually I also have to take in two files - the first being a text file, like a book. The second will be a list of words, I have to count how many times those words appear in the first file. Obviously I'll have to work with lists but I would love some help on where to being. Here is the code that I have so far (and works)
(define filestats
(lambda (srcf wordcount linecount charcount )
(if (eof-object? (peek-char srcf ) )
(begin
(close-port srcf)
(display linecount)
(display " ")
(display wordcount)
(display " ")
(display charcount)
(newline) ()
)
(begin
(read-char srcf)
(filestats srcf 0 0 (+ charcount 1))
)
)
)
)
(define filestatistics
(lambda (src)
(let ((file (open-input-file src)))
(filestats file 0 0 0)
)
)
)
How about 'tokenizing' the file into a list of lines, where a line is a list of words, and a word is a list of characters.
(define (tokenize file)
(with-input-from-file file
(lambda ()
(let reading ((lines '()) (words '()) (chars '()))
(let ((char (read-char)))
(if (eof-object? char)
(reverse lines)
(case char
((#\newline) (reading (cons (reverse (cons (reverse chars) words)) lines) '() '()))
((#\space) (reading lines (cons (reverse chars) words) '()))
(else (reading lines words (cons char chars))))))))))
once you've done this, the rest is trivial.
> (tokenize "foo.data")
(((#\a #\b #\c) (#\d #\e #\f))
((#\1 #\2 #\3) (#\x #\y #\z)))
The word count algorithm using Scheme has been explained before in Stack Overflow, for example in here (scroll up to the top of the page to see an equivalent program in C):
(define (word-count input-port)
(let loop ((c (read-char input-port))
(nl 0)
(nw 0)
(nc 0)
(state 'out))
(cond ((eof-object? c)
(printf "nl: ~s, nw: ~s, nc: ~s\n" nl nw nc))
((char=? c #\newline)
(loop (read-char input-port) (add1 nl) nw (add1 nc) 'out))
((char-whitespace? c)
(loop (read-char input-port) nl nw (add1 nc) 'out))
((eq? state 'out)
(loop (read-char input-port) nl (add1 nw) (add1 nc) 'in))
(else
(loop (read-char input-port) nl nw (add1 nc) state)))))
The procedure receives an input port as a parameter, so it's possible to apply it to, say, a file. Notice that for counting words and lines you'll need to test if the current char is either a new line character or a white space character. And an extra flag (called state in the code) is needed for keeping track of the start/end of a new word.