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The piano keyboard has 88 keys: enter image description here

I would like to encode one note or one chord in 88 bit array. I do this, for example: A4 is 0000000000000000000000000000000000000000000000000000000001000000000000000000000000000000

Next I convert this number to decimal. When I have chord the final number is really big. The problem is when I put this big number into array from NumPy I gets: 

note_data = np.array(note_data, dtype=np.int64)
OverflowError: Python int too large to convert to C long

My point is to feed recurrent neural netowrk this kind of data or maybe is better representation on music notes into numbers?

EDIT:

Another example of encode C3 note:

C3 = np.array([0000000000000000000000000001000000000000000000000000000000000000000000000000000000000000], dtype=int)

and error sitll the same:

OverflowError: Python int too large to convert to C long

It's possible to reduce this error?

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1 Answer 1

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That seems like a bad way to encode the information. A C chord and a C7 chord would be very different numbers even though they are similar conceptually.

I would take your idea, but instead of smashing the information into a single int, make your input a binary vector of length 88 created in the way you describe. In this way, a C and C7 will have similar inputs, differing in only one location of the vector.

As a simplified one octave example going from C to B:

C : [1,0,0,0,1,0,0,1,0,0,0,0]
C7: [1,0,0,0,1,0,0,1,0,0,1,0]

This kind of encoding is making use of “dummy variables”. It is like the “Bag of Words” encoding used a lot in Natural Language Processing.

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  • $\begingroup$ Ok, Thank you for reply. I known One hot encoding method. But in which way can I encode chord like C, E, G on one hot encoding method? $\endgroup$
    – lukassz
    Jan 30, 2019 at 22:15
  • $\begingroup$ binary vector of length 88 is not too long as I desctibed? Yes, C and C7 is similar but C7 have more information it also have E and G. $\endgroup$
    – lukassz
    Jan 30, 2019 at 22:19
  • $\begingroup$ No, a vector of length 88 should be no problem at all. $\endgroup$
    – kbrose
    Jan 31, 2019 at 13:37
  • $\begingroup$ I made an example in one octave. $\endgroup$
    – kbrose
    Jan 31, 2019 at 13:46
  • $\begingroup$ ok I got your point I did this before asking on datascience. If we have 88 binary vector representing single chord. C chord and C7 would be very diffrent numbers. If we have input vector of length 12 C is 2192 and C7 2194 its close. How about octaves? I decide to test with 88 input because I can place chord in diffrent octaves. Now I known it's bad idea. In which way can I represent octave? As concatenation 12 bits for notes and 8 bits of octave? $\endgroup$
    – lukassz
    Feb 2, 2019 at 13:50

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