音符字符串(C#-4、F-3等)到MIDI音符值，在Python中

Question

下面答案中的代码将字符串(如C#-4或F-3 )中的音符转换为其相应的MIDI音符值。

我张贴这篇文章是因为我厌倦了每次我需要的时候都在网上挖掘它。我相信我不是唯一能找到用武之地的人。我刚刚写了这篇文章--它是经过测试的，是正确的。它是用Python编写的，但我觉得它几乎是所有人都能理解的。

Answer 1

#Input is string in the form C#-4, Db-4, or F-3. If your implementation doesn't use the hyphen, 
#just replace the line :
#    letter = midstr.split('-')[0].upper()
#with:
#    letter = midstr[:-1]
def MidiStringToInt(midstr):
    Notes = [["C"],["C#","Db"],["D"],["D#","Eb"],["E"],["F"],["F#","Gb"],["G"],["G#","Ab"],["A"],["A#","Bb"],["B"]]
    answer = 0
    i = 0
    #Note
    letter = midstr.split('-')[0].upper()
    for note in Notes:
        for form in note:
            if letter.upper() == form:
                answer = i
                break;
        i += 1
    #Octave
    answer += (int(midstr[-1]))*12
    return answer

Answer 2

NOTES_FLAT = ['C', 'Db', 'D', 'Eb', 'E', 'F', 'Gb', 'G', 'Ab', 'A', 'Bb', 'B']
NOTES_SHARP = ['C', 'C#', 'D', 'D#', 'E', 'F', 'F#', 'G', 'G#', 'A', 'A#', 'B']

def NoteToMidi(KeyOctave):
    # KeyOctave is formatted like 'C#3'
    key = KeyOctave[:-1]  # eg C, Db
    octave = KeyOctave[-1]   # eg 3, 4
    answer = -1

    try:
        if 'b' in key:
            pos = NOTES_FLAT.index(key)
        else:
            pos = NOTES_SHARP.index(key)
    except:
        print('The key is not valid', key)
        return answer

    answer += pos + 12 * (int(octave) + 1) + 1
    return answer