# Help needed Binary Length values

## Recommended Posts

Dear Friends,

1) How to convert a given length series in a binary series ? For example:

((1/16 -3/16 1/16 -1/8 1/16 -1/4)) with 1/16 as a base could be transformed in binary like:

(1 0 0 0 1 0 0 1 0 0 0 0)

and/or

2) How to convert a length  3/16 in 1/16 -1/16 -1/16, i.e. a kind of length conversion

based on quantize.

3/16 could be converted in

1/16 -1/16 -1/16

or

1/32 -1/32 -1/32 -1/32 -1/32 -1/32

depending on the value regarded as the reference (1/16 in the first case or 1/32 in the second)

Thanks !

Julio

Code example

```(setf ccpa1 (omn :length (length-staccato 1/16 (time-point-system (pitch-rotate 0 (pcs '3-11b :pitch))'s :start 0))))

;EXTRA FUNCTION NEEDED

;;;LENGTH-LEGATO (by ANDRE MEIER)

(defun length-staccato (n alist)
(let ((newlengths)
(new-omn (omn-merge-ties (flatten  alist)))
(time-sign (get-time-signature alist)))
(progn
(setf newlengths (loop for i in (omn :length new-omn)
when (> i 0)
append (if (= n i)
(list i)
(list n (* -1 (abs (- i n)))))

else collect i))
(if (omn-formp alist)
(omn-to-time-signature (make-omn :length newlengths
:pitch (omn :pitch new-omn)
:velocity (omn :velocity new-omn)
:articulation (omn :articulation new-omn))
time-sign)
newlengths))))```

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something like that?

```(defun length-to-binary (lengthlist n)
(let ((newlist (loop for i in (omn :length lengthlist)
collect (/ i n))))
(loop for x in newlist
when (> x 0)
append (append (list 1) (gen-repeat (1- x) '0))
else append  (gen-repeat (abs x) '0))))

(length-to-binary '(-q q e) 1/16)
=> (0 0 0 0 1 0 0 0 1 0)

(length-to-binary '(-q s s q e) 1/16)
=> (0 0 0 0 1 1 1 0 0 0 1 0)```

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Thanks a lot, André !!!

Exactly. That's great !

Julio

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• By terekita
Hello,

For various reasons, I need to start with rhythms defined in one time signature and remap them. I'm having trouble doing this because of what happens to ties when you use omn :length (though I realize they are being collected under :articulation).

In this case, I'm taking bars of 4/4 regrouping into bars of 2/4.

(setf myr '( (s s -s s -e s s -s s -e t t t t s s) (e e_q -h))) (setf tr (omn-to-time-signature myr '(2 4)))
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(setf foo '((s c4 c4 - c4 -e s c4 c4) (-s c4 -e t c4 c4 c4 c4 s s) (e c4 c4 tie q) (-h)))
As you can see, there is a tie in the third bar.

Now, how can I extract that rhythm so that I can apply to another set of pitches?

If I do the following, I no longer have the tie represented in the rhythm:
(omn :length foo) ;; results in ;; ((1/16 1/16 -1/16 1/16 -1/8 1/16 1/16) (-1/16 1/16 -1/8 1/32 1/32 1/32 1/32 1/16 1/16) (1/8 1/8 1/4) (-1/2)) ;; i.e., no tie in the third bar
Also, if I try to span the rhythm onto another set of pitches, I again get a result that is missing the tie:
(setf my-pitches '(c4 d4 e4)) (make-omn :length foo :pitch my-pitches)
What I'd really love is to get back a list of lengths that is like my original list, just regrouped in new time signatures. So, ideally something like:
(setf myr '( (s s -s s -e s s -s s -e t t t t s s) (e e_q -h))) ;; imaginary function (regroup-to-two-four myr) ;;would result in the following: ;; ((s s -s s -e s s)(-s s -e t t t t s s)(e e_q)(-h))
Finally, I should add that in a related problem, I need to extract the number of attacks in a bar after it has been regrouped. So, for this reason again, I need to figure out how to extract the rhythm while preserving ties and not re-attacking tied notes.

I'd very appreciate and thank you kindly for any assistance in doing this.

Thanks very much, Michael
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